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How to create a "save and add another" button and show on Wagtail admin model page

Was wondering if it is possible to add a custom button within a Wagtail model page that will allow me to save (create) the current data fields and move on to another page. The "save and add another" button was already available in django and I want to have something like that on the Wagtail model page.
Thanks.

Wagtail has a few ways to customise content that is shown in the various Wagtail modeladmin views.
Code Example
Step 1 - create a custom CreateView
This will override the get_context_data method to provide the create_url, we could do this simpler (the context already has access to the instance) but it is nice to be explicit.
Override the get_success_url method to allow for a URL param of next to be set, if that exists the next URL will be this instead of the default behaviour.
products/wagtail_hooks.py
from wagtail.contrib.modeladmin.views import CreateView
# ... other imports
class CustomCreateView(CreateView):
def get_context_data(self, **kwargs):
context = super().get_context_data(**kwargs)
context['create_url'] = self.create_url
return context
def get_success_url(self):
next = self.request.GET.get('next')
if next:
return next
return super().get_success_url()
class ProductsModelAdmin(ModelAdmin):
create_view_class = CustomCreateView
Step 2 - Override the template to add a new button
Wagtail modeladmin uses a template path override approach, you will need to create a new template that aligns with your desired override (for example we may just want to override the product model create view only.
templates/modeladmin/products/products/create.html -> this says to override the create template for the products model within the products app.
Add the code below and check it is working, sometimes the templates path can be a bit tricky so ensure you can see the new button AND that the button has the data-next attribute that should be your create URL.
templates/modeladmin/products/products/create.html
{% extends "modeladmin/create.html" %}
{% load i18n wagtailadmin_tags %}
{% block form_actions %}
{{ block.super }}
<div class="dropdown dropup dropdown-button match-width">
<button type="submit" class="button button-secondary action-save button-longrunning" data-next="{{ create_url }}" data-clicked-text="{% trans 'Saving…' %}">
{% icon name="spinner" %}<em>{% trans 'Save & add another' %}</em>
</button>
</div>
{% endblock %}
Step 3 - Add JS to modify the form action URL on click
The next step requires a bit of JavaScript, we want to attach a different behaviour to the 'save and add another' button.
The JS used here does not require jQuery and should work in IE11
Once the DOM is loaded (which means JS can do things), find all the buttons with the data-next attribute.
Add a listener to the 'click' of each of those buttons which will dynamically update the form action attribute with the extra URL part.
This extra URL part gets read by the get_success_url method in the custom create class but only on a successful save, otherwise you will stay on the page and be able to fix errors.
{% block extra_js %}
{{ block.super}}
<script>
document.addEventListener('DOMContentLoaded', function() {
document.querySelectorAll('[data-next]').forEach(function(button) {
button.addEventListener('click', function(event) {
var form = document.querySelector('.content-wrapper form');
form.action = form.action + '?next=' + button.dataset.next;
});
});
})
</script>
{% endblock %}

View function is not being called after submitting form in django

I have made a simple form inside a html file whose path is www.site.com/posts/5. Whenever the form is submitted, it redirects back to the same page i.e www.site.com/posts/5 displaying a message given by user in the form.
However, whenever the form is submitted it doesn't call the foobar view.
The urls.py, views.py and html files are as follows:-
urls.py
urlpatterns = [
path('posts/<int:foo>',user_views.display, name="display",
path('posts/<int:foo>',user_views.foobar, name="makefoo"),
]
views.py
def foobar(request, foo):
#do something
html file
<form name="fooform" action= "{% url 'makefoo' 5 %}" method = "post">
{% csrf_token %}
<input type="text" name="FOO_BODY" maxlength="300" required>
<input type="submit" value="comment">
<input type="reset" value="clear">
</form>
Edit : user_views is just from user import views as user_views

You can not attach two views to the same URL. The {% url ... %} template tag, only generates a URL for that path. But if there is a "url clash", then it is possible that the requests ends up in the other view.
You thus should define another URL, or encode the post logic in the display view. In case of a POST request, you can thus first take the necessary steps, and then for example return a redirect to the page, such that we can again render the page:
def display(request, foo):
if request.method == 'POST':
# do something
return redirect(display, foo=foo)
#do something else (original code)
return HttpResponse(..)
This is the famous Post/Redirect/Get web development design pattern [wiki]. This is usually better than returning a HTTP response directly in the POST, since if the user performs a refresh, the POST will be performed a second time.

As mentioned in the comment by #williem, you have two path() defined in the urls.py.
Always First matching route will be picked up from the url route table. So whenever r^'posts/' is requested it will call the display() from the user_views, so it will never go to foobar(). Either remove the route with display() or change the sequence. Also, I assume you imported the user_views.
Reference:
https://docs.djangoproject.com/en/2.1/topics/http/urls/

How can I add a form made by formbuilder to every page in Wagtail?

Is there any way to add form (for example feedback form) to every page in CMS? I really like to use Wagtail FormBuilder so Editor guy can change fields.
My first idea is to create custom form page (inherited from AbstractEmailForm) as site root child and load it to base.html trough template tag. I can access page properties this way but I cant render the form.
Here is my template tag:
#register.assignment_tag(takes_context=True)
def get_feedback_form(context):
return context['request'].site.root_page.get_children().type(FeedbackFormPage).first()
And this is how I use it from base.html:
{% get_feedback_form as feedback_form %}
...
{{ feedback_form.specific.title }} <-- this works
{{ feedback_form.specific.form.as_p }} <-- this doesnt work
It would be nice somehow to create a form as snippet or add it to Site Settings, but I didnt find how to do that.

The main issue is how you are generating the form in the template with .form.as_p.
You will need to generate the form with the .get_form function, but you are best to do it within your template as the current user and page needs to be past in as arguments like this.
form = feedback_form_page.get_form(
page=feedback_form_page, user=request.user)
You can see how the form is built for the AbstractForm model here:
https://github.com/wagtail/wagtail/blob/master/wagtail/wagtailforms/models.py#L278
Full detailed example below, along with how you could work the form selection into the Site Settings module.
Link to a Form in Site Settings
Assuming you are referring to the Site Settings contrib module:
http://docs.wagtail.io/en/v1.13/reference/contrib/settings.html
The 'Edit Handlers' section of the documentation explains a great way to link to a page inside of your site settings.
http://docs.wagtail.io/en/v1.13/reference/contrib/settings.html?highlight=site%20settings#edit-handlers
Example (in models.py):
from wagtail.contrib.settings.models import BaseSetting, register_setting
# ...
#register_setting
class MyCustomSettings(BaseSetting):
feedback_form_page = models.ForeignKey(
'wagtailcore.Page', null=True, on_delete=models.SET_NULL)
panels = [
# note the page type declared within the pagechooserpanel
PageChooserPanel('feedback_form_page', ['base.FormPage']),
]
Once you set this model up, you will need to do makemigration and migrate for the changes to work in admin. You will then see inside the settings menu a sub-menu titled 'My Custom Settings'
Adding linked Form to every page
Add a block (so it can be overridden in templates) that has an include in your base template (eg. myapp/templates/base.html).
<!-- Footer -->
<footer>
{% block feedback_form %}{% include "includes/feedback_form.html" %}{% endblock feedback_form %}
{% include "includes/footer.html" %}
</footer>
Create an include template (eg. myapp/templates/includes/feedback_form.html)
{% load feedback_form_tags wagtailcore_tags %}
{% get_feedback_form as feedback_form %}
<form action="{% pageurl feedback_form.page %}" method="POST" role="form">
<h3>{{ feedback_form.page.title}}</h3>
{% csrf_token %}
{{ feedback_form.form.as_p }}
<input type="submit">
</form>
Build a Template Tag to get the form and page
Your template tag needs to build the form with the page's self.get_form() function. Eg. you your template tag (base/templatetags/feedback_form)
from django import template
from myapp.models import MyCustomSettings
register = template.Library()
# https://docs.djangoproject.com/en/1.9/howto/custom-template-tags/
#register.assignment_tag(takes_context=True)
def get_feedback_form(context):
request = context['request']
my_custom_settings = MyCustomSettings.for_site(request.site)
feedback_form_page = my_custom_settings.feedback_form_page.specific
form = feedback_form_page.get_form(
page=feedback_form_page, user=request.user)
return {'page': feedback_form_page, 'form': form}

This still works in wagtail 2.3 just need to replace
#register.assignment_tag(takes_context=True)
with
#register.simple_tag(takes_context=True) to conform with django 2.2
Also {% load feedback_form_tags wagtailcore_tags %} assumes your file inside of templates tags is named feedback_form_tags.py. I also added an __init__.py in the template tags folder although I'm not sure that was actually necessary.

Add a custom button to a Django application's admin page

I have an application in Django with a routine which would be available only to the admin. What I want to do is add a button to perform the routine in this application's section of the admin app.
Am I supposed to make a template for it, and if that's the case, how do I add a html template for an app in the admin. Or maybe there's a command to simply add a button?

Messing with the admin forms can be complicated but i've commonly found that adding links, buttons, or extra info is easy and helpful. (Like a list of links to related objects witout making an inline, esp for things that are more viewed than edited).
From Django docs
Because of the modular design of the admin templates, it is usually
neither necessary nor advisable to
replace an entire template. It is
almost always better to override only
the section of the template which you
need to change.
This will add a list over the top of the form.
Place in templates/admin/[your_app]/[template_to_override]:
{% extends "admin/change_form.html" %}
{% block form_top %}
{% for item in original.items %}
{{ item }}
{% endfor %}
{% endblock %}

Django1.10:
1) Override admin/submit_line.html:
{% load i18n admin_urls %}
<div class="submit-row">
{% if extra_buttons %}
{% for button in extra_buttons %}
{{ button }}
{% endfor %}
{% endif %}
{% if show_save %}<input type="submit" value="{% trans 'Save' %}" class="default" name="_save" />{% endif %}
{% if show_delete_link %}
{% url opts|admin_urlname:'delete' original.pk|admin_urlquote as delete_url %}
<p class="deletelink-box">{% trans "Delete" %}</p>
{% endif %}
{% if show_save_as_new %}<input type="submit" value="{% trans 'Save as new' %}" name="_saveasnew" />{% endif %}
{% if show_save_and_add_another %}<input type="submit" value="{% trans 'Save and add another' %}" name="_addanother" />{% endif %}
{% if show_save_and_continue %}<input type="submit" value="{% trans 'Save and continue editing' %}" name="_continue" />{% endif %}
</div>
This assumes, of course, that button's string representation is an appropriate browser input or button element, and is marked safe with django.utils.safestring.mark_safe. Alternatively, you could use the safe template filter or access the attributes of button directly to construct the <input>. In my opinion, it's better to isolate such things to the python level.
2) Override MyModelAdmin.change_view:
def change_view(self, request, object_id, form_url='', extra_context=None):
extra_context = extra_context or self.extra_context()
return super(PollAdmin, self).change_view(
request, object_id, form_url, extra_context=extra_context,
)
This method enables you to add buttons to any ModelAdmin easily. Alternatively to step (1), you could extend admin/change_form.html and override block submit_row. This would be slightly more verbose due to extra tags required in the template.
If you want the extra action available across all of your models (or a specific subset) then subclass ModelAdmin with the desired functionality (an example would be to add archiving to your models. You could even add an override for delete--and the other default buttons--so that the mode is archived instead of deleted; this would require some template modifications)

You can also use django-admin-tools, which allows you to easily customize the admin front page like a dashboard. Using a LinkList, you can point to some view method and check if the user is authenticated. It goes like thies:
# dashboard.py (read more about how to create one on django-admin-tools docs)
class CustomIndexDashboard(Dashboard):
"""
Custom index dashboard for captr.
"""
def init_with_context(self, context):
self.children.append(modules.LinkList(
_('Tasks'),
children=[
['Your task name', '/task']
]
))
# urls.py (mapping uri to your view function)
urlpatterns += patterns('yourapp.views',
(r'^task$', 'task'),
)
# views.py
def task(request):
if request.user.is_authenticated():
update_definitions_task.delay() # do your thing here. in my case I'm using django-celery for messaging
return redirect('/admin')

You might consider adding a custom admin action for this kind of object (similar to the built in 'delete'), if appropriate. Some benefits include: "pure Django", not having to mess with templates, and being able to act on multiple objects at once.
Django’s admin lets you write and register “actions” – simple
functions that get called with a list of objects selected on the
change list page. If you look at any change list in the admin, you’ll
see this feature in action; Django ships with a “delete selected
objects” action available to all models.
https://docs.djangoproject.com/en/dev/ref/contrib/admin/actions/
I got the idea from this article on how to add a custom action button, which is another answer all together. I was able to get by with the simpler built-in actions though.
https://medium.com/#hakibenita/how-to-add-custom-action-buttons-to-django-admin-8d266f5b0d41

Don't mess with the admin pages.
Create an "application" for this. Yes, your function is just a "routine". That's okay. Many smaller applications are a good thing.
This application has nothing new in models.py. No new model. Zero lines of code.
This application has a useful URL in urls.py. Something that can be used to display this admin page. One URL. Not many lines of code (less than a dozen.)
This application has one view function in views.py. On "GET", this view function presents the form. On "POST", this view function does the "routine". This is the "heart" of your application. The GET -- of course -- simply returns the template for display. The POST does the real work, and returns a final status or something.
This view function is protected with a decorator so that only an admin can execute it.
See http://docs.djangoproject.com/en/1.2/topics/auth/#django.contrib.auth.decorators.user_passes_test. You want to write a test for being an admin. lambda u: u.is_staff is probably it.
This application has one template, presented by the GET and POST. That template has your form with your button. The one you can't add to admin easily.
The tests.py is a test case with two users, one who is an admin and one who is not an admin.
No messing with built-in admin pages.

The "next" parameter, redirect, django.contrib.auth.login

I'm trying to redirect users to custom url "/gallery/(username)/" after successfully logging in. It currently redirects to the default "/account/profile/" url While I know what I can override the redirect url in my settings.py, my url is dynamic thus it will not work.
Documentation states that I need to use the "next" parameter and context processors. I have the {{next}} in my template, but I'm confused on how to actually pass the "/gallery/(username)". Any help would be greatly appreciated.
p.s: I'm trying to steer away from writing my own login view.

Django's login view django.contrib.auth.views.login accepts a dictionary named extra_context. The values in the dictionary are directly passed to the template. So you can use that to set the next parameter. Once that is done, you can set a hidden field with name next and value {{ next }} so that it gets rendered in the template.

I confess I usually use 2 redirects in order to get something like this to work.
First, Make your own registration/login.html page. You can copy-and-paste the html example in this section of the authentication docs to make the process a little easier. Instead of using the dynamic '{{ next }} variable from the context, however, hardwire the value of next to go to a generic landing view of logged-in users
<input type="submit" value="login" />
<input type="hidden" name="next" value="/gallery/" />
Then, in the view that you map to the /gallery/ URL, extract the User object from the request (since the user will now be logged in, especially if the gallery view is wrapped in a #permission_required or #login_required decorator. Use that view to redirect to the appropriate user-specific gallery page:
#login_required
def gallery(request):
url = '/gallery/%s/' % request.user.username
return HttpResponseRedirect(url)

If you already have the custom template for login form you need to add the following inside your <form> tag:
<input type="hidden" name="next" value="{{next}}" />
BTW, you don't have to create your own login view. django.contrib.auth.views.login works fine. You only need to create a template for it (registration/login.html)

being an newbie to django and stumbling over this somewhat older thread i found a differing solution for the problem of dynamically (=override a custom default only if needed) setting the next-param that i'd like to share (working fine with django 1.5, earlier versions untested):
just as django-d i wanted avoid repetition and a custom login-view, so i used the stock django.contrib.auth.views.login-view by adding the line of
url(r'^login/$', 'django.contrib.auth.views.login', {'template_name': 'myapp/login.html',}, name='login'),
to my urls.py and within the login.html-templates form-element:
{% if not next or not next.strip %}
{# avoid django.contrib.auth.views.login s default of /account/profile/ #}
{% url 'afterlogindefaultview' as next %}
{% endif %}
<input type="hidden" name="next" value="{{ next }}" />
which to my understanding follows the decoupling-practice of the url-configurations from the views.
so for views that should redirect to my apps login and afterwards head to a non-default view
i use
return HttpResponseRedirect('%s?next=%s' % (reverse('login'), reverse('mycustomnext')) )
from the view where i want to have the user to log in. i use this to get back to the view where i left off for logging the user in.

You can use a static redirect to /loggedin/ and then associate the url to a view that makes the correct redirect.
Login takes an extra step but if you want to use django's view it does the job.

create your own view for logging in, with it's own url, don't use the admin's one.
you can store the next page in the session, or pass it as a GET parameter to the login view
(i.e. /login?next=gallery) just don't forget to sanitize and validate that value before redirecting to it.