CODEWITHSUNDEEP
pythonregex
dinesh2707
dinesh2707

Reputation: 33

Return next line of matched search string

I have the following variable in a string

some_var = ".... \n
... \n
Hello Subject \n
12:34:56:78:90 \n
... \n"

I' m trying to get just the value 123456789. I tried the following code but gives me next two lines from the line matched.

re.search(r'Subject((.*\n){2})', some_var).group()

Output of above code:

Hello Subject
12:34:56:78:90

Expected output:

12:34:56:78:90

Upvotes: 3

Views: 116

Answers (4)

SilleBille
SilleBille

Reputation: 639

A small modification to Tim's answer:

some_var = ".... \n... \nHello Subject \n12:34:56:78:90 (0x44) \n... \n"
print re.search(r'Subject.*\n(\S+)', some_var).group(1)

Explanation: \S+ = gets the first string and avoids (0x44)

Demo

Upvotes: 1

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626903

You may do without a regex at all if you do not need to match Subject as a whole word, and if you do not care about the type of symbols you match on the line below the line having Subject substring.

Use

some_var = ".... \n    ... \n    Hello Subject \n    12:34:56:78:90 \n    ... \n"
lst = some_var.split("\n")              # Split at newline
cnt = len(lst)                          # Get the item count
for idx, line in enumerate(lst):        # Enumerate lst to access index + item
    if "Subject" in line and idx < cnt - 1: # If not at end and line has "Subject"
        print(lst[idx+1].strip())       # Strip from whitespace and print next line

See the Python demo.

Upvotes: 0

Priya
Priya

Reputation: 1554

This might work.

import re;
some_var = ".... \n... \nHello Subject \n12:34:56:78:90 \n... \n";
# you might want to try \r too if its required with \n
s = re.search('Subject[\ ]*\n([\d:]+)', some_var);
if s:
    print(s.group(1));

Upvotes: 0

Tim Biegeleisen
Tim Biegeleisen

Reputation: 521419

I don't know what prompted you to choose the pattern you are using, but it looks wrong for extracting that number. Instead, use this pattern:

Subject.*\n(.*?)\n

And then access the matched number using group(1) which is the first (and only) matched capture group.

some_var = ".... \n... \nHello Subject \n12:34:56:78:90 \n... \n"
print re.search(r'Subject.*\n(.*?)\n', some_var).group(1)

Demo

Upvotes: 3

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