Regex find non digit and/or end of string

Question

How do I include an end-of-string and one non-digit characters in a python 2.6 regular expression set for searching? I want to find 10-digit numbers with a non-digit at the beginning and a non-digit or end-of-string at the end. It is a 10-digit ISBN number and 'X' is valid for the final digit. The following do not work: ``` is10 = re.compile(r'\D(\d{9}[\d|X|x])[$|\D]') is10 = re.compile(r'\D(\d{9}[\d|X|x])[\$|\D]') is10 = re.compile(r'\D(\d{9}[\d|X|x])[\Z|\D]') ``` The problem arises with the last set: [\$|\D] to match a non-digit or end-of-string. Test with: ``` line = "abcd0123456789" m = is10.search(line) print m.group(1) line = "abcd0123456789efg" m = is10.search(line) print m.group(1) ```

Accepted Answer

You have to group the alternatives with parenthesis, not brackets:

```
r'\D(\d{9}[\dXx])($|\D)'
```

`|` is a different construct than `[]`. It marks an alternative between two patterns, while `[]` matches one of the contained characters. So `|` should only be used inside of `[]` if you want to match the actual character `|`. Grouping of parts of patterns is done with parenthesis, so these should be used to restrict the scope of the alternative marked by `|`.

If you want to avoid that this creates match groups, you can use `(?: )` instead:

```
r'\D(\d{9}[\dXx])(?:$|\D)'
```

Regex find non digit and/or end of string

Solution

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