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.
Related
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 !
I have an app with a Restaurant model. I'd like to understand what is the best way to put together a view that displays the list of restaurant objects, but also has a search form above that a user could enter parameters to filter the results displayed. If no parameters are entered, all the restaurants should be shown. I'm already using haystack and have a search form, but currently it is on a standalone search.html template. I also have an ListView on a separate template, and I guess I'm looking for an end result that combines these.
I did some reading on line and it's unclear what the best way to do it is:
using just listview from Django with some queryset filtering
combining haystack SearchView with django class based views?
this is my best bet so far - creating a customized version of SearchView from Haystack
ideally, ultimately the search capabilities would include things like allow autocompleting the user's inputs, and dynamically filter the results as the user types.
any thoughts on what the best way is to go about this and any examples out there?
There probably is out there some package that gives you everything automatically, displaying the queryset list and allowing simple adding of a search bar and so on (like the admin site). But I take it you are still a beginner to web development, so I'd strongly suggest you drop the listviews, drop haystack, drop everything, and do it all yourself.
See, they're not bad approaches or anything, but they're like shortcuts. And shortcuts are good only if they shorten the original way, which in your case isn't all that long.
To give you an example, here's a simple approach to displaying a list of items:
views.py
def restaraunt_list(request):
restaraunts = Restaraunt.objects.all().order_by('name')[:100] # Just for example purposes. You can order them how you'd like, and you probably want to add pagination instead of limiting the view number arbitrarily
context = {'restaraunts': restaraunts}
return render_to_response('index.html', context)
index.html:
<ul>
{% for rest in restaraunts %}
<li>{{ rest }}</li>
{% endfor %}
</ul>
That's it. Now it displays it in a list. Now to add a search filter all you need to do is this:
index.html (add this anywhere you want it)
<form>
<input type='text' name='q' id='q'></input>
<input type='submit' value='search!'></input>
</form>
When a user sends the data from 'q' it is sent through GET, and then on server side you add this:
def restaraunt_list(request):
restaraunts = Restaraunt.objects.all()
# filter results!
if request.method == 'GET':
q = request.GET['q']
if q != '':
restaraunts = restaraunts.filter(name__contains=q)
# ordering is the last thing you do
restaraunts = restaraunts.order_by('name')[:100]
context = {'restaraunts': restaraunts}
return render_to_response('index.html', context)
Now, this will work, but from what you wrote I understand you want the search to be live the moment a key is pressed. For that, you'd need to use AJAX (go learn how it works, I'm not gonna delve into it here). As for autocomplete, you should check out jQuery UI.
But, like I said, before jumping ahead and using all those stuff, I suggest you first learn the basics. Go through the django tutorial (if you haven't already), and use the amazingly detailed django-docs every step of the way. When some specific things won't work and you're stuck, come here to Stackoverflow and someone will surely help you. good luck!
I have a (what I think) is a pretty basic question about appending a get variable to url in django.
I have a filter sidebar on the page which creates a list of available classes based on the selection the user makes in the sidebar. For example, user may click on Aquatics class as a filter(other filters include things like location,price, time of day,etc) it will produce a list of aquatics classes available. Obviously, you can revise the search by choosing a specific location and now it should show a list of aquatics classes in that location.
what I would like is to make it so that every time use makes a choice, it adds a get variable to url like so: example.com/search?category=1&location=3. I'm not sure how I should go about it in the template. I know I need to get the list of variables by doing something like this
{% for k, v in request.GET.iteritems %}
{{ k }}={{ v }}&
{% endfor %}
But I dont really know what to do next... Could someone point me in the right direction?
You need to use this. request.GET.urlencode
That code automatically encode your GET parameter into URL friendly.
I could not find an answer for this.. So here my question. For a new project i'd like to use django-taggit.
Does someone have a suggestion on how to create a tag-cloud based on the current queryset?
The desired behavior is to 'start' with an unfiltered list - then allow narrowing town the results with applying filters and tags. At the beginning the tag-cloud shows e.g. the 50 most common tags. After choosing a tag (or other criteria) the tag-cloud should only display the remaining possibilities.
I know that django-tagging offers Tag.objects.usage_for_queryset() for this situation. But I would prefer to use '-taggit' over '-tagging'.
django-taggit-templatetags appears to be the 'go-to' place for a tagcloud for django-taggit.
It doesn't appear to handle querysets though. :(
So, I've added:
#register.inclusion_tag('taggit_templatetags/tagcloud_include_qs.html')
def include_tagcloud_qs(queryset):
try:
queryset = queryset.annotate(num_times=Count('taggeditem_items'))
except FieldError:
queryset = queryset.annotate(num_times=Count('taggit_taggeditem_items'))
num_times = queryset.values_list('num_times', flat=True)
weight_fun = get_weight_fun(T_MIN, T_MAX, min(num_times), max(num_times))
queryset = queryset.order_by('name')
for tag in queryset:
tag.weight = weight_fun(tag.num_times)
return {"tags": queryset}
to
templatetags/taggit_extras.py
And this to a new file at taggit_templatetags/tagcloud_include_qs.html
<div>
{% for tag in tags %}
<font size={{tag.weight|floatformat:0}}>{{tag}}</font>
{% endfor %}
</div>
I'm using it like this in my templates:
{% include_tagcloud_qs my_queryset %}
I haven't spent much time looking at the django-taggit-templatetags code, so feel free to update this with a better solution!
PS:
I'm getting a queryset in my view like this:
my_queryset = Tag.objects.filter(foo__bar=baz).distinct()
This answer shows how to build a tag cloud. You'd create a queryset in your view according to your parameters, generate a dictionary, and render it in your templates as shown in that answer.
I would suggest using django-tagging. It is well documented. I have created tag clouds with it. You can access tag clouds via model, model instance, etc via template tags that are easy to load. This is a little hackish but using the .counts method you can hack up some css to increase the size of each font as you would see in a real tag cloud. Django-tagging actually excels in this area as it has a default template tag with formatting options for everything you have described.
I've added a TagBase.get_for() function in https://github.com/twig/django-taggit/commit/42cd4e04f00496103f295c0afd8297074be50dcf
This basically fetches the Tags used for a given queryset, and from there you can do what you need to do.
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!