Python regex get substring after preceding substring match

Question

Text has no spaces, so I cannot split at all and use indexing on a list of strings.

The pattern I am looking for is:

check=

It is followed by a number and encoded querystring items (apache logfile) and is on every line of the file twice. I want output that gives me just what follows check=

For instance, the string in a line looks like:

11.249.222.103 - - [15/Aug/2016:13:17:56 -0600] "GET /next2/120005079807?check=37593467%2CCOB&check=37593378%2CUAP&box=match&submit=Next HTTP/1.1" 500 1633 "https://mvt.squaretwofinancial.com/newmed/?button=All&submit=Submit" "Mozilla/5.0 (Windows NT 6.1; WOW64; rv:45.0) Gecko/20100101 Firefox/45.0"

And I need to fetch 37593467 and 37593378 in this case.

re.findall(r'(?<=check=)\d+', text)?

Aran-Fey
– Aran-Fey

2016-08-20 16:18:54 +00:00
Commented Aug 20, 2016 at 16:18 — Aran-Fey
– Aran-Fey, Commented Aug 20, 2016 at 16:18

Dinesh Pundkar · Accepted Answer · 2016-08-20 16:24:53Z

2

Please check this code.

import re

text = '''11.249.222.103 - - [15/Aug/2016:13:17:56 -0600] "GET /next2/120005079807?check=37593467%2CCOB&check=37593378%2CUAP&box=match&submit=Next HTTP/1.1" 500 1633 "https://mvt.squaretwofinancial.com/newmed/?button=All&submit=Submit" "Mozilla/5.0 (Windows NT 6.1; WOW64; rv:45.0) Gecko/20100101 Firefox/45.0"'''


for match in re.findall("check=(\d+)",text):
    print 'Found "%s"' % match

Output:

C:\Users\dinesh_pundkar\Desktop>python demo.py
Found "37593467"
Found "37593378"

Couple of URLs for help :

edited Aug 20, 2016 at 16:24

answered Aug 20, 2016 at 16:19

Dinesh Pundkar

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1 Comment

tadamhicks Over a year ago

ok, to modify, how do I also search on the text after...i.e. how do I get the match to display the number based on whether it found %2CCOB or %2CUAP?

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