Python regex multiline replacement

Question

I searched existing questions but they do not seem to answer this specific question.

I have the following python program

description = """\
before

{cs:id=841398|rep=myrepo}: after
"""
pattern = re.compile(r"(.*)\{cs:id=(.*)\|rep=(.*)\}(.*)")

and I need to replace the regex in the description to look like the below but I can't get the pattern and replacement syntax right

description="""\
before

<a href="http://crucible.app.com:9090/myrepo?cs=841398">841398</a> : after
"""

The crucible.app.com:9090 is a constant that I have beforehand so I basically need to substitute the pattern with my replacement.

Can someone show me what is the best python regex find and replace syntax for this?

Have you looked into re.sub?

Jerry
– Jerry

2013-08-27 14:41:26 +00:00
Commented Aug 27, 2013 at 14:41 — Jerry
– Jerry, Commented Aug 27, 2013 at 14:41

Martin Ender · Accepted Answer · 2013-08-27 14:43:54Z

2

There is no need for the first and last (.*) in your pattern. To write back captured groups in the replacement string, use \1 and \2:

description = re.sub(pattern, "<a href=\"http://crucible.app.com:9090/\2?cs=\1\">\1</a>", description)

By the way, another way to improve your pattern (performance- and robustness-wise) is to mkae the inner repetitions more explicit so that they cannot accidentally go past the | or }:

pattern = re.compile(r"\{cs:id=([^|]*)\|rep=([^}]*)\}")

You can also use named groups:

pattern = re.compile(r"\{cs:id=(?P<id>[^|]*)\|rep=(?P<rep>[^}]*)\}")

And then in the replacement string:

"<a href=\"http://crucible.app.com:9090/\g<repo>?cs=\g<id>\">\g<id></a>"

answered Aug 27, 2013 at 14:43

Martin Ender

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falsetru · Accepted Answer · 2013-08-27 14:51:47Z

2

Use re.sub / RegexObject.sub:

>>> pattern = re.compile(r"{cs:id=(.*?)\|rep=(.*?)}")
>>> description =  pattern.sub(r'<a href="http://crucible.app.com:9090/\1?cs=\2">\1</a>', description)
>>> print(description)
before

<a href="http://crucible.app.com:9090/841398?cs=myrepo">841398</a>: after

\1, \2 refer to matched group 1, 2.

I modified the regular expression slightly.

No need to escape {, }.
Removed capturing group before, after {..}.
Used non-greedy match: .*?

edited Aug 27, 2013 at 14:51

answered Aug 27, 2013 at 14:45

falsetru

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