I am a newbee in Django, And I am trying to alter one of my view to order by the user signed in. For example, If 'test#example.com' is signed in, then the Tview should give data related to Test#example.com in the first row. At the moment it is in no specific order.
Current order: for a user signed as test#example.com
GEL_TEACT trea#example.com
TREAT_ACT test#example.com
I want to order this as:
Product OWNER
TREAT_ACT test#example.com
GEL_TEACT trea#example.com
my View:
class PListView(ListView):
model = Product
template_name = "app/product_list.html"
project_list_view = PListView.as_view()
my template:
<tbody>
{% for product in object_list %}
<tr>
<td>{{ product.title }}</td>
<td>
<a href="{% url "product-update" project_id=project.pk %}">
<i class="fi-pencil"></i>
</a>
</td>
</tr>
{% endfor %}
</tbody>
Thanks for your time.
These are three different options, depending on what you need. Just use one of them.
In your example, a logged in user should be displayed first. That is just one item to put in front of the list. So its easiest to do that right in the template
<ul>
<li>{{ user.username }}</a></li>
{% for item in users %}
{% if item != user %}
<li>{{ item.username }}</a></li>
{% endif %}
{% endfor %}
</ul>
However, if you want to get the list in a particular order, there are two ways to achieve that. If the order is just for this one ListView, then use the get_queryset() method of ListView()
class PListView(ListView):
model = Product
template_name = "app/product_list.html"
get_queryset(self):
return Product.objects.all().order_by('name')
However, if the data should always be ordered in this way, on all lists of the Product model, you can order it right on the model instead
class Product(models.Model):
name = models.CharField(max_length=50)
class Meta:
ordering = ['name', ]
Related
I know I can change the display title for a model in Django Admin using
class Meta:
verbose_name='Custom Model Name Here'
However, is there a way to display which app heading a model is displayed under?
For example, if I create a custom user model Users in a new app also called users then the default user model goes from Authentication and Authorization > Users to Users > Users.
I would like to retain it under the original heading Authentication and Authorization > Users.
I have read this answer which suggests changes the app verbose_name, however it only changes the verbose name of the app associated with the model. I want to show the model in a different group on the admin panel. You can see the issue that approach takes here:
You could set up a proxy model
In app_1.models.py
class MyModel(models.Model):
...
In app_2.models.py
from app_1.models import MyModel
class MyModelProxy(MyModel):
class Meta:
proxy = True
Then register the MyModelProxy in admin as normal
There is no simple way of doing what you're intending, check out the Django code/template that generates the view:
#never_cache
def index(self, request, extra_context=None):
"""
Display the main admin index page, which lists all of the installed
apps that have been registered in this site.
"""
app_list = self.get_app_list(request)
context = {
**self.each_context(request),
'title': self.index_title,
'app_list': app_list,
**(extra_context or {}),
}
request.current_app = self.name
return TemplateResponse(request, self.index_template or 'admin/index.html', context)
templates/admin/index.html:
{% for app in app_list %}
<div class="app-{{ app.app_label }} module">
<table>
<caption>
{{ app.name }}
</caption>
{% 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 %}
{% if model.view_only %}
<td>{% trans 'View' %}</td>
{% else %}
<td>{% trans 'Change' %}</td>
{% endif %}
{% else %}
<td> </td>
{% endif %}
</tr>
{% endfor %}
</table>
</div>
{% endfor %}
As you can see, Django admin template is looping your app_list directy, so the only way of doing it would be to override the admin/index.html template to place your models in your desired order.
From Django docs:
For those templates that cannot be overridden in this way, you may still override them for your entire project. Just place the new version in your templates/admin directory. This is particularly useful to create custom 404 and 500 pages.
https://docs.djangoproject.com/en/2.2/ref/contrib/admin/#overriding-vs-replacing-an-admin-template
In firstapp.models.py
class ActualModel(models.Model):
class DisplayModel(ActualModel):
class Meta:
proxy = True
in secondapp.admin.py
from firstapp.models import DisplayModel
admin.site.register(DisplayModel)
Easier way to do that:
In model class Meta add
app_label = 'app2'
db_table = 'app1_modelname'
so you assigning model to another app with app_label and assigning db_table with old (current) table name to prevent making migrations and table renami
I am trying to split a list from my model across two columns, using this html code in the template:
< div class ="col-md-6" >
{%for value in object_list %}
<ul>< ahref="/sites/{{value.url}}/">{{value.Site}}</a></ul>
{% endfor %}
I was planning to achieve this with the slice tag to filter the list, e.g.:
{%for value in object_list|slice:"10:20" %}
It does not work however, and I think it might be because I have context data i.e. {{value.Site}}, instead of just {{Site}} for example. This is the corresponding view:
class homeview(ListView):
template_name = 'annual_means/home.html'
def get_queryset(self):
return AnnualMean.objects.values("Site", "url").distinct()
What do I need to do to get the slice to work?
I think, what you need is this:
<table>
<tr>
<th>URL</th>
<th>SITE</th>
</tr>
{% for value in object_list %}
<tr>
<td>{{value.url}}</td>
<td>{{value.Site}}</td>
</tr>
{% endfor %}
</table>
URLs and Sites will be displayed as a table.
I've got a DetailView to display a Menu (restaurant), with the following structure:
Menu > Courses > Course_Categories > Dishes
i.e. (Theather Menu) > (Starters, Entrees, Desserts) > (Fish Soup, Ribs, etc.)
I want to prefetch the course_categories and dishes, in addition to that I want to set a property on each dish to be used in the template. This property (the price) is dependent on time of day, therefore it's not simple stored as a value on the Dish.
I tried the following:
class MenuView(generic.DetailView):
template_name = "cms/detail/menu.html"
model = Menu
def get_object(self, queryset=None):
pkValue = pk = self.kwargs.get(self.pk_url_kwarg, None)
menu = Menu.objects.get(id=pkValue)
courses = menu.courses.all().prefetch_related('course_categories').prefetch_related('dishes')
for course in courses.all().iterator():
for course_category in course.course_categories.all().iterator():
for dish in course_category.dishes.all().iterator():
dish.price = "0.00"
return menu
When I iterate the data in my template, it shows everything, but no value for the dish.price property. What I think happens is that the related sets are re-retrieved, and therefore my custom set property doesn't show.
Template:
{% for course in menu.courses.iterator %}
<tr>
<td><strong>{{ course.name }}</strong></td>
</tr>
{% for course_category in course.course_categories.iterator %}
<tr>
<td><em> {{ course_category.name }}</em></td>
</tr>
{% for dish in course_category.dishes.iterator %}
<tr>
<td> {{ dish.name }} {{ dish.price}}</td>
</tr>
{% endfor %}
{% endfor %}
{% endfor %}
Any suggestions?
You are using iterator() which explicitly does not cache the results. When you do {% for course in menu.courses.iterator %}, the results are fetched from the database again.
Use simply .all() in your view instead of .all().iterator(), and stop using .iterator in your template, and you shouldn't have the problem.
I have a model called Subtopic. One of my templates runs a forloop on an object, returning a different field for each cell of a table row.
Two of the table cells look up a field which is a ManytoMany foreign key, both to the same foreign model, Resource. I want each to display different results, based on the value of a boolean field within the Resource model.
What you see below is currently working fine, but doesn't attempt to filter by the boolean field.
models.py:
class ICTResourceManager(models.Manager):
def get_query_set(self):
return super(ICTResourceManager, self).get_query_set().filter('is_ict': True)
class NonICTResourceManager(models.Manager):
def get_query_set(self):
return super(NonICTResourceManager, self).get_query_set().filter('is_ict': False)
class Resource(models.Model):
subtopics = models.ManyToManyField(Subtopic)
external_site = models.ForeignKey(ExternalSite)
link_address = models.URLField(max_length=200, unique=True, verify_exists=False)
requires_login = models.BooleanField()
is_ict = models.BooleanField()
flags = models.ManyToManyField(Flag, blank=True)
comment = models.TextField()
def __unicode__(self):
return u'%s %s' % (self.external_site, self.link_address)
objects = models.Manager()
ict_objects = ICTResourceManager()
nonict_objects = NonICTResourceManager()
class Meta:
ordering = ['external_site', 'link_address']
views.py:
def view_ks5topic(request, modulecode, topicshortname):
listofsubtopics = Subtopic.objects.filter(topic__module__code__iexact = modulecode, topic__shortname__iexact = topicshortname)
themodule = Module.objects.get(code__iexact = modulecode)
thetopic = Topic.objects.get(module__code__iexact = modulecode, shortname__iexact = topicshortname)
return render_to_response('topic_page.html', locals())
My template:
{% for whatever in listofsubtopics %}
<tr>
<td>
{{ whatever.objective_html|safe }}
<p>
{% if request.user.is_authenticated %}
{% with 'objective' as column %}
{% include "edit_text.html" %}
{% endwith %}
{% else %}
{% endif %}
</td>
<td>
{% regroup whatever.resource_set.all by external_site.name as resource_list %}
{% for external_site in resource_list %}
<h4>{{ external_site.grouper }}</h4>
<ul>
{% for item in external_site.list %}
<li>{{ item.comment }}</li>
{% endfor %}
</ul>
{% endfor %}
</td>
</tr>
{% endfor %}
As you can see, I've added extra managers to the model to do the filtering for me, but when I replace the appropriate lines in the template, I just get blanks. I have tried: for external_site.ict_objects in resource_list and for item.ict_objects in resource_list and <a href="{{ item.ict_objects.link_address }}">. If this were in the view I could probably do the filter just by .filter('is_ict': True), but with this being inside a forloop I don't know where to do the filtering.
I also tried writing regroup whatever.resource_set.filter('is_ict': True) in the template, but the syntax for regrouping seems to use resource_set.all rather than resource_set.all() (and I don't know why) so the filter text doesn't work here.
Turns out it was possible to do it using a custom template filter. The original efforts to filter within the template weren't working, given that as is well documented the template language is not a fully-fledged python environment. My original question remains open for anyone who knows an alternative method that more directly addresses the question I was originally asking, but here's how I did it:
myapp_extras.py:
from django import template
register = template.Library()
def ict(value, arg):
"filters on whether the is_ict Boolean is true"
return value.filter(is_ict=arg)
register.filter('ict', ict)
My template, note the use of the custom filter in line 2:
<td>
{% regroup whatever.resource_set.all|ict:1 by external_site.name as resource_list %}
{% for external_site in resource_list %}
<h4>{{ external_site.grouper }}</h4>
<ul>
{% for item in external_site.list %}
<li>{{ item.comment }}</li>
{% endfor %}
</ul>
{% endfor %}
</td>
After this I was able to remove the additional custom managers from the model.
One further question, when filtering for the boolean is_ict field, I found that I had to use filter(is_ict=1) and filter(is_ict=0). Is that the only way to refer to a True or False value?
I am displaying the database field names and values in html table in django templae with this code
<table id="listTable" >
<tr>
{% for fieldname in object_fields %}<th>{{ fieldname }}</th>{% endfor %}
</tr>
{% for object in object_list %}
<tr class="{% cycle 'row1' 'row2' as rowcolors %}">
{% for fieldvalue in object %}<td>{{ fieldvalue }}</td>{% endfor %}
</tr>
{% endfor %}
</table>
Now the problem i have one field called image_name and i want that for that there should be hyperlink inserted with the text but how can do that so that other columns don't get affected.
I use this code for all the tables
What I would do in this case is add a property to the class that wraps the value of the field in an anchor tag and add that property to your object_list instead of the default value of the field. Here's a quick and dirty example to get you going.
#models.py
class MyClass(models.Model):
name = models.CharField(max_length=50)
email = models.EmailField()
#property
def email_link(self):
return u'%(email)s' % {'email' : self.email}