Question

Code Iteration #2

Changing var1 to a raw string by using the stringVar = r'string' worked great. With the code below I am now getting an exception of:

Traceback (most recent call last):
  File "regex_test.py", line 8, in <module>
    pattern = re.compile(var2 + "(.*)")
  File "/usr/lib/python2.7/re.py", line 190, in compile
    return _compile(pattern, flags)
  File "/usr/lib/python2.7/re.py", line 242, in _compile
    raise error, v # invalid expression
sre_constants.error: unbalanced parenthesis

—

#!/usr/bin/python

import re

var1 = r'\\some\String\to\Match'
var2 = '\\\\some\\String\\'

pattern = re.compile(var2 + "(.*)")
found = pattern.match(var1, re.IGNORECASE)

if found:
    print "YES"
else:
    print "NO"

---

I am trying to include a variable in my regular expression. This question is related to this other question, but differs slightly by using a compiled pattern vs the variable within the match. According to everything I’ve read, the example code below should work.

#!/usr/bin/python

import re

var1 = re.escape('\\some\String\to\Match') # A windows network share
var2 = "\\\\some\\String\\"

print var1 # Prints \\some\\String\ o\\Match
print var2 # Prints \\some\String\

pattern = re.compile(var2)
found = pattern.match(var1 + "(.*)", re.IGNORECASE)

if found:
    print "YES"
else:
    print "NO"

When I print out my variables I am seeing some weird behavior. I thought the re.escape would escape all needed chars within a string.

When I execute the code in Python 2.7 on Ubuntu 12.4.1 I get the following exception

Traceback (most recent call last):
  File "regex_test.py", line 11, in <module>
    pattern = re.compile(var2)
  File "/usr/lib/python2.7/re.py", line 190, in compile
    return _compile(pattern, flags)
  File "/usr/lib/python2.7/re.py", line 242, in _compile
    raise error, v # invalid expression
sre_constants.error: bogus escape (end of line)

What am I missing that is causing the exception to be thrown?

Answer 1

re.escape is used for escaping a string to use as a regular expression, but you escape var1 and then use var2 as the regular expression.

I think the following is what you are trying to accomplish:

var1 = r'\\some\String\to\Match'
var2 = re.escape('\\\\some\\String\\')
pattern = re.compile(var2 + '(.*)', re.IGNORECASE)
found = pattern.match(var1)

Note that the r'\\some\String\to\Match' is a raw string literal, but you cannot use it for var2 since it needs to end in a backslash.