I have a simple use case.
I have 3 models foo, foo1 and foo2. I have created model forms for them.
Now in a page I have these forms together on that page.
PROBLEM : The user has the option to click on a "ADD" button. Once he clicks on it he will be able to add 1 more foo2 i.e. 1 more foo2 form will open or be displayed. How should I do this ? i.e. how do I add a modelform dynamically ?
Edit:
This can be done using jquery. But my question is how do I differentiate it in the view ? Should I use prefix ? i.e. should I add dynamic prefix to the forms that are being created ?
As Rohan says in the comments, if you have more than one of the same kind of form on a page, you should be using formsets. Each form within the formset has its own prefix which includes an ID which you can simply increment with your Javascript.
Related
I am missing an idea or a way to handle this in django.
I have a list of instances (class Sample) and I want to present the last 10 of them in a form with a checkbox before... I want to have a button "work on these" to submit the checked instances to a view. This view should present a detailView of the selected instances.
I don't know, how to handle this exactly. Can somebody give me a hint? All I did so far was just creating a form of a instance.
Thanks in advance.
You can use a ModelChoiceField or a ModelMultipleChoiceField.
class MyForm(forms.Form):
my_list_of_instances = forms.ModelMultipleChoiceField(queryset = MyModel.objects.all().order_by('-id')[:10], widget=forms.CheckboxSelectMultiple)
This should generate a list of checkbox with your last 10 instances created.
With this you have your form ready to receive your input...just point your action to your view.
Edit:
In case you want to have several forms, one for each instance, you can work with Formsets. You would not get the checkboxes tho...you would get any change applied to your instances. If you want to have both (checkboxes and formsets) you can use both suggestions, using the checkbox values sent to the server to filter the formsets you are going to save.
I am trying to achieve this : I want one parent form which have a "submit" button and it should contain the two child forms in which one of the child (here suppose have two child child-1 and child-2 ) child-1 gets added to form dynamically when user click "add child-1". How to achieve this in django forms ?
I think you can create form with hidden fields(child-1) and show them with jQuery after pressing on "add child-1" button.
If your children are of the same origin (ex. a form asking if you have chlidren; you add one child and then another), then you need formsets:
https://docs.djangoproject.com/en/dev/topics/forms/formsets/
If you want a single form to be broken in parts/steps (ex. step 1: fill in your address; step 2: provide information about your company; step 3: etc...), you need a form wizard:
https://docs.djangoproject.com/en/dev/ref/contrib/formtools/form-wizard/
I am using jquery autocomplete with my generic views on Django. I am getting the list via AJAX, but there is one problem. When the user get the category he wants, he will select that value. Thus, if I append the value on that field $("#id_category").val(ui.item.value); it will show the pk on that field instead of value, so if I append $("#id_category").val(ui.item.label); django will complain it should be the instance. Django is completely right. How to make this work?
Edit:
I have made id_category hidden field and added category_display to append the text value. It is working. But while editing the category_display would be empty. Again I am using generic view. How to solve this one on the edit?
I had similar issue and I suggest you to think about question: 'What will happen if user enters not valid category name(category with this name doesn`t exist)?'
if you need to accept entered value and add new category, you have to make clean_category method in form:
def clean_category(self):
name = self.data['category']
return Category.objects.get_or_create(name=name)[0]
and display ui.item.label for user
if new category isn`t acceptable you can raise error in clean_category method or use something like select2(http://ivaynberg.github.io/select2/) for your field. I choose second variant.
I have a couple of fields in my model that to which I wish to add a link that will allow the user to search for names/files (external from the application's database).
So what I would like is:
Field name: [text box] - LINK
Is there a straightforward django way of achieving this?
Cheers.
You need to change the widget that the form field uses to display the models information. You basically add some html after the input to link to where you want.
Here's some code I put together to create a widget that displays how many characters are left for a CharacterField so it's similar to what you are looking to do:
https://github.com/pastylegs/django-widget-charsleft/blob/master/widget_charsleft/widgets.py
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