CODEWITHSUNDEEP
regexpython-3.x
coder3521
coder3521

Reputation: 2646

Regular expression to get after character a specific pattern

Input String :

-line[8qWWQ5-swd-WER-DWDS]]<-SUCCESS#[14][MY_SAMPLE_TEST]
-line[8qWWQ5-swd-WER-DWDS]]<-SUCCESS#[4][MY_SAMPLE_TEST2]
-line[8qWWQ5-swd-WER-DWDS]]<Failed#[17][[14]SERVERERROR(TYPE-241)

Expected output :

MY_SAMPLE_TEST
MY_SAMPLE_TEST2
SERVERERROR

My regular expression : (?<=#).*

The above regular expression I can get everything after # , also I tried :

rex = (?<=#\[...\[).*(?=])

which gives me correct output for the first line i.e : MY_SAMPLE_TEST but as 2nd line has only one digit i.e 4 so it doesn't matches , similar problem with 3rd line

It is possible to write a single expression which could give the expected out put ? , Any help would be great

Upvotes: 1

Views: 113

Answers (2)

Siddharth Singh
Siddharth Singh

Reputation: 1

Made an assumption that the matched string should be ending just before it finds a closing ] or starting ( characters. Here is the working regex : 

#(?:\[+\d+\]+)*\[?([^\(\]]+)(?:\(.+\))?\]?

It worked on the samples provided without assuming spaces or underscores in the text to be extracted. Here is a demo link : https://regexr.com/47muk

Upvotes: 0

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626738

You may capture these values using

#(?:\[+\d+]+)*\[*([^][()]+)

See the regex demo

Details

  • # - a hash sign
  • (?:\[+\d+]+)* - 0 or more repetitions of:
    • \[+ - 1+ [ chars
    • \d+ - 1+ digits
    • ]+ - 1+ ] chars
  • \[* - 0+ [ chars
  • ([^][()]+) - Group 1: one or more chars other than (, ), [ and ]

Python demo:

import re
strs = ['-line[8qWWQ5-swd-WER-DWDS]]<-SUCCESS#[14][MY_SAMPLE_TEST]', '-line[8qWWQ5-swd-WER-DWDS]]<-SUCCESS#[4][MY_SAMPLE_TEST2]', '-line[8qWWQ5-swd-WER-DWDS]]<Failed#[17][[14]SERVERERROR(TYPE-241)']
rx = re.compile(r'#(?:\[+\d+]+)*\[*([^][()]+)')
for s in strs:
    m = rx.search(s)
    if m:
        print(m.group(1))

Output:

MY_SAMPLE_TEST
MY_SAMPLE_TEST2
SERVERERROR

Upvotes: 1

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