Saturday, August 17, 2013

Capturing a search query + params in Django 1.5

Hi,

I have a simple web application in which I have a search box. When a user enters a query in the box and hits "search", they get a list of results via a popular search API.

My issue is that I am using django-endless-pagination to paginate (no kidding!) the API response - after some formatting, of course. When I click next, the pagination appends url parameters. I am wondering which url pattern to use to capture said parameters. The url scheme is as follows:

http://127.0.0.1:8000/search/ -> works fine, goes to my search view.
http://127.0.0.1:8000/search/?page=2 -> after a user clicks on the pagination links. I have no url pattern to handle this.

I am no regex expert. I also do not have an XKCD tee :-/

SOS?

Thanks.

--
Regards,
Sithu Lloyd Dube

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