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Home/ Questions/Q 6369659
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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T00:50:00+00:00 2026-05-25T00:50:00+00:00

Inside this code I have an ‘args’ value which is ‘title.title’, this value displays

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Inside this code I have an ‘args’ value which is ‘title.title’, this value displays in my URL.

How do I make the url read like this: “index.html/something_something” instead how how it displays now, which is: “index.html/something%20something”.

 def livesearch():
    '''Auto completes the search query'''
    partialstr = request.vars.partialstr
    query = db.listing.title.like('%'+partialstr+'%')
    titles = db(query).select(db.listing.ALL)
    items = []

    for title in titles:
        items.append(DIV(A(title.title, _id="resultLiveSearch", _href=URL('search', args='%s' % title.title))))

    return TAG[''](*items)

I tried making it:

_href=URL('search', args='%s' % title.title.replace("%20","_")

Which I thought would work, but it didn’t, what am I doing wrong? How can I fix this?

Thank you.

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  1. Editorial Team
    Editorial Team
    2026-05-25T00:50:01+00:00Added an answer on May 25, 2026 at 12:50 am

    The %20 is a replacement for a single white space, so you should replace spaces with underscores before passing them to URL:

    title.title.replace(' ', '_')
    
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