CODEWITHSUNDEEP
python-3.xurl-parametersurl-parsing
Enrike Fry
Enrike Fry

Reputation: 21

Parsing a URL param changes the value in Python

I´m trying to get a param from a URL using the urlparse function, but when i tried to decode the param the string changes and gives me an error. like this:

    url = request.url
    parsed = urlparse.urlparse(url)
    log.info(parsed)
    url_code = urlparse.parse_qs(parsed.query)['id_compra']
    log.info(url_code)

the output looks like this

INFO     http://c34f30683db9.ngrok.io/payment?id_compra=%2FYkv86B1UdMds9CuYr%2FbMCD3BpKL%2FzElNtVRc8KiALXWgtB343EjrUsmTObrBzv48FshzRu%2FD3Www4B+5SAxcb2xx78=
INFO     ParseResult(scheme='http', netloc='c34f30683db9.ngrok.io', path='/payment', params='', query='id_compra=%2FYkv86B1UdMds9CuYr%2FbMCD3BpKL%2FzElNtVRc8KiALXWgtB343EjrUsmTObrBzv48FshzRu%2FD3Www4B+5SAxcb2xx78=', fragment='')
INFO     ['/Ykv86B1UdMds9CuYr/bMCD3BpKL/zElNtVRc8KiALXWgtB343EjrUsmTObrBzv48FshzRu/D3Www4B 5SAxcb2xx78=']

I want to obtain the param like it is in the URL, but the result changes the character '+' to an empty space and I don't want that to happen.

Upvotes: 0

Views: 177

Answers (1)

Kyle Hurst
Kyle Hurst

Reputation: 252

you can probably quote the query to get back to the urlsafe format. urlparse some some simple unquoting of the query and you just need to reverese that process.

from urllib.parse import quote
urlsafequery = quote(urlquery)

Upvotes: 0

Related Questions