How to generate a regular expression that matches one specific string

Question

Here, I've tried to create a regular expression that matches one particular string:

#This is the string that the regex is intended to match
theString = "..!-+!)|(!+-!.."

print(re.compile(theString).match(theString))

This produces an error instead of matching the string:

raise error, v # invalid expression
sre_constants.error: unbalanced parenthesis

Is there any way to generate a regular expression that matches just one specific string, like this one?

Not sure I understand the question. For example, the regex "apple" will only match the string "apple". — Tim Jones, Mar 04 '14 at 01:27
Why do you want to use a regular expression? You could just use a simple search function if it is really just a specific string. — zchrykng, Mar 04 '14 at 01:27
@TimJones `re.compile("..!-+!)|(!+-!..")` does not match the string `"..!-+!)|(!+-!.."`. — Anderson Green, Mar 04 '14 at 01:28
The problem with that string is that it contains lots of characters that are "special" in regex syntax. — zchrykng, Mar 04 '14 at 01:28
@zchrykng I'm using a regular expression because I need to generate a regex that I can use to [split another string without removing separators](http://stackoverflow.com/questions/4416425/how-to-split-string-with-some-seperator-but-without-removing-that-seperator-in-j). — Anderson Green, Mar 04 '14 at 01:29
@AndersonGreen then pasztorpisti's answer is probably what you are looking for. — zchrykng, Mar 04 '14 at 01:30
possible duplicate of [Escaping regex string in Python](http://stackoverflow.com/questions/280435/escaping-regex-string-in-python) — Anderson Green, Mar 04 '14 at 02:23

score 6 · Accepted Answer · answered Mar 04 '14 at 01:28

6

import re
your_string = "..!-+!)|(!+-!.."
your_regex_string = "^" + re.escape(your_string) + "$"
your_regex = re.compile(your_regex_string)

answered Mar 04 '14 at 01:28

pasztorpisti

1 Answers1