Flask -- pass arguments to outgoing url.

Question

Some of my endpoints of my flask api use values from an html form to determine what data to return. I'd like to pass the parameters that I use to generate the data I'm returning back to the URL rendered in the browser. Is there a way to do that with flask?

Example:

http://example.com/search

On the page, the user types the word "duck" into a search box, hits go, and in addition to the results, Flask returns url showing all the parameters I passed into the api underneath the application... e.g:

http://example.com/search?&search_term=duck&page=0&per_page=20

atupal · Accepted Answer · 2014-02-16 08:29:42Z

3

You can use url_for to generate the url:

from flask import redirect, url_for, request ... # duck = request.form.get('search_term') return redirect(url_for('search', search_term=duck, page=0, per_page=20)) ...

answered Feb 16, 2014 at 8:29

atupal

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2 Comments

kelorek Over a year ago

Currently I'm ending the method with render_template() which has a form, and that form POSTs to a url_for(). I don't seem to be able to pass parameters into the url_for() in the jinja template. Is that expected?

atupal Over a year ago

@kelorek You can use url_for in jinja too, see stackoverflow.com/questions/9023488/… and stackoverflow.com/questions/11124940/…

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