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'WSGIRequest' object has no attribute 'htmx'

Hi just looking for some help at solving this error in Django whilst trying to call a view that to accept a htmx request. The final result is to display a popup Modal of images from a Gallery when a thumbnail is clicked.
HTMX installed via script in head.
View
if request.htmx:
slug = request.GET.get('slug')
context = {'pictures': Media.objects.filter(slug=slug)}
return render(request, 'main/gallery-detail.html', context=context)
context = {'objects_list': Albums.objects.all()}
return render(request, 'main/gallery.html', context=context)
Relevant html with the button to open gallery of images.
<a class="btn btn-primary" hx-post="{{ request.path }}?slug={{ img.slug }}" hx-target="#modal">
{{ img.slug }}</a>
{% endfor %}
<div id="modal">{% include "main/gallery-detail.html" %}</div>

This error mostly occurs if you haven't included django-htmx in the settings.py.
Try making the below changes and see if it works :
Add "django_htmx.middleware.HtmxMiddleware" to the MIDDLEWARE.
Add "django_htmx" to the INSTALLED_APPS.

Additional action button doesn't work on flask-admin

I'm trying to add one more action to flask-admin forms.
It has to increment rating (+1) and it works with batch action, but not with single. Please help me find the bug, I've spent a lot of time trying to make this thing work properly.
Here's the code:
I made an html template in templates folder - custom_lists.html
{% extends 'admin/model/list.html' %}
{% block list_row_actions %}
{{ super() }}
<form class="icon" method="POST" action="/admin/user/action/">
<input id="action" name="action" value="approve" type="hidden">
<input name="rowid" value="{{ get_pk_value(row) }}" type="hidden">
<button onclick="return confirm('Are you sure you want to approve selected items?');" title="Approve">
<span class="fa fa-ok glyphicon glyphicon-ok"></span>
</button>
</form>
{% endblock %}
this succeeded with an icon on the list, but if i click to it - it says
Not Found
The requested URL was not found on the server. If you entered the URL
manually please check your spelling and try again.
added to templates folder and added to AdidasView class this:
list_template = 'custom_list.html'
#action('approve', 'Approve', 'Are you sure you want to approve selected items?')
def action_approve(self, ids):
try:
query = Adidas.query.filter(Adidas.id.in_(ids))
count = 0
for image in query.all():
image.rating += 1
count += 1
db.session.commit()
flash(ngettext('Item was successfully approved.',
'%s items were successfully approved.'%count,count))
except Exception as ex:
if not self.handle_view_exception(ex):
raise
flash(gettext('Failed to approve items. %(error)s', error=str(ex)), 'error')

I have not changed the template but I have done it differently as following by setting the column_extra_row_actions variable and defining the action_play function
column_extra_row_actions = [
EndpointLinkRowAction('glyphicon glyphicon-play', 'event.action_play')
]
#expose('/action/play', methods=('GET',))
def action_play(self, *args, **kwargs):
return self.handle_action()

This solution does not seem to apply to this example, but I also struggled with a case where I received a 404 when I using an action on one item via the button, while the batch action worked as expected.
After taking a look at the JS for the batch action I realized that the two HTML forms for individual actions and batch actions are practically identical. The only difference is that when using batch actions there may be more input fields in the form. That implies that if you get a 404 on one, but not the other, there must be an error in your HTML.
In my case I was not aware that Flask-Admin addresses models_with_underscores_in_their_name as modelswithunderscoresintheirname. Therefore instead of
<form class="icon" method="POST" action="/admin/mymodel/action/">
my erroneous code was
<form class="icon" method="POST" action="/admin/my_model/action/">
Note the difference in the action field.
With this change I was able to use the #action API as explained in the Flask-Admin docs.

what is the best way to use less Django template syntax as possible with Django Form?

I am just beginner of Django.
At first, It seemed very convenient to use Django Form. I can get all which i already I defined in Model with ModelForm class. I can just put { form } in template. There are already pre-defined Field type and I can use it conveniently.
But it is tricky if you need to work with one who doesn't know Django or Python and are charge of making front-end part.
For example, I make template like this.
example.html
<form action="." method="post"> {% csrf_token %}
{{ form }}
<input type="submit" value="ok">
views.py
class TestView(FormView):
template_name = 'example.html'
form_class = TestForm
success_url = '/success/'
def form_valid(self, form):
print('data is validated! you can do work with cleaned data!')
return super(TestView, self).form_valid(form)
Q1 : A front-developer can't work with {{ form }}. right?
So I can change just example.html because it works properly if you set name attribute and type attribute correctly.
example.html
<form action="." method="post"> {% csrf_token %}
<input type="text" name="name">
<input type="text" name="calorie">
<input type="submit" value="Ok">
</form>
models.py
class Cereal(models.Model):
name = models.CharField(max_length=50)
calorie = models.CharField(max_length=50)
Q2 : I think it would be fine for a front developer to work with it if I change example.html above? right?
Well, if my questions are all fine, It seems fine to use ModelForm and write html code instead of {{ form }} django template syntax. But if you use ModelChoiceField, it seems nasty. You set queryset argument and it shows with select widget on template. for example queryset reqsult is { A, B, C }. it will turn to like below
<select id="id_something" name="name of something">
<option>A</option>
<option>B</option>
<option>C</option>
</select>
As i told you earlier, I don't want to use Django template syntax like {{ form }} as much as I can. so I am going to write like below instead.
views.py
render(request, "example.html", {queryset : something.objects.all()})
example.html
<select id="id_something" name="name of something">
{% for element in queryset %}
<option>{{ element }}</option>
{% endfor %}
</select>
Q3: does this approach seem alright? Isn't it better way to use less Django template syntax when you use Django Form?
well, I concluded that using DRF(Django RestFramework) is the best way to separate tasks for front-end and back-end developers independently. But In current situation I am facing, Template(html, css, js) is already made. I need to change it to use with Django framework. Front end developer will change little bit with the Template which is already made.
When I watched some pycon video in Youtube, some people said that Djang framework is kind of too much coupled.
I hope I can get Idea more efficiently to use Django Model form with front-end developer.
Thanks for reading. My question could be ridiculous for an expert. Please ask me again if it is not clear.

Sending POST data from inside a Django template 'for loop'

With this HTML:
...
{% for thing in things %}
<form method="post">
{% csrf_token %}
{{ thing.name }}
{{ form.value }}
<input type="submit" value="Submit" />
</form>
{% endfor %}
...
My website lists multiple 'things' from my database, so there can be many forms generated on the one page. How can I somehow determine in my views.py, which 'thing's' form is being submitted?
More elaboration:
Imagine you have a page of objects listed one after the other, and each object has a like button associated with it, that adds a like to the object it is next to. That's essentially what I'm trying to do here.
The problem is, I have a form that can process the like, but how do I take that like and add it to the object that it's displayed next to on the page? (by the aforementioned 'for loop')
I'm completely confused on how to go about this, am I looking at the problem the wrong way, or is there a standard idiom around this problem that I don't know about?
Thank you :)

The most common design pattern for model instance updates is to provide the primary key of an object in the url where you are submitting your post data.
# urls.py
from django.conf.urls import *
from library.views import UpdateThing
urlpatterns = patterns('',
url('^update_thing/(?P<pk>[\w-]+)$', UpdateThing.as_view(), name='update_thing'),
# views.py
def my_view(request, pk=None):
if pk:
object = thing.objects.get(pk=pk)
form = MyModelForm(data=request.POST or None, instance=object)
if form.is_valid():
...
Now, let's specify (using Django's url template tag) that we want to submit post data for each object to the correct url.
{% for thing in things %}
<form method="post" action="{% url 'update_thing' thing.pk %}">
{% csrf_token %}
{{ thing.name }}
{{ form.value }}
<input type="submit" value="Submit" />
</form>
{% endfor %}
The url tag does a reverse lookup through your urls for the name kwarg supplied for a given url, and accepting positional arguments (such as, in this case, thing.pk) and, when needed, keyword arguments.

The standard way to handle multiple forms of the same kind on one page with Django is to use Formsets.
It handles the annoying details like displaying errors on one form while preserving the input on others etc.
However, in your specific case that might be overkill. If you just want to create a like for an object, there isn't really any user input that needs to be validated, so you don't really need a form. Just perform a POST to a specified URL, maybe with Javascript. If the user messes with the URL, you display a 404.

form action not working in django

I have django 1.4 and I am following a tutorial which uses an older version of django. Its a simple tutorial which creates a wiki app with Page as model.
The problem is that the view function corresponding to a POST method in a form is not getting invoked.
This is the content in the urls.py:
url(r'^wikicamp/(?P<page_name>[^/]+)/edit/$', 'wiki.views.edit_page'),
url(r'^wikicamp/(?P<page_name>[^/]+)/save/$', 'wiki.views.save_page'),
url(r'^wikicamp/(?P<page_name>[^/]+)/$', 'wiki.views.view_page'),
This is the content of the template edit.html:
<from method = "get" action="/wikicamp/{{page_name}}/save/">
{% csrf_token %}
<textarea name = "content" rows="20" cols="60">
{{content}}
</textarea>
<br/>
<input type="submit" value="Save Page"/>
</form>
this is link to save
And this is the content in views.py:
def edit_page(request, page_name):
try:
page = Page.objects.get(pk=page_name)
content = page.content
except Page.DoesNotExist:
content = ""
return render_to_response("edit.html", {"page_name":page_name, "content":content}, context_instance=RequestContext(request))
def save_page(request, page_name):
return HttpResponse("You're looking at the page %s." % page_name)
I initially I was getting csrf related error and I then tried all the fixes provided in https://docs.djangoproject.com/en/dev/ref/contrib/casrf/ and followed many many stackoverflow question related to POST and django. Now nothing happens when I click the 'Save Page' button, nothing! Not even any request being sent from the form (Using firebug to track the HTTP request and response)

You have a typo in your HTML: from instead of form.
You may realize this, but that code won't really save anything. I'm not sure what blog you are following, but you would be better-off following the official Django tutorial in the documentation, then reading the forms docs.

You may need to change method to "POST" in your form.
<from method = "get" action="/wikicamp/{{page_name}}/save/">
to
<form method = "post" action="/wikicamp/{{page_name}}/save/">

There are some spelling mistakes, such as from instead of form.
Also the form is malformed.
Change:
this is link to save
to
<input type="submit" value="Save Page" />
And thirdly, change the method= "get"to method="POST".
The entire form should look like this
<form method = "POST" action="/wikicamp/{{page_name}}/save/">
{% csrf_token %}
<textarea name = "content" rows="20" cols="60">
{{content}}
</textarea>
<br/>
<input type="submit" value="Save Page"/>
</form>
Also what #DanielRoseman said. But hey, it might come further down the road.