[FIXED] Replace An Exact Grouped Part in RegEx Python

Issue

I have a template that I need to replace a part of that using Regex in Python. Here is my template: (Note that there is at least a new line between two comments)

hello
how's everything

<!--POSTS:START-->
some text
<!--POSTS:END-->

Some code here

I want to replace everything between <!--POSTS:START--> and <!--POSTS:END--> in Python. So I made <!--POSTS:START-->\n([^;]*)\n<!--POSTS:END--> pattern but it includes <!--POSTS:START--> and <!--POSTS:END--> too.

Here is what I want:

re.sub('...', 'foo', message)

# expected result:
hello
how's everything

<!--POSTS:START-->
foo
<!--POSTS:END-->

Some code here

Thanks.

Solution

You can use a capture group for the start and end markers and reference those as \1, \2, etc in the target replacement string.

If the text has multiple occurrences of <!--POSTS:START-->...<!--POSTS:END--> then the regexp with .*? will replace each of those groups. If the '?' is removed the regexp then it will remove all text from the start of the first group to the end of the last group.

Try this:

import re

s = '''
hello
how's everything

<!--POSTS:START-->
some text
<!--POSTS:END-->

Some code here
'''

# for multi-line matching need extra flags in the regexp
s = re.sub(r'(<!--POSTS:START-->\n).*?(\n<!--POSTS:END-->)', r'\1foo\2', s, flags=re.DOTALL)

# this inlines the DOTALL flag in the regexp for same result
# s = re.sub(r'(?s)(<!--POSTS:START-->\n).*?(\n<!--POSTS:END-->)', r'\1foo\2', s)

print(s)

Output:

hello
how's everything

<!--POSTS:START-->
foo
<!--POSTS:END-->

Some code here

Answered By - CodeMonkey