How to make Django model just view (read-only) in Django Admin? - django

This question is continuation of How to remove Add button in Django admin, for specific Model?
I have realized that my first question was not formulated good, so I tough that it is better that I start new question, that to fix the old one. Because there was already some answers.
So question is how to make Django Model that will be view read only. So that you can not add new, delete old, change current, but that you also do not have button for that on web admin UI.
Solution from first question are all related to fields, but not to whole model.
They all work, in sense that you will not be able to edit those field, but I am not satisfied how they do it.
Current solution are:
Use readonly_fields on all fields from model -> I do not like it because you can click on row to change it, but you can not edit fields.
editable=False on filed definition -> This will not show the field on web admin UI. But you can still click on row, but you will just not see anything, and still have Save button.
def has_add_permission(self, request): -> same as 2
don't give anyone the add permission for this model -> same as 2
Any thoughts ?

You need to set list_display_links attribute of your ModelAdmin class to (None,). But this can be done only in __init__ after standard ModelAdmin __init__ call otherwise it will throw ImproperlyConfigured exception with text ... list_display_links[0]' refers to 'None' which is not defined in 'list_display'. And you should define has_add_permisssion anyway to hide add button:
class AmountOfBooksAdmin(admin.ModelAdmin):
actions = None # disables actions dropbox with delete action
list_display = ('book', 'amount')
def has_add_permission(self, request):
return False
def __init__(self, *args, **kwargs):
super(AmountOfBooksAdmin, self).__init__(*args, **kwargs)
self.list_display_links = (None,)
# to hide change and add buttons on main page:
def get_model_perms(self, request):
return {'view': True}
To hide 'view' and 'change' buttons from madin admin page you must place index.html from django/contrib/admin/templates/admin/ to you templates dir /admin and change it:
{% for model in app.models %}
...
{% if model.perms.view %}
<th scope="row">{{ model.name }}</th>
{% else %}
{% if model.admin_url %}
<th scope="row">{{ model.name }}</th>
{% else %}
<th scope="row">{{ model.name }}</th>
{% endif %}
{% endif %}
....

Related

Can't make Django ModelForm Select Choice Field work in Template

I tried multiple solutions around but couldn't make my select field work in Django Template. I am beginner and humbly apologize if my question is not up to standard. My select button is not showing any options in django form. I only have two options to show. Is there anyway I can get form field without having iterate through it in template? Using Django 3 and Python 3.8. Help will be much appreciated.
Model:
class User(AbstracUser):
ACC_TYPE = (('Student', 'Student'), ('Teacher', 'Teacher'))
role = models.CharField(max_length=100, choices=ACC_TYPE
.....
Form:
class UserForm(UserCreationForm): #Using django's default form
account_type = forms.CharField(widget=forms.Select(choices=User.ACC_TYPE))
model = User
fields =('.............,
'account_type')
def __init__(self, *args, **kwargs):
super(UserCreationForm, self).__init__(*args, **kwargs)
self.fields['account_type'].choices = [(each[0], each[1]) for each in User.ACC_TYPE]
self.fields['account_type'].required = True
Template:
.......
<select>
{% for opt, val in form.account_type.choices %}
<option value="{{ opt }}">{{ value }}</option>
{% endfor %}
</select>
......
Your logic is fine. You shouldn't be having this issue. Only reason could be and its a long shot...that you are using some bootstrap theme with customized CSS classes in it. Now either you load those custom classes per field or you can use django-crispy-fields that inherits all classes with form-group. You will need to load the form with {{ form|crispy }}. For details please check settings for django-crispy-forms. Remember to load the tag at top of page where you want to use it as per settings {% load crispy_forms_tags %}. Read Docs here.

Django - Open create/add form with fields precompiled based on other object model

In my project I use Django admin with Django 1.11, so I have a modelA and a modelB and users can create both of them, but sometimes I need to create a modelA object with some values field based on modelB object.
To do this I have override my /templates/my_app_name/my_modelB/change_form.html template and add a button like this:
{% extends "admin/change_form.html" %}
{% load i18n admin_urls %}
{% block object-tools-items %}
<div class="form-row form-row-field-quick-view">
Create ObjA from this ObjB
</div>
{% endif %}
{% endblock %}
In this way when I open some ObjB change form I get a link, based on a view whose task is redirect users to the URL /admin/app_name/modelA/add/ and show the proper form with some fields precompiled based on ObjB values which the form came from.
So I created a view like this:
class create_object_A_view(View):
model = ModelA
url = '/admin/app_name/modelA/add/'
def get(self, request, *args, **kwargs):
#get data from objB, say it foo and bar....
form = ModelAForm(initial={'field1': 'foo' , 'field2':'bar'})
return redirect(self.url,
{'form':form,
'opts': self.model._meta,
'change': False,
'is_popup':False,
'save_as':False,
'has_delete_permission':False,
'has_add_permission': False,
'has_change_permission': False})
Doing this I get the correct redirect to /admin/app_name/modelA/add/ but the fields that I want to be precompiled are empty.
Is this the proper way to achieve this result?
You can make that happen, by redirecting to the modelA add url with the form data in url itself
for example:
return redirect(self.url+"?"+form.data.urlencode())

Add a link to custom django admin view

I've created a custom admin view as documented here.
class MyAdmin(admin.ModelAdmin):
def get_urls(self):
urls = super().get_urls()
my_urls = [
path('stats/', self.admin_site.admin_view(self.stats)),
]
return my_urls + urls
def stats(self, request):
request.current_app = self.admin_site.name
context = dict(
# Include common variables for rendering the admin template.
self.admin_site.each_context(request),
# Anything else you want in the context...
key='blah',
)
return TemplateResponse(request, "sometemplate.html", context)
The URL is working and the template is loading.
But how, can I get a link to my new custom view into the overview of the Django admin?
There is already a similar question How do you add a new entry into the django admin index?, but all of the provided answers were not very satisfying for me.
Therefore I ran through the source code and was looking for a possibility which wouldn't involve overriding any templates, neither index.html nor app_index.html.
The file django/contrib/admin/sites.py holds the code responsible for rendering index.html and app_index.html, the first is the template that displays what is illustrated in the question.
The method index renders the template index.html and displays the available apps with the registered model admins. It uses the method get_app_list to get the apps. Within this method the method _build_app_dict is called, which gets the models and the model admins.
The method app_index renders the template app_index.html and displays the registered model admins for a single app. It uses the method _build_app_dict, mentioned before.
Thus I decided to override this method in my custom admin. Based on the example in the question it can look like this (differences to the original example are shown in bold):
class MyAdmin(admin.AdminSite):
def get_urls(self):
urls = super().get_urls()
my_urls = [
path('stats/', self.admin_site.admin_view(self.stats), name='stats'),
]
return my_urls + urls
def stats(self, request):
request.current_app = self.admin_site.name
context = dict(
# Include common variables for rendering the admin template.
self.admin_site.each_context(request),
# Anything else you want in the context...
key='blah',
)
return TemplateResponse(request, "sometemplate.html", context)
def _build_app_dict(self, request, label=None):
# we create manually a dict to fake a model for our view 'stats'
# this is an example how the dict should look like, i.e. which keys
# should be present, the actual values may vary
stats = {
'name': 'Stats',
'admin_url': reverse('my_admin:stats'),
'object_name': 'Stats',
'perms': {'delete': False, 'add': False, 'change': False},
'add_url': ''
}
# get the app dict from the parent method
app_dict = super(MyAdmin, self)._build_app_dict(request, label)
# check if there is value for label, then the app_index will be rendered
if label:
# append the manually created dictionary 'stats'
app_dict['models'].append(stats)
# otherwise the index will be rendered
# and we have to get the entry for our app,
# which is in this case 'traffic'
# using TrafficConfig.name or TrafficConfig.label
# might be better than hard coding the value
else:
app = app_dict.get('traffic', None)
# if an entry for 'traffic' has been found
# we append our manually created dictionary
if app:
app['models'].append(stats)
return app_dict
my_admin = MyAdmin(name='my_admin')
my_admin.register(Traffic)
Now when we open our custom admin we'll see something like this:
TRAFFIC
---------------------------------
Traffics + Add \ Change
Stats \ Change
This is because we manipulated the dictionary used to render the template and it uses the values we specified, in this case the most relevant is the name Stats, the admin_url which will call the custom view stats. Since we left add_url empty, there will be no + Add link displayed.
Important is also the penultimate line, where we call our custom admin and pass it a name, it will be used as a url namespace.
EDIT:
Unfortunately I noticed only after posting the answer that the question is asking how to display a link for a custom view created in a ModelAdmin, whereas my answer explains how to do this for a custom admin AdminSite. I hope it still of some help.
I know this is a bit old but I had the same issue and I found a simpler (yet maybe messier) way of doing it, that doesn't involve overriding templates or methods.
I just created a proxy model in models.py such as:
class Proxy(models.Model):
id = models.BigAutoField(db_column='id', primary_key=True)
def __str__(self):
return "<Label: id: %d>" % self.id
class Meta:
managed = False
verbose_name_plural = 'proxies'
db_table = 'proxy'
ordering = ('id',)
Which is just a mysql view that a created from am existing table
create view proxy
as select id
from samples
LIMIT 10;
And finally in admin.py
#admin.register(Proxy)
class LabelAdmin(admin.ModelAdmin):
change_list_template = 'label_view.html'
def changelist_view(self, request, extra_context=None):
...
return render(request, "label_view.html", context)
Although not a good way, but you can try overriding the default index.html of the django admin as :
{% for model in app.models %}
<tr class="model-{{ model.object_name|lower }}">
{% if model.admin_url %}
<th scope="row">{{ model.name }}</th>
{% else %}
<th scope="row">{{ model.name }}</th>
{% endif %}
{% if model.add_url %}
<td>{% trans 'Add' %}</td>
{% else %}
<td> </td>
{% endif %}
{% if model.admin_url %}
<td>{% trans 'Change' %}</td>
{% else %}
<td> </td>
{% endif %}
</tr>
{% endfor %}
<!-- extra link you want to display -->
<tr>
<th scope="row">link_name</th>
<td>{% trans 'Change' %}</td>
</tr>
After overriding this block put index.html inside your templates/admin directory of your project. Hope this helps.

How to auto-generate simple HTML with model fields? [duplicate]

I'm trying to create a basic template to display the selected instance's field values, along with their names. Think of it as just a standard output of the values of that instance in table format, with the field name (verbose_name specifically if specified on the field) in the first column and the value of that field in the second column.
For example, let's say we have the following model definition:
class Client(Model):
name = CharField(max_length=150)
email = EmailField(max_length=100, verbose_name="E-mail")
I would want it to be output in the template like so (assume an instance with the given values):
Field Name Field Value
---------- -----------
Name Wayne Koorts
E-mail waynes#email.com
What I'm trying to achieve is being able to pass an instance of the model to a template and be able to iterate over it dynamically in the template, something like this:
<table>
{% for field in fields %}
<tr>
<td>{{ field.name }}</td>
<td>{{ field.value }}</td>
</tr>
{% endfor %}
</table>
Is there a neat, "Django-approved" way to do this? It seems like a very common task, and I will need to do it often for this particular project.
model._meta.get_all_field_names() will give you all the model's field names, then you can use model._meta.get_field() to work your way to the verbose name, and getattr(model_instance, 'field_name') to get the value from the model.
NOTE: model._meta.get_all_field_names() is deprecated in django 1.9. Instead use model._meta.get_fields() to get the model's fields and field.name to get each field name.
You can use Django's to-python queryset serializer.
Just put the following code in your view:
from django.core import serializers
data = serializers.serialize( "python", SomeModel.objects.all() )
And then in the template:
{% for instance in data %}
{% for field, value in instance.fields.items %}
{{ field }}: {{ value }}
{% endfor %}
{% endfor %}
Its great advantage is the fact that it handles relation fields.
For the subset of fields try:
data = serializers.serialize('python', SomeModel.objects.all(), fields=('name','size'))
Finally found a good solution to this on the dev mailing list:
In the view add:
from django.forms.models import model_to_dict
def show(request, object_id):
object = FooForm(data=model_to_dict(Foo.objects.get(pk=object_id)))
return render_to_response('foo/foo_detail.html', {'object': object})
in the template add:
{% for field in object %}
<li><b>{{ field.label }}:</b> {{ field.data }}</li>
{% endfor %}
Here's another approach using a model method. This version resolves picklist/choice fields, skips empty fields, and lets you exclude specific fields.
def get_all_fields(self):
"""Returns a list of all field names on the instance."""
fields = []
for f in self._meta.fields:
fname = f.name
# resolve picklists/choices, with get_xyz_display() function
get_choice = 'get_'+fname+'_display'
if hasattr(self, get_choice):
value = getattr(self, get_choice)()
else:
try:
value = getattr(self, fname)
except AttributeError:
value = None
# only display fields with values and skip some fields entirely
if f.editable and value and f.name not in ('id', 'status', 'workshop', 'user', 'complete') :
fields.append(
{
'label':f.verbose_name,
'name':f.name,
'value':value,
}
)
return fields
Then in your template:
{% for f in app.get_all_fields %}
<dt>{{f.label|capfirst}}</dt>
<dd>
{{f.value|escape|urlize|linebreaks}}
</dd>
{% endfor %}
In light of Django 1.8's release (and the formalization of the Model _meta API, I figured I would update this with a more recent answer.
Assuming the same model:
class Client(Model):
name = CharField(max_length=150)
email = EmailField(max_length=100, verbose_name="E-mail")
Django <= 1.7
fields = [(f.verbose_name, f.name) for f in Client._meta.fields]
>>> fields
[(u'ID', u'id'), (u'name', u'name'), (u'E-mail', u'email')]
Django 1.8+ (formalized Model _meta API)
Changed in Django 1.8:
The Model _meta API has always existed as a Django internal, but wasn’t formally documented and supported. As part of the effort to make this API public, some of the already existing API entry points have changed slightly. A migration guide has been provided to assist in converting your code to use the new, official API.
In the below example, we will utilize the formalized method for retrieving all field instances of a model via Client._meta.get_fields():
fields = [(f.verbose_name, f.name) for f in Client._meta.get_fields()]
>>> fields
[(u'ID', u'id'), (u'name', u'name'), (u'E-mail', u'email')]
Actually, it has been brought to my attention that the above is slightly overboard for what was needed (I agree!). Simple is better than complex. I am leaving the above for reference. However, to display in the template, the best method would be to use a ModelForm and pass in an instance. You can iterate over the form (equivalent of iterating over each of the form's fields) and use the label attribute to retrieve the verbose_name of the model field, and use the value method to retrieve the value:
from django.forms import ModelForm
from django.shortcuts import get_object_or_404, render
from .models import Client
def my_view(request, pk):
instance = get_object_or_404(Client, pk=pk)
class ClientForm(ModelForm):
class Meta:
model = Client
fields = ('name', 'email')
form = ClientForm(instance=instance)
return render(
request,
template_name='template.html',
{'form': form}
)
Now, we render the fields in the template:
<table>
<thead>
{% for field in form %}
<th>{{ field.label }}</th>
{% endfor %}
</thead>
<tbody>
<tr>
{% for field in form %}
<td>{{ field.value|default_if_none:'' }}</td>
{% endfor %}
</tr>
</tbody>
</table>
Ok, I know this is a bit late, but since I stumbled upon this before finding the correct answer so might someone else.
From the django docs:
# This list contains a Blog object.
>>> Blog.objects.filter(name__startswith='Beatles')
[<Blog: Beatles Blog>]
# This list contains a dictionary.
>>> Blog.objects.filter(name__startswith='Beatles').values()
[{'id': 1, 'name': 'Beatles Blog', 'tagline': 'All the latest Beatles news.'}]
You can use the values() method of a queryset, which returns a dictionary. Further, this method accepts a list of fields to subset on. The values() method will not work with get(), so you must use filter() (refer to the QuerySet API).
In view...
def show(request, object_id):
object = Foo.objects.filter(id=object_id).values()[0]
return render_to_response('detail.html', {'object': object})
In detail.html...
<ul>
{% for key, value in object.items %}
<li><b>{{ key }}:</b> {{ value }}</li>
{% endfor %}
</ul>
For a collection of instances returned by filter:
object = Foo.objects.filter(id=object_id).values() # no [0]
In detail.html...
{% for instance in object %}
<h1>{{ instance.id }}</h1>
<ul>
{% for key, value in instance.items %}
<li><b>{{ key }}:</b> {{ value }}</li>
{% endfor %}
</ul>
{% endfor %}
I used https://stackoverflow.com/a/3431104/2022534 but replaced Django's model_to_dict() with this to be able to handle ForeignKey:
def model_to_dict(instance):
data = {}
for field in instance._meta.fields:
data[field.name] = field.value_from_object(instance)
if isinstance(field, ForeignKey):
data[field.name] = field.rel.to.objects.get(pk=data[field.name])
return data
Please note that I have simplified it quite a bit by removing the parts of the original I didn't need. You might want to put those back.
You can have a form do the work for you.
def my_model_view(request, mymodel_id):
class MyModelForm(forms.ModelForm):
class Meta:
model = MyModel
model = get_object_or_404(MyModel, pk=mymodel_id)
form = MyModelForm(instance=model)
return render(request, 'model.html', { 'form': form})
Then in the template:
<table>
{% for field in form %}
<tr>
<td>{{ field.name }}</td>
<td>{{ field.value }}</td>
</tr>
{% endfor %}
</table>
Below is mine, inspired by shacker's get_all_fields.
It gets a dict of one model instance, if encounter relation field, then asign the field value a dict recursively.
def to_dict(obj, exclude=[]):
"""生成一个 dict, 递归包含一个 model instance 数据.
"""
tree = {}
for field in obj._meta.fields + obj._meta.many_to_many:
if field.name in exclude or \
'%s.%s' % (type(obj).__name__, field.name) in exclude:
continue
try :
value = getattr(obj, field.name)
except obj.DoesNotExist:
value = None
if type(field) in [ForeignKey, OneToOneField]:
tree[field.name] = to_dict(value, exclude=exclude)
elif isinstance(field, ManyToManyField):
vs = []
for v in value.all():
vs.append(to_dict(v, exclude=exclude))
tree[field.name] = vs
elif isinstance(field, DateTimeField):
tree[field.name] = str(value)
elif isinstance(field, FileField):
tree[field.name] = {'url': value.url}
else:
tree[field.name] = value
return tree
This function is mainly used to dump a model instance to json data:
def to_json(self):
tree = to_dict(self, exclude=('id', 'User.password'))
return json.dumps(tree, ensure_ascii=False)
There should really be a built-in way to do this. I wrote this utility build_pretty_data_view that takes a model object and form instance (a form based on your model) and returns a SortedDict.
Benefits to this solution include:
It preserves order using Django's built-in SortedDict.
When tries to get the label/verbose_name, but falls back to the field name if one is not defined.
It will also optionally take an exclude() list of field names to exclude certain fields.
If your form class includes a Meta: exclude(), but you still want to return the values, then add those fields to the optional append() list.
To use this solution, first add this file/function somewhere, then import it into your views.py.
utils.py
#!/usr/bin/env python
# -*- coding: utf-8 -*-
# vim: ai ts=4 sts=4 et sw=4
from django.utils.datastructures import SortedDict
def build_pretty_data_view(form_instance, model_object, exclude=(), append=()):
i=0
sd=SortedDict()
for j in append:
try:
sdvalue={'label':j.capitalize(),
'fieldvalue':model_object.__getattribute__(j)}
sd.insert(i, j, sdvalue)
i+=1
except(AttributeError):
pass
for k,v in form_instance.fields.items():
sdvalue={'label':"", 'fieldvalue':""}
if not exclude.__contains__(k):
if v.label is not None:
sdvalue = {'label':v.label,
'fieldvalue': model_object.__getattribute__(k)}
else:
sdvalue = {'label':k,
'fieldvalue': model_object.__getattribute__(k)}
sd.insert(i, k, sdvalue)
i+=1
return sd
So now in your views.py you might do something like this
from django.shortcuts import render_to_response
from django.template import RequestContext
from utils import build_pretty_data_view
from models import Blog
from forms import BlogForm
.
.
def my_view(request):
b=Blog.objects.get(pk=1)
bf=BlogForm(instance=b)
data=build_pretty_data_view(form_instance=bf, model_object=b,
exclude=('number_of_comments', 'number_of_likes'),
append=('user',))
return render_to_response('my-template.html',
RequestContext(request,
{'data':data,}))
Now in your my-template.html template you can iterate over the data like so...
{% for field,value in data.items %}
<p>{{ field }} : {{value.label}}: {{value.fieldvalue}}</p>
{% endfor %}
Good Luck. Hope this helps someone!
Instead of editing every model I would recommend to write one template tag which will return all field of any model given.
Every object has list of fields ._meta.fields.
Every field object has attribute name that will return it's name and method value_to_string() that supplied with your model object will return its value.
The rest is as simple as it's said in Django documentation.
Here is my example how this templatetag might look like:
from django.conf import settings
from django import template
if not getattr(settings, 'DEBUG', False):
raise template.TemplateSyntaxError('get_fields is available only when DEBUG = True')
register = template.Library()
class GetFieldsNode(template.Node):
def __init__(self, object, context_name=None):
self.object = template.Variable(object)
self.context_name = context_name
def render(self, context):
object = self.object.resolve(context)
fields = [(field.name, field.value_to_string(object)) for field in object._meta.fields]
if self.context_name:
context[self.context_name] = fields
return ''
else:
return fields
#register.tag
def get_fields(parser, token):
bits = token.split_contents()
if len(bits) == 4 and bits[2] == 'as':
return GetFieldsNode(bits[1], context_name=bits[3])
elif len(bits) == 2:
return GetFieldsNode(bits[1])
else:
raise template.TemplateSyntaxError("get_fields expects a syntax of "
"{% get_fields <object> [as <context_name>] %}")
Yeah it's not pretty, you'll have to make your own wrapper. Take a look at builtin databrowse app, which has all the functionality you need really.
This may be considered a hack but I've done this before using modelform_factory to turn a model instance into a form.
The Form class has a lot more information inside that's super easy to iterate over and it will serve the same purpose at the expense of slightly more overhead. If your set sizes are relatively small I think the performance impact would be negligible.
The one advantage besides convenience of course is that you can easily turn the table into an editable datagrid at a later date.
I've come up with the following method, which works for me because in every case the model will have a ModelForm associated with it.
def GetModelData(form, fields):
"""
Extract data from the bound form model instance and return a
dictionary that is easily usable in templates with the actual
field verbose name as the label, e.g.
model_data{"Address line 1": "32 Memory lane",
"Address line 2": "Brainville",
"Phone": "0212378492"}
This way, the template has an ordered list that can be easily
presented in tabular form.
"""
model_data = {}
for field in fields:
model_data[form[field].label] = eval("form.data.%s" % form[field].name)
return model_data
#login_required
def clients_view(request, client_id):
client = Client.objects.get(id=client_id)
form = AddClientForm(client)
fields = ("address1", "address2", "address3", "address4",
"phone", "fax", "mobile", "email")
model_data = GetModelData(form, fields)
template_vars = RequestContext(request,
{
"client": client,
"model_data": model_data
}
)
return render_to_response("clients-view.html", template_vars)
Here is an extract from the template I am using for this particular view:
<table class="client-view">
<tbody>
{% for field, value in model_data.items %}
<tr>
<td class="field-name">{{ field }}</td><td>{{ value }}</td>
</tr>
{% endfor %}
</tbody>
</table>
The nice thing about this method is that I can choose on a template-by-template basis the order in which I would like to display the field labels, using the tuple passed in to GetModelData and specifying the field names. This also allows me to exclude certain fields (e.g. a User foreign key) as only the field names passed in via the tuple are built into the final dictionary.
I'm not going to accept this as the answer because I'm sure someone can come up with something more "Djangonic" :-)
Update: I'm choosing this as the final answer because it is the simplest out of those given that does what I need. Thanks to everyone who contributed answers.
Django 1.7 solution for me:
There variables are exact to the question, but you should definitely be able to dissect this example
The key here is to pretty much use the .__dict__ of the model
views.py:
def display_specific(request, key):
context = {
'question_id':question_id,
'client':Client.objects.get(pk=key).__dict__,
}
return render(request, "general_household/view_specific.html", context)
template:
{% for field in gen_house %}
{% if field != '_state' %}
{{ gen_house|getattribute:field }}
{% endif %}
{% endfor %}
in the template I used a filter to access the field in the dict
filters.py:
#register.filter(name='getattribute')
def getattribute(value, arg):
if value is None or arg is None:
return ""
try:
return value[arg]
except KeyError:
return ""
except TypeError:
return ""
I'm using this, https://github.com/miracle2k/django-tables.
<table>
<tr>
{% for column in table.columns %}
<th>{{ column }}</th>
{% endfor %}
</tr>
{% for row in table.rows %}
<tr>
{% for value in row %}
<td>{{ value }}</td>
{% endfor %}
</tr>
{% endfor %}
</table>
This approach shows how to use a class like django's ModelForm and a template tag like {{ form.as_table }}, but have all the table look like data output, not a form.
The first step was to subclass django's TextInput widget:
from django import forms
from django.utils.safestring import mark_safe
from django.forms.util import flatatt
class PlainText(forms.TextInput):
def render(self, name, value, attrs=None):
if value is None:
value = ''
final_attrs = self.build_attrs(attrs)
return mark_safe(u'<p %s>%s</p>' % (flatatt(final_attrs),value))
Then I subclassed django's ModelForm to swap out the default widgets for readonly versions:
from django.forms import ModelForm
class ReadOnlyModelForm(ModelForm):
def __init__(self,*args,**kwrds):
super(ReadOnlyModelForm,self).__init__(*args,**kwrds)
for field in self.fields:
if isinstance(self.fields[field].widget,forms.TextInput) or \
isinstance(self.fields[field].widget,forms.Textarea):
self.fields[field].widget=PlainText()
elif isinstance(self.fields[field].widget,forms.CheckboxInput):
self.fields[field].widget.attrs['disabled']="disabled"
Those were the only widgets I needed. But it should not be difficult to extend this idea to other widgets.
Just an edit of #wonder
def to_dict(obj, exclude=[]):
tree = {}
for field in obj._meta.fields + obj._meta.many_to_many:
if field.name in exclude or \
'%s.%s' % (type(obj).__name__, field.name) in exclude:
continue
try :
value = getattr(obj, field.name)
except obj.DoesNotExist as e:
value = None
except ObjectDoesNotExist as e:
value = None
continue
if type(field) in [ForeignKey, OneToOneField]:
tree[field.name] = to_dict(value, exclude=exclude)
elif isinstance(field, ManyToManyField):
vs = []
for v in value.all():
vs.append(to_dict(v, exclude=exclude))
tree[field.name] = vs
else:
tree[field.name] = obj.serializable_value(field.name)
return tree
Let Django handle all the other fields other than the related fields. I feel that is more stable
Take a look at django-etc application. It has model_field_verbose_name template tag to get field verbose name from templates: http://django-etc.rtfd.org/en/latest/models.html#model-field-template-tags
I just tested something like this in shell and seems to do it's job:
my_object_mapped = {attr.name: str(getattr(my_object, attr.name)) for attr in MyModel._meta.fields}
Note that if you want str() representation for foreign objects you should define it in their str method. From that you have dict of values for object. Then you can render some kind of template or whatever.
Django >= 2.0
Add get_fields() to your models.py:
class Client(Model):
name = CharField(max_length=150)
email = EmailField(max_length=100, verbose_name="E-mail")
def get_fields(self):
return [(field.verbose_name, field.value_from_object(self)) for field in self.__class__._meta.fields]
Then call it as object.get_fields on your template.html:
<table>
{% for label, value in object.get_fields %}
<tr>
<td>{{ label }}</td>
<td>{{ value }}</td>
</tr>
{% endfor %}
</table>
If you model name is Client and you are getting client object by id then proceed like this
client = Client.objects.get(id=id)
fields = Client._meta.get_fields()
for field in fields:
value = getattr(client, field.name)
print(field.name)
print(value)
<table border='1'>
<tr>
{% for mfild in fields%}
<td>{{mfild}}</td>
{% endfor%}
</tr>
{%for v in records%}
<tr>
<td>{{v.id}}</td>
<td>{{v.title}}</td>
<td class="">{{v.desc}}</td>
</tr>
{% endfor%}
</table>
enter code here

Output Django Model as Table

I have a view definition that (attempts to) outputs a model as a table. This is what I have so far:
def output_table(request):
output = My_Model()
return render_to_response('outputtable.html', {'output': output})
Here's the HTML for outputtable.html:
<html>
<table>
{{ output.as_table }}
</table>
</html>
What am I doing wrong? It doesn't work. Now, it's passing the model correctly, because if I change My_Model() to My_Model.objects.all() and then output it as simply {{ output }} then it shows me what I would see in the Django shell.
So what you need to do is the following:
1) add
from django.forms import ModelForm to your models.py
2) add
class My_Model_Form(ModelForm):
class Meta:
model = My_Model
3) in your views.py, change output = My_Model() to output = My_Model_Form()
Now you are all set. So the trick is to inherit your Form from your original model.
If you're just looking to output one specific model, change your template to something like
<tr>
<td>Name:</td>
<td>{{ output.name}}</td>
</tr>
for all the fields you care about. If this is something you want to be able to do for any arbitrary model in your app, take a look at this snippet. That would provide you with a fields collection to loop over.
The models.Model class doesn't have a method as_table() like the forms.ModelForm class does.
My solution was to also use template tags. I went with an inclusion tag.
myapp/templatetags/model_helpers.py
from django import template
register = template.Library()
#register.inclusion_tag('myapp/model_table.html', takes_context=True)
def model_as_table(context, model_key=None, model_table_attrs_key=None):
if model_key is None:
model_key = 'object'
if model_table_attrs_key is None:
model_table_attrs_key = 'model_table_attrs'
try:
attrs = context[model_table_attrs_key]
except KeyError:
attrs = context[model_key]._meta.get_all_field_names()
table_context = {'rows': []}
for attr in attrs:
try:
value = str(getattr(context[model_key], attr))
if value:
table_context['rows'].append({'attr': attr,
'value': context[model_key][attr]})
except AttributeError:
pass
# Needs a way to display many_to_many fields.
except StopIteration:
pass
return table_context
myapp/templates/myapp/model_table.html
{% for row in rows %}
<tr>
<td class="name">{{ row.attr }}</td>
<td class="field">{{ row.value }}</td>
</tr>
{% endfor %}
myapp/templates/myapp/outputtable.html
{% load model_helpers %}
<table>
{% model_as_table %}
</table>
With this implementation you can pass which model's attributes you want to display and in which order.
myapp/views.py
def output_table(request):
output = My_Model()
return render_to_response('outputtable.html',
{'output': output, 'model_table_attrs': ['attr1', 'attr2']})
I also like this cleanly separate html and python code.
There is no method as_table on a model instance (MyModel())or on Querysets (MyModel.objects.all()). You may have seen as_table as the forms method as_table. You don't have a form there.
If you want to print a model instance/ a queryset as a table, you will have to design it yourself.
OMG this is old but Generic Views seem to be a good fit for this problem, for me at least. A List detail view should help me get running faster. :)
I came up with a solution that worked for my specific need which renders any simple model's data into table rows. Add the following filter into templatetags/your_tags_file.py (don't forget the __init__.py file in the templatetags folder):
from django import template
register = template.Library()
#register.filter()
def as_table(model):
ret = ""
for name in model._meta.get_all_field_names():
try:
field = str(getattr(model, name))
if field:
ret += '<tr><td class="name">'+name+'</td><td class="field">'+field+'</td></td>'
except AttributeError:
pass
return ret
In the template now you can just do:
{% load your_tags_file %}
<table>
{{output|as_table|safe}}
</table>
This will render the model in a simple table for you. You can easily add any desired thead and tbody logic as you see fit by modifying the logic of how ret is generated. Hope this helps someone.