I made a simple django form, with a list of choices (in radio buttons):
class MyForm(forms.Form):
choices=forms.ChoiceField( widget=forms.RadioSelect(), choices=[(k,k) for k in ['one','two','three']],label="choose one")
I would like the form to submit automatically when a user selects one of the options. In straightforward HTML I would've done it as
<select name='myselect' onChange="FORM_NAME.submit();">
....
</select>
But I do not know how to integrate this into the form class without writing a template. Specifically, I would need to know FORM_NAME so I can call FORM_NAME.submit() in the above snippet.
Can it be done without using a template?
I think you do not need to know the form name. This should work as well:
<select name='myselect' onChange="this.form.submit();">
A quick solution to integrate this into your form would involve adding a attribute to your widget.
widget=forms.RadioSelect(attrs={'onchange': 'this.form.submit();'})
Now one could argue if this isn't better separated from your form definition (separating definition, style and behaviour), but that should do it.
Related
I have a relatively complicated form that's used in multiple places on my website (in fact, it's a form from which many other form classes inherit). In the templates, the inherited part of this form is always formatted identically—but that formatting is somehwat involved; each field is rendered and positioned manually in the template.
This means that every template which displays this form has a lot of identical HTML markup that renders the form appropriately.
I would like to create a custom output that can be called, similar to the as_table() methods. I'm aware that one can override the normal_row, error_row, etc. attributes—but the formatting of this form goes beyond that (for example, three of the form's five fields should be printed side-by-side, with a combined title). All of the tutorials/answered-questions I've seen either refer to overriding the above-mentioned attributes, or give instructions on how to manually render forms.
Originally, I was thinking something like this:
Class StrangeForm(form.Forms):
....
def as_table_custom():
html_string = "\
<tr><td>Title 1:</td><td>self.fields['field1']</td><tr>\
<tr><td>Title 2:</td><td>self.fields['field2']</td><tr>\
<tr><td>Titles 3, 4, 5:</td><td>self.fields['field3']\
</td><td>self.fields['field4']</td><td>self.fields['field5']</td></tr>\
"
return html_string
But, after reading through the _html_output() and as_table() methods of Django's forms.py file, it doesn't look like it'll be that easy. If I write this from scratch, have to somehow account for errors, help text, etc. I think.
Is there an easy way to override something such that the form's HTML output can be defined like above? Or do I have to re-write things from scratch. If the latter, how can I account for all of the things I need to account for?
I wouldn't take this approach. You're better off creating the form in an HTML template that you include in the various templates where you have a form you want to display that way.
So create a my_strange_form.html template where you assume a 'form' object is passed in the context with the right number of fields. In that template just create the HTML, using things like {{ field.label_tag }} and {{ field }}. You can loop through the fields with a {% for field in form %} and check the counter of your loop with {{ forloop.counter }}. I foresee a lot of {% if forloop.counter... %} statements to generate the combined row, and it will look ugly, but you'll only have to look at it once :-)
Then in your main templates {% include 'my_strange_form.html' with form=form %}.
I like Django form render(‘form’: form) and template {{ form }}, but it doesn’t allow you to arrange fields. I was just wondering if there was an easier way to do it, without rendering the whole form manually.
What Django does:
First Name
Middle Name
Last Name
What I want to do:
First Name Middle Name
Last Name
What makes Django {{ form }} so great is that it puts field labels and help text and all that jazz in the right places. Where as manually, you have to put all that in. If there isn’t an easier way, I’m ok with that, but I thought I would at least ask.
You could try to render individual fields of the form, like this part of the docs shows.
When I render a form, it generates something like this for each Field:
<div class="field_content">
<label>...</label>
<div class="field">...</class>
</div>
I would like to be able to uniquely identify each Field in my stylesheet. Is there a way to add another class to the outer div (in addition to field_content), or an outer div (parent to field_content)?
If it's enough to set the class of the field you can specify it as an attribute of your widget:
name = forms.CharField(widget=forms.TextInput(attrs={'class':'special'}))
If you need more control it's probably best to render your form manually or if you don't mind relying on an external app there are even more flexible solutions:
In django-floppyforms you could use form layouts and in django-crispy-forms you could use layout objects and to achieve what you want.
I need to place a form in all my subsites and home page. It's placed on right column. It's action is "/". If there are some errors I must display them.
I made Form class etc. I display it via {{ form.as_p }} and designed all.
The problem is - how to put it on all sites? I don't want to make form instance in all my views functions.
What is the best solution for that?
Looks like a work for a context processor. :-)
def my_form_processor(request):
return {'my_form': MyForm()}
I want to do the following:
Having a model (p.e. a model which handles data about photographic reports) create a section which has a preview of an specific flickr album. The URL will be provided by an URLField (until the first save the preview will not be available).
After the first save, it'll show previews of all the images inside that album, and make them selectable (through jQuery for example). Then again, when the images are selected and the object is saved (I think I can use django signals for this) it will notify a specific user telling him a selection has been made.
Is there any plugins available, or any easy way to implement this in django-admin?
Update: 22 days and no anwers... does that mean it can't be done in django-admin?
I personally can't think of any easy way to implement this in the Django admin, simply because I doubt many people who've done it have thought to open source it. I can imagine that it would be very specific to a certain user's / programmer's needs.
In any case, if you wanted to solve this issue, I'd say that your best bet would be overriding the Django admin templates in your django/contrib/admin/templates/admin folder. I believe you'd be best off by editing change_form.html.
My basic approach would be this:
Check the name of the model using opts.verbose_name. For example, if you wanted to do this processing for a model whose verbose name is "Gallery", you would do
{% ifequal opts.verbose_name "Gallery" %}
<!-- neat gallery view -->
{% else %}
<!-- regular form -->
{% endifequal %}
Make a custom template tag that will display the gallery view / form given the object_id and the type of object. This way you can replace the <!-- neat gallery view --> with a {% show_gallery object_id %}. See the Django Docs for more info on creating custom template tags. It's pretty straightforward.
Add whatever Javascript or custom stuff in your template tag template. What you choose to do is up to you.
Sorry you haven't gotten many more answers to your question. Hope this helps!