I've built a site where I can create new posts (essays) by the admin panel. The output is visible to users. But when I place some HTML as content in the form it doesn't render itself on the page.
example:
Output on the page (with marked unrendered HTML):
I would like to know how to fix it and also, how to name the topic I want to know ( I couldn't find anything related to my problem, probably because I don't know how to express it).
Additionally, I just start to wonder if there is one more problem nested inside. How to link CSS from the static folder having this HTML mentioned above?
Django offer the autoescape template in the builtins tags
{% autoescape off %}
{{ myhtml }}
{% endautoescape %}
But your logic seems wrong, you don't need to create a new page with the doctype, just create a base template and use the block content tag to insert your article.
In your base template replace the description and title of your page by variables that will be populated by the article data.
You need to learn the basic of Django https://docs.djangoproject.com/en/4.1/ trust me you won't regret it !
Related
In my base.html template I have a link to a website that I would like displayed at the top of each page that extends it:
base.html
Results
Except for results.html, when I load that page I would like the link loaded as:
results.html
Home
I'm under the impression that the template language could solve my problem with an if statement:
if currentTemplate/urlRoute != results.html:
button = Results
else:
button = Home
Please help point me in the right direction to implement this if possible :)
Thanks :)
You can get current URL in template by using request.path. The request variable automatically gets passed into each template context if you use Django's RequestContext as recommended (you probably do, since it's the default way).
Then you can just do {% if "reports" in request.path %}......{% endif %}.
That said, a cleaner approach would be to put the link in your base.html in a {% block %} template tag, like this:
{% block top_link %}Results{% endblock %}
Then this URL will be the same in all the pages, and you will be able to override it in your reports page by just specifying another content for the block.
I want to add a common sidebar for all my templates on my site. Let's say, I want a picture and some text, that the final user could modify whenever she feels like without having to mess with my base.html template (the base class of all my templates).
To be clear, I know that I can put plugin place holders with the template tag:
{% placeholder sidebar %}
My problem is that if I have five 5 templates and the content of the sidebar is the same for all of them, the user have to go through all the pages and change them one by one.
On the other hand, it cannot be static because I want the user to be able to modify the content through the admin.
Specifically, I am trying to do this with the cmsplugin-contact which saves me the troubles of configuring forms and emails.
You can create a special page that isn't published and add a "sidebar" placeholder to the template. You then use the {% show_placeholder %} template tag to render that sidebar placeholder in the base template that each of your 5 other pages are using
My problem
With Django CMS 2.3.3, when creating a Page I use cmsplugin_picture* next to a couple of other cmsplugins. In my cms template, instead of doing:
{% placholder "content" %} //calling the Django Page including all plugins...
I would like to call each cmsplugin seperately, but how would I do that?
I looked at Django tag template (filters) here and also studied Django CMS template tags here, but neither seem to suggest that possibility. I have to say I am a beginner so I might not have connected the dots...
What I try to achieve:
In my template I have a IMG tag (outside of the {% placeholder "content" %} tag) which I want to populate with an image url that I define in my Page/cmsplugin_picture. So I am looking for a placeholder tag that allows me to grab that image. In my wildest dreams I would name it:
{% show_placeholder "content" request.current_page.get_cmsplugin_picture %}
Obviously the above doesn't work, but does something like this exist?
**I have also tried cmsplugin_filer, but to me it isn't necessarely more beneficial to fix this particular problem.*
EDIT:
What I mean by Page/cmsplugin_picture -> In a Django CMS Page you can select between your installed cmsplugins to add to a Page. In my case I select cmsplugin_picture and upload an image (within that plugin). This image I want to 'call' in my Django Template. So it is a not a static url, but dynamic.
You should make a second placeholder where your img tag is (and optionally limit the types and amount of plugins using CMS_PLACEHOLDER_CONF (http://docs.django-cms.org/en/2.3.3/getting_started/configuration.html#cms-placeholder-conf).
I've started using Django and am going right to generic views. Great architecture! Well, the documents are great, but for the absolute beginner it is a bit like unix docs, where they make the most sense when you already know what you're doing. I've looked about and cannot find this specifically, which is, how do you set up an object_list template so that you can click on an entry in the rendered screen and get the object_detail?
The following is working. The reason I'm asking is to see if I am taking a reasonable route or is there some better, more Djangoish way to do this?
I've got a model which has a unicode defined so that I can identify my database entries in a human readable form. I want to click on a link in the object_list generated page to get to the object_detail page. I understand that a good way to do this is to create a system where the url for the detail looks like http://www.example.com/xxx/5/ which would call up the detail page for row 5 in the database. So, I just came up with the following, and my question is am I on the right track?
I made a template page for the list view that contains the following:
<ul>
{% for aninpatient in object_list %}
<li><a href='/inpatient-detail/{{ aninpatient.id }}/'>{{ aninpatient }}</a></li>
{% endfor %}
</ul>
Here, object_list comes from the list_detail.object_list generic view. The for loop steps through the object list object_list. In each line I create an anchor in html that references the desired href, "/inpatient-detail/nn/", where nn is the id field of each of the rows in the database table. The displayed link is the unicode string which is therefore a clickable link. I've set up templates and this works just fine.
So, am I going in the right direction? It looks like it will be straightforward to extend this to be able to put edit and delete links in the template as well.
Is there a generic view that takes advantage of the model to create the detail page? I used ModelForm helper from django.forms to make the form object, which was great for creating the input form (with automatic validation! wow that was cool!), so is there something like that for creating the detail view page?
Steve
If you're on django < 1.3 then what you are doing is basically perfect. Those generic views are quite good for quickly creating pages. If you're on django 1.3 you'll want to use the class based generic views. Once you get a handle on those they are are crazy good.
Only note I have is that you should use {% url %} tags in your templates instead of hardcoding urls. In your urls.conf file(s) define named urls like:
url('inpatient-detail/(?P<inpatient_id>\d+)/$', 'your_view', name='inpatient_detail')
and in your template (for django < 1.3):
...
In 1.3 a new url tag is available that improves life even more.
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!