Where's the error?

[Brian McErlean]

"Bror Johansson" <bror.johansson at avionics.saab.se> wrote in message news:<9uibrc$199$1 at newstoo.ericsson.se>...
> I have tried, retried, and retried again to understand why Python 2.1 tells
> me this:
> 
> Traceback (most recent call last):
>   File "D:\Src\Python\renamevobtag.py", line 12, in getlsvobresults
>     regline =
> re.compile('^(\*)?\s+\\([^\s]+)\s+\\\\([^\s]+)\s+(p(rivate)|(ublic))\s*$')
>   File "d:\python21\lib\sre.py", line 90, in compile
>     return _compile(pattern, flags)
>   File "d:\python21\lib\sre.py", line 136, in _compile
>     raise error, v # invalid expression
> error: unbalanced parenthesis
> 
> when approaching this line:
> 
>     regline =
> re.compile('^(\*)?\s+\\([^\s]+)\s+\\\\([^\s]+)\s+(p(rivate)|(ublic))\s*$')
> 
> Can anyone help me clear my eyesight?
> 
> I can of course switch over to 'split', but I want to get rid of the extra
> backslashes.
> 
> /BJ

When using regexps, remember that python interprets "\" characters for
the strings, so by the time it reaches re.compile, the string reads:


'^(*)?s+\([^s]+)s+\\([^s]+)s+(p(rivate)|(ublic))s*$'

The first \\ has become a single \ before the "(", which is treated as
a literal "(" for the regex, rather than a group, hence the unbalanced
parameters.

The easiest way to fix it is to use rawstrings.  Prepend an "r" before
the string to bypass python's string escaping.

ie.
  re.compile(r'^(\*)?\s+\\([^\s]+)\s+\\\\([^\s]+)\s+(p(rivate)|(ublic))\s*$')

this should match lines like:    r"* \abc123! \\xyz  private
...anything..."

Hope this helps.

Brian McErlean