I have been trying to getting to drips with django, and I am experimenting with the tutorials. So, what I have at the moment is a View class which spits out a queryset - which I access in the template as follows:
<a id="{{i.0}}" .....##whatever goes here
which works great. What I would like to do is to prepend the character x before the value of {{i.o}} - but I am unsure how to do this. So, I am able to append using:
<a id="{{i.0|add:"x"}}" .....##whatever goes here
but, how do I prepend?
Any guidance would be great on this.
<a id="x{{ i.0 }}" .....##whatever goes here
Related
Trying to use {{ post|truncatewords:"100" }} for a teaser in Cactus. The post is using the markdown filter and truncate words is showing the html. I also tried {{ post|truncatewords:"100"|markdown }} which does seem like it is attempting to format correctly, however line breaks and body tags still show.
Example Output:
{'body': u'\n\n
Cactus templates are set up using the Django Template Language. Django\u2019s template language is both powerful and easy to use. A template is simply a HTML file. Let\'s take a look at how to use the blog template.
\n\n
General file str...
Seems like this should be pretty straight forward but I cannot figure out what I am doing wrong.
Any help would be greatly appreciated.
Thanks.
Looks like post is a dictionary, and the post body you want to show is under the body key. You probably want something like:
{{ post.body|markdown|truncatewords_html:100 }}
I'm really new to Django. I'm having difficulty displaying images based on their name and according to the url pattern.
So basically the url consists of several variables within them and I want to be able to use that to fetch a particular image that is named with those variables.
Example:
localhost:8080/farm/chicken
this would fetch an image inside of my ../static/images/ folder and get:
farm_chicken.jpg
Another example:
localhost:8080/zoo/alligator
would get:
zoo_alligator.jpg
I can fetch the url parameters. So, should I make these image names within my views.py file and pass the names (zoo_alligator) into a context to be retrieved by the template later on? Would this be the correct way?
Thank you for your advice everybody! I appreciate all the help!
You won't actually do something like that, in general. What you should do is to send the image as a context variable from your view function to your template.
If you are using your url node to determine which picture to show, then in your corresponding view function, you are actually using "alligator" or "chicken" to load up the corresponding Animal class.
Once the correct animal object is instantiated, you could send this animal object to your django template and load in the image using a snippet similar to this:-
Like this:-
{% if animal.get_latest_medium_url %}
<img id="animal_image" class="img-rounded left" src="/media/{{ animal_image }}" />
{% endif %}
The get_latest_medium_url is a method in my Merchant class and it computes the url there.
So, should I make these image names within my views.py file and pass
the names (zoo_alligator) into a context to be retrieved by the
template later on? Would this be the correct way?
Sure, this is one way to do it. Something like this:
(r'show/(?P<in_path>.*)$','someapp.image_view')
Then in image_view:
def image_view(request,in_path):
img = in_path.replace('/','_')+'.jpg'
return render(request,'some_template.html',{'path':img})
However, as your view is very simple - you can pass the path directly to the template from urls.py, using direct_to_template:
from django.views.generic.simple import direct_to_template
(r'show/(?P<in_path>.*)$',direct_to_template,{'template':'some_template.html'})
In some_template.html:
<img src="{{ params.in_path }}">
The problem is that you won't get your string formatting done as the default filters do not have a "replace" function. You can easily write a custom filter:
#register.filter
#stringfilter
def format_path(the_path):
return the_path.replace('/','_')+'.jpg'
Then modify the template:
<img src="{{ params.in_path|format_path }}">
You should read the documentation on writing custom filters and tags for more details including where to store the filter code to make sure django can find it.
I am having a great deal of difficulty getting my head round displaying secveral resources on one page with Ditto. I cant seem to get TV's to show along with my content.
Heres how I have set it out:
I have a page with my Ditto call:
[!Ditto? &parents='134' &orderBy='createdon ASC' &tpl='temp'!]
I have a simple chunk called temp set up as such:
<div id="content">
[*articlename*]
[+content+]
</div>
And I have a template with the TV articlename assigned to all the resource under parent 134.
The content shows fine but none of the TV's do. Can anyone point me in the right direction? thanks!
I think the problem is in your syntax. You need to use a placeholder tag in the chunk for your TV:
Try this:
<div id="content"> [+articlename+] [+content+] </div>
I have found the answer: You are meant to use [+articlename+] for 'chunk TVs' rather then [*articlename*]. This is different to getResources.
I have a django template page, and want a link from this page, containing current URL, for example, I am on /article/11 and want link to /article/11/remove
I tried the following construction:
Remove article
But I get link to /article/remove instead of /article/11/remove
However when I change it to
<a href="{{ request.path }}">
I get link to /article/11
How can I get URL not trimmed?
I don't see why it doesn't point you to /article/11remove, which is what it sounds like it should do, but either way, you're missing a slash. Try <a href="{{ request.path }}/remove"> instead.
However, that's really not the right way to do it. You shouldbe using {% url 'name_of_remove_view' %} to get the url, not assuming it's going to be wherever you are plus /remove.
Edit: In that case, your problem is probably that {{ request.path }} is not outputting anything at all. That would explain why just having "remove" would take you to /article/remove, and having "" would take you to where you currently are, due to the way that relative URLs work. You might want to make sure that you have a request object at all in your template environment.
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.