我如何使用正则表达式从字符串中获取文件名

Question

I have this string here:

"['\r\n                    File: FLO_JIUWASOKLDM_05_HetR_IUSJA_&_Cracks.mp4', <br/>, '\r\n                    Size: 48.14 MB                ']"

and I have this regex \\w+\\.\\w+

And I want the regex to get the filename FLO_JIUWASOKLDM_05_HetR_IUSJA_&_Cracks.mp4

But it breaks at the ampersand, which returns _Cracks.mp4 what do I need to do to fix it? I'm super new to Regex.

Answer 1

There are many options to exercise here, one for example, would be:

([^\s]+\.[a-z][a-z0-9]+)

Demo

Test

# coding=utf8
# the above tag defines encoding for this document and is for Python 2.x compatibility

import re

regex = r"([^\s]+\.[a-z][a-z0-9]+)"

test_str = "\"['\\r\\n                    File: FLO_JIUWASOKLDM_05_HetR_IUSJA_&_Cracks.mp4', <br/>, '\\r\\n                    Size: 48.14 MB                ']\"
"

matches = re.finditer(regex, test_str, re.MULTILINE | re.IGNORECASE)

for matchNum, match in enumerate(matches, start=1):

    print ("Match {matchNum} was found at {start}-{end}: {match}".format(matchNum = matchNum, start = match.start(), end = match.end(), match = match.group()))

    for groupNum in range(0, len(match.groups())):
        groupNum = groupNum + 1

        print ("Group {groupNum} found at {start}-{end}: {group}".format(groupNum = groupNum, start = match.start(groupNum), end = match.end(groupNum), group = match.group(groupNum)))

# Note: for Python 2.7 compatibility, use ur"" to prefix the regex and u"" to prefix the test string and substitution.

Answer 2

\\w is the shorthand for "word character", meaning letters, numbers, and underscore. Note the lack of ampersand. To include the ampersand, you could use the character class [\\w&] . Your regex would then be

[\w&]+\.\w+

• BTW note that this may also match 48.14 depending on the regex function you use.

But maybe you want to include more characters than just ampersand. How about all non-whitespace characters?

\S+\.\w+

• This uses \\S , which is the inversion of the whitespace shorthand \\s .

Answer 3

Instead of figuring out what characters the file name may contain (note it may even contain spaces usually), you mayleverage the context: you know it starts after File: and space(s) and runs up to the ' .

So, you may achieve what you need using

m = re.search(r"File:\s*([^']+)", s)
if m:
    print(m.group(1))

See the online Python demo .

See also the regex demo and the regex graph :

Details

• File: - a literal substring
• \\s* - 0+ whitespaces
• ([^']+) - Capturing group 1 ( match_object.group(1) ): 1 or more chars other than ' .