According to the docs:
http://django-grappelli.readthedocs.org/en/latest/customization.html#rearrange-inlines
The two classes for the placeholder are important. First, you need a class placeholder. The second class has to match the id of the inline–group.
All's well and good, I was able to set up my inlines fine, my issue is now - where does grappelli get the "id of the inline group" I can't find any reference, and pouring through the source code is offering me no solace.
Simply, I want to change the element-id that grappelli is using. Currently, it looks to me that it is taking the object name itself and converting to a lowercase name and appending set to the end. Do we have access to override the "id of the inline-group"?
Also, I am not 100% sure exactly how (or where) grappelli is doing this, it is definitely not documented... at all in fact.
Any help would be much appreciated.
It is the id of the inline element on HTML page. You can check the id of the default HTML inline element.
<div id="[related_name of ForeignKey]-group">
For example:
If in model "MyModel2", you have a ForeignKey like this:
my_model_1 = models.ForeignKey(MyModel1, related_name='my_model_2')
Then the id should be "my_model_2-group".
The id of the inline group is set in grappelli/templates/admin/edit_inline, in stacked.html line 5, or tabular.html line 6 (depending on which type of inline you're usng):
id="{{ inline_admin_formset.formset.prefix }}-group" >
You can override this by copying the file (stacked.html or tabular.html) into your template directory and setting the variable "template" to the file's new location e.g.:
# admin.py
class MyModelInline(admin.StackedInline):
template = 'path/to/stacked.html'
...
Then edit whatever you want in e.g. stacked.html.
I don't know if this is the best-practices way of doing this, but it's similar to what's done in the django tutorial.
Related
The scenario is that there are some dynamic texts on some templates, that will contain hyperlinks.
For this, I have a SiteDataKeyValue model, in which the dynamic texts for different parts of the template are inputted. This is the model:
class SiteDataKeyValue(models.Model):
key = models.CharField(
max_length=200, verbose_name="نام متن مورد نظر", unique=True
)
value = models.TextField(verbose_name="متن")
def __str__(self):
return self.key
A solution that I've found already, is Django urlize template tag. As mentioned in the docs, this tag converts texts like https://www.google.com to www.google.com, which is nice but not what I'd like to achieve. I want to be able to change the hyperlink text, so the output would be something like: Click Here!.
I searched for a bit, came across modules like bleach, which is a fine module, but I couldn't find the answer I was looking for (I skimmed through the docs and there was nothing about the hyperlink text).
Also I saw a comment somewhere telling that this could be achieved by writing a custom Django template tag, but although I tried to do this regarding the custom template filters docs, I didn't have a clue to how to achieve this.
I'm not asking for the code, although it would be really appreciated if you provide instructions for writing this custom template tag, or better, if you could point me to something like this that is already out there.
First of all you can extend urlize tag like the answer in this
or you can change the main code which you can find it in django.utils.html and override its url variable to change it.
But I think the best method is extending the urlize tag
like this:
{% text | urlize | change_a_text_filter:{{ dome_new_a_text }} %}
then you can scrape the text and use regex to find >sample-text</a> then you can change it to the argument that defines in your tag
from django import template
register = template.Library()
#register.simple_tag
def change_a_text_filter(format_string, arg):
# find the url that made in urlize with regex
# change it with arg
# return the result
I was on a completely wrong road to solve this problem. I was trying to urlize a link from TextField, but didn't consider the fact that I only needed to implement html code as Visit link.com! in the TextField, and then use safe template tag to render html directly as below:
{{ text.value|safe }}
So in this solution, there is no need to urlize, and of course there is no need to extend this tag neither.
NOTE: As commented by #rahimz (link to comment) I understand that there are safety concerns regarding safe tag, So I should emphasize that only me and a company-trusted admin will have access to admin panel and there is no worries that this admin will send malicious code through this TextField.
I'm using Django's standard comment system and I would like to extend its anti-spam honeypot capability.
I thought of changing the default "name" and "id" of the field to something more alluring for spam-bots such as "website". I checked the html and this looks like this:
<p style="display:none;">
<label for="id_honeypot">Never send a human to do a machine's job</label>
<input type="text" name="honeypot" id="id_honeypot" />
</p>
Am I correct in thinking that changing the defaults of this element would boost its anti-spam capabilities? I tried modifying it in the django/contrib/comments/forms.py like this:
class CommentForm(CommentDetailsForm):
#use to be honeypot = forms.CharField(...
website = forms.CharField(required=False,
label=_('Never send a human to do a machines job')
def clean_honeypot(self):
"""Check that nothing's been entered into the honeypot."""
value = self.cleaned_data["website"]
if value:
raise forms.ValidationError(self.fields["website"].label)
return value
And this successfully changes the name and id in the html generated by django BUT then the whole mechanism stops working - I tried populating this invisible field, submitted and the comment was added.
I have a few other ideas as well, but first I'd really like to get this working - is it possible to modify the default honeypot name and id AND have it working like it should?
P.S I believe a more elegent way of doing this would be to extend django.contrib.comments and code the modification there instead of working on actual django code - what would be the best way of accomplishing this?
Given a bit more time to tinker around I found the answer to both of my questions:
In order to modify the standard honeypot or to create your own, you have to extend the CommentForm class by adding a clean_NAME_OF_HONEYPOT function as well as a NAME_OF_HONEYPOT variable both of which look similar to the standard ones and you also have to override the security_errors function to include the name of your new/modified honeypot in the dictionary.
The best way to do this is to create your custom comments app as described here: https://docs.djangoproject.com/en/dev/ref/contrib/comments/custom/ .
I hope this answer helps anyone else in my situation.
I am using .mo files for localization in Django.
Also, in my database, I store some translated text in different fields, such as:
name_en, name_es, name_de (they are all columns in each row).
What will be the best method to choose the correct field inside a template?
i.e.:
{{ name.some_method }} will generate the correct translation based on the current localization.
Thanks,
Meit
You should look at http://goodcode.io/articles/django-multilanguage/ Here’s a simple solution that may fit your use case and is easy to implement and understand.
You should look at Django Transmeta, it work the same way as what you've done (DB fields with language code) but it's a more complete solution. It already deal with the template stuff, etc.
You can check Model Internationalization and Django Packages for more info and ideas in this domain.
I can see two method for doing this, one in your view and the other one is in the template...
In view:
Probably you keep the user language information somewhere so,
user_lang = 'es'
obj = Somemodel.objects.get(pk=123434)
obj.local_name = getattr(obj, 'name_%s'%user_lang)
So, you keep local translation in a specific variable of the instance and in your template you can use is as:
{{obj.local_name}}
But that might be costly if you wish to pass the template a queryset instead of a single instance. For a such usege you have to evaluate that value for each object in your queryset.
In template:
That is a more complex way of solving the porblem in the template...
Define a template tag and pass object_id, and local language information and get the translated text using a similar getattr function. But in that point, if you wish to use this for more than one model, you probably have to pass a content type information for your template tag too, such as:
{% get_translation <object_id> <content_type_id> <local_language> %}
And in your template tag function, do something like:
from django.contrib.contenttypes.models import ContentType
....
cont_obj = Content_type.objects.get_for_id(<cotent_type_id>) #get the related model
obj = cont_obj.get_object_for_this_type(pk=<object_id>) # get your object
return getattr(obj, 'name_%s'%<local_language>)
I have a generic inline which I am attaching to certain models. It works fine, but the admin interface looks kind of ugly:
http://i.stack.imgur.com/dI3UH.png
As you can see, the same heading is repeated several times. In this instance I am only ever going to add one entry to the inline, which I have setup like so:
class PageMetaInline(generic.GenericStackedInline):
model = PageMeta
extra = 1
max_num = 1
Is there any way to remove the unnecessary header "Page Meta: #1" ?
You could override the admin template for stacked inlines for the PageMeta model.
Copy the original template (view current version in trunk on the Django site) to 'admin/yourapp/pagemeta/stacked.html', then customise by removing the html you do not wish to display.
Is there a straightforward, common way to apply custom styling on admin change list element depending on its properties?
update
To be more precise: let's say I have a simple model object.
Foo
field1
field2
field3
#property
property1()
#property
property2()
ModelAdmin.list_display is defined as a subset of the available fields, so not every attribute (field/property) is displayed in the change list table.
I'd like to apply custom CSS class to the object's row when certain condition is fulfilled, for example: if foo_instance.property1 is True then add class bar to the corresponding tr element.
Now copy the template admin/base_site.html from within the default Django admin template directory (django/contrib/admin/templates) into an admin subdirectory of whichever directory you're using in TEMPLATE_DIRS. For example, if your TEMPLATE_DIRS includes "/home/my_username/mytemplates", as above, then copy django/contrib/admin/templates/admin/base_site.html to /home/my_username/mytemplates/admin/base_site.html. Don't forget that admin subdirectory.
Note that any of Django's default admin templates can be overridden. To override a template, just do the same thing you did with base_site.html -- copy it from the default directory into your custom directory, and make changes.
from django's tutorial
What exactly do you mean by "change list element" and "it's properties"? Using CSS 2 or CSS 3 selectors you can do some things. Otherwise, you might be able to do it easily using jQuery (or whatever). Since it is merely presentation related, I think this would be the cleanest solution.
Old question but if you stumble across it, the following tips might be helpful.
You can use django-liststyle to customise your admin changelist rows.
It's quite simple to implement your example:
class FooAdmin(admin.ModelAdmin, ListStyleAdminMixin):
...
def get_row_css(self, obj, index):
if obj.property1:
return 'bar'
return ''
Django Suit (not free) also offers "List row and cell attributes" style customisation