Passing # character in url with python

Question

I have the following code:

@app.route('/famcall/Audio/AudioMng_PlayDTMF/<string:output_stream>/<string:dtmf>/<string:duration>/<string:gap>', methods=['GET'])
    def fam_audio_play_dtmf(output_stream, dtmf, duration, gap):
        params = [self.mapAudioOutputStream.business_to_tech(output_stream), dtmf, duration, gap]
        return self.call_fam_function_and_wait_response("AudioMng_PlayDTMF", params)

and I want to use the character # in my url because it´s a value that can be used for dtmf. Is there an encode function in python which can help me in this? Or should I forget this scenario?

mrCarnivore · Accepted Answer · 2017-12-04 16:52:37Z

1

You can use a replacement: replace # with %23.

answered Dec 4, 2017 at 16:52

mrCarnivore

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Peter DeGlopper · Accepted Answer · 2017-12-04 16:59:09Z

1

Use urllib.quote (Python 2) or urllib.parse.quote (Python 3) to escape your dtmf value or anything else that might contain reserved characters.

answered Dec 4, 2017 at 16:59

Peter DeGlopper

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2 Comments

H.A. Over a year ago

How can I use this with my code? I´ve used it like this and it doesn´t work: params = [self.mapAudioOutputStream.business_to_tech(output_stream), dtmf, duration, gap] data = quote(params)

Peter DeGlopper Over a year ago

@H.Avram - it takes one string, not a list. urllib.quote(dtmf). The point is that it covers all reserved characters, not just #. EG urllib.quote('foo#bar:baz') returns 'foo%23bar%3Abaz'.

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