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First time working with Python Regex and I just need a little tip on matching strings.

I have a url like this: url = "https://www.youtube.com/api/timedtext?xorp=True&xoaf=1&v=UloIw7dhnlQ&signature=C2AF3C2887A37043353A86AAAACFA796659B56CB.E736B7146447843F2D3311234744DC0D9937AF7B&asr_langs=fr%2Cru%2Ces%2Cnl%2Cit%2Cde%2Cko%2Cen%2Cpt%2Cja&sparams=asr_langs%2Ccaps%2Cv%2Cxoaf%2Cxorp%2Cexpire&expire=1541769991&key=yttt1hl=&encaps=asrlang=enfmt=srv3"

I am trying to match everything except the part that begins with expire=1541769991 (2nd to last line). This is what I have come up with:

matchObj = re.match( r'(.*)expire=(.*)&(.*?)', url)

The problem is the third group includes the text after the last occurrence of &. I want the text following the first occurrence of & after expire=. I tried adding a ? after & to make it non-greedy too. How would I go about doing this?

Try this regex,

matchObj = re.match( r"(.*)expire=[^&]*(&.*)", url)

You could do something like this:

import re

url = "https://www.youtube.com/api/timedtext?xorp=True&xoaf=1&v=UloIw7dhnlQ&signature=C2AF3C2887A37043353A86AAAACFA796659B56CB.E736B7146447843F2D3311234744DC0D9937AF7B&asr_langs=fr%2Cru%2Ces%2Cnl%2Cit%2Cde%2Cko%2Cen%2Cpt%2Cja&sparams=asr_langs%2Ccaps%2Cv%2Cxoaf%2Cxorp%2Cexpire&expire=1541769991&key=yttt1hl=&encaps=asrlang=enfmt=srv3"

match = re.match("(.+?)(expire=.+?&)(.+$)", url)
print(match.group(1) + match.group(3))

Output

https://www.youtube.com/api/timedtext?xorp=True&xoaf=1&v=UloIw7dhnlQ&signature=C2AF3C2887A37043353A86AAAACFA796659B56CB.E736B7146447843F2D3311234744DC0D9937AF7B&asr_langs=fr%2Cru%2Ces%2Cnl%2Cit%2Cde%2Cko%2Cen%2Cpt%2Cja&sparams=asr_langs%2Ccaps%2Cv%2Cxoaf%2Cxorp%2Cexpire&key=yttt1hl=&encaps=asrlang=enfmt=srv3

Or if you simply want the text without the expire=, you can remove it:

result = re.sub("expire=d+?&", "", url)

Note that assumes that the value of expire are all digits.