12

I've been messing around with the python re modules .search method. cur is the input from a Tkinter entry widget. Whenever I enter a "\" into the entry widget, it throws this error. I'm not all to sure what the error is or how to deal with it. Any insight would be much appreciated.

cur is a string

tup[0] is also a string

Snippet:

se = re.search(cur, tup[0], flags=re.IGNORECASE)

The error:

Exception in Tkinter callback
Traceback (most recent call last):
  File "C:\Python26\Lib\Tkinter.py", line 1410, in __call__
    return self.func(*args)
  File "C:\Python26\Suite\quidgets7.py", line 2874, in quick_links_results
    self.quick_links_results_s()
  File "C:\Python26\Suite\quidgets7.py", line 2893, in quick_links_results_s
    se = re.search(cur, tup[0], flags=re.IGNORECASE)
  File "C:\Python26\Lib\re.py", line 142, in search
    return _compile(pattern, flags).search(string)
  File "C:\Python26\Lib\re.py", line 245, in _compile
    raise error, v # invalid expression
error: bogus escape (end of line)
rectangletangle
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4 Answers4

18

"bogus escape (end of line)" means that your pattern ends with a backslash. This has nothing to do with Tkinter. You can duplicate the error pretty easily in an interactive shell:

>>> import re
>>> pattern="foobar\\"
>>> re.search(pattern, "foobar")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/re.py", line 142, in search
    return _compile(pattern, flags).search(string)
  File "/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/re.py", line 241, in _compile
    raise error, v # invalid expression
sre_constants.error: bogus escape (end of line)

The solution? Make sure your pattern doesn't end with a single backslash.

Bryan Oakley
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  • Is it possible to treat a \ like an ordinary character? Similar to the r"string" method. – rectangletangle Dec 13 '10 at 23:09
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    @Anteater7171: backslash is special to regular expressions. You have two choices: don't user regular expressions or modify the string so that the special meaning is removed. For the latter, adding an additional backslash does the trick (ie: the pattern '\\' means a literal backslash). – Bryan Oakley Dec 14 '10 at 03:08
  • +1 You should have added your last comment to the answer. To expand on the comment, in the context of e.g. re.search that would mean making ``pattern = "foobar\\\\"``, i.e. you need 4 backslashes so the pattern argument for re.search ends up having two, which would make for a legal regexp. – ThomasH Sep 19 '12 at 08:41
  • This is actually somewhat unfortunate. It means that for writing save code, you need to check your variable beforehand every time. And in particular, you need to be aware of this case. Do you think that the pattern ending on backslash is the only case where re.find does not return None or a match object? – Radio Controlled Nov 28 '19 at 07:58
  • I would prefer if it printed a warning and either ignored the backslash or returned None – Radio Controlled Nov 28 '19 at 07:59
  • @RadioControlled: no, re.find will throw errors for any type of invalid expression, and there are many ways to write invalid expressions. – Bryan Oakley Nov 28 '19 at 15:46
  • @BryanOakley, Yeah, I guess if one just wants to find a string in a string, one should use python's native `.find()` method... Or is there an equivalent for `re`, such that the output is either `None` or a match object instead of the index or `-1` returned by `.find()`? – Radio Controlled Nov 29 '19 at 11:20
12

The solution to this issue is to use a raw string as the replacement text. The following won't work:

re.sub('this', 'This \\', 'this is a text')

It will throw the error: bogus escape (end of line)

But the following will work just fine:

re.sub('this', r'This \\', 'this is a text')

Now, the question is how do you convert a string generated during program runtime into a raw string in Python. You can find a solution for this here. But I prefer using a simpler method to do this:

def raw_string(s):
    if isinstance(s, str):
        s = s.encode('string-escape')
    elif isinstance(s, unicode):
        s = s.encode('unicode-escape')
    return s

The above method can convert only ascii and unicode strings into raw strings. Well, this has been working great for me till date :)

rtnpro
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3

The first parameter to re is the pattern to search for, thus if 'cur' contains a backslash at the end of the line, it'll be an invalid escape sequence. You've probably swapped your arguments around (I don't know what tup[0] is, but is it your pattern?) and it should be like this

se = re.search(tup[0], cur, flags=re.IGNORECASE)

As you very rarely use user input as a pattern (unless you're doing a regular expression search mechanism, in which case you might want to show the error instead).

HTH.

EDIT:
The error it is reporting is that you're using an escape character before the end of line (which is what bogus escape (end of line) means), that is your pattern ends with a backslash, which is not a valid pattern. Escape character (backslash) must be followed by another character, which removes or adds special meaning to that character (not sure exactly how python does it, posix makes groups by adding escape to parentheses, perl removes the group effect by escaping it). That is \* matches a literal asterix, whereas * matches the preceding character 0 or more times.

falstro
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  • I'm using user input to search the contents of tup[0]. cur is my pattern. – rectangletangle Dec 13 '10 at 09:09
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    @Anteater; well then, if cur is the pattern, i.e. the user is entering the pattern, it is obviously incorrect and the user needs to be made aware of it..? – falstro Dec 13 '10 at 09:35
  • that depends on whether the user thinks they are entering a regular expression. If they think they are entering a fixed string the code needs to be fixed. If they think they are entering a regular expression they should be informed that the expression is invalid (but hopefully in the spirit of informing or teaching rather than chastising). – Bryan Oakley Dec 14 '10 at 03:11
  • @Bryan; good point. Then of course you should not use regular expressions to search the string but rather something equivalent of C's `strstr`, which ought to be faster and not prone to these problems. @Anteater said "`cur` is [his] pattern" though, so I assumed he would like to have the user enter a regular expression. – falstro Dec 14 '10 at 08:12
3

If you are trying to search for "cur" in "tup[0]" you should do this through "try:... except:..." block to catch invalid pattern:

try :
    se = re.search(cur, tup[0], flags=re.IGNORECASE)
except re.error, e:
    # print to stdout or any status widget in your gui
    print "Your search pattern is not valid."
    # Some details for error:
    print e
    # Or some other code for default action.
Taras
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