How to extract text between certain patterns using regular expression (RegEx)?

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My text:

27/07/18, 12:02 PM - user_a: https://www.youtube.com/
 Watch this
27/07/18, 12:15 PM - user_b: <Media omitted>
27/07/18, 12:52 PM - user_b: Read this fully
some text
some text
.
some text
27/07/18, 12:56 PM - user_c: text ..

Here I want to extract the messages sent by the users. I tried two regex. But I didn't get the answer I wanted

First regex:

re.findall(r''+user_name+ ':(.*)', data)

Here I couldn't able to extract the text multi lines

Second regex:

re.findall(r''+ user_name + ':[^(:)]*', data)

Here I couldn't able to extract the full text having a hyper link .i.e., I could able to get only "https". It considers the symbol ":" as an endpoint.

How do I handle this ? Any kind of suggestions would be really great & helpful

text-extraction regex-greedy python-3.x regex
2 Answers

I believe your regex should be: user_b: (.*?)^[0-9]. After your user is found, match the rest of the line until a number as the first character of a line is found (the next entry). Make sure to turn on multi-line.

See a demo here.

user_25434160

You may use the following pattern:

user_b: (.*?)(?=^[0-9]{2}/[0-9]{2}/[0-9]{2})

Regex demo here.

Note the usage of re.MULTILINE and re.DOTALL. The first flag is needed to match beginning of line patterns over multiline text, whereas re.DOTALL is needed to enable the . to match newlines too.

In Python:

import re
data = '''
27/07/18, 12:02 PM - user_a: https://www.youtube.com/
 Watch this
27/07/18, 12:15 PM - user_b: <Media omitted>
27/07/18, 12:52 PM - user_b: Read this fully
some text
some text
.
some text
27/07/18, 12:56 PM - user_c: text ..
'''
usern = 'user_b'

pattern = re.compile(r""+usern+r": (.*?)(?=^[0-9]{2}/[0-9]{2}/[0-9]{2})",re.DOTALL|re.MULTILINE)
print(re.findall(pattern,data))

Prints:

['<Media omitted>\n', 'Read this fully\nsome text\nsome text\n.\nsome text\n']
user_33904190
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