I'm building a Blog in Django (using Generic Views) and I use the same template for both my date based and list detail views. I'm trying to setup pagination, but I want to do so with URL patterns rather than using an ugly ?page=1 url suffix.
The problem is in the actual html template, I cannot find a way to determine which view was used to render the page, so while I have access to all the pagination stuff, I have no way to generate the appropriate URL.
In other words, if the view was rendered by my archive_month(request, month, year, page=0) view, I would need to structure the URL for the next and previous pages as /blog/dec/2009/PageX/, versus the blog index, which would mean the URL would be /blog/pageX/.
Well I just realized that date_based generic views don't support pagination, so problem solved.
Related
I'm creating a web page on Django,
I want to create next and previous post in the bottom of blog page.
is there any way to use a simple tag like {{post.title}} in HTML and refer it to view page to find the index of current post ?
This is where you should use pagination. If you are using DRF you might need this pagination.
I have a django view that lists several urls on external sites.
When I render them I would like to add a few url parameters to each.
These urls are to an external system and thus not listed in my urls.py. Furthermore, some of the links have a hash '#' so it is not as easy as appending a few parameters to the end of the string.
Based on these requirements it seems the url template tag will not be a good fit. I was wondering if there is a custom filter out that to do this.
You don't need Django's url tag here. The url tag is to resolve to URLs that belong to your application.
However, there is the nice django-spurl library. It allows you to handle query parameters via template tags.
An example from the documentation to add query parameters:
{% spurl base="http://example.com/?foo=bar" add_query="bar=baz" %}
<!--
will result in: http://example.com?foo=bar&bar=baz
-->
I did pagination in my django projet. Everything works just perfect, but my urls looks terrible, like
host:8000/?page=1
How to create nice urls like
host:8000/page/2/ or host:8000/2/
I use standard Paginator class via ListView
How to do this w/o third party code ?
If you define url pattern like this:
url(r'^/page/(?P<page>\d+)/$', 'myapp.views.list_view'),
then ListView will pass page url keyword into paginator.
Notice:
Each path segment is supposed to be a valid resource, so it's not clear what you will display on /path/ URL.
Django pagination system assumes that webpages will default to using the URL query, so it's recommended to keep it as a URL query and it's more revealing.
I have a form with 6-7 fields. After user input, my webapp searches for those fields in a database and displays the results.
Now the issue is, that the URL ends up having all the form field names and their values in it.
result/?name=lorem&class=arc&course=ipsum
Now with the form having 7-8 fields the url ends up looking ugly.
Is there a Django technique to 'hide' these from the URL? Quotes around hide because I'd be okay with a completely different way to pass the objects to my database from the form as well.
Use a POST request. Here's the django docs on forms and a specific example using POST>. HTML-wise, all you need to do is change the method on the form tag.
I do not recommend to use POST requests for search. If you'll use GET it will be easer for user, he can just bookmark a link and save search or share search results with friends.
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.