regex of backslash in python

Question

I'm trying to get the regex of a backslash, if the backslash is not present in the test string and not in the regex expression it is working, I tried \\ and \\\\

This is my code:

def getString():
    test=',"string":"x\u002fg06HaX0M\u002fuTJh7Q",Bla'
    m=re.match(r",\"string\":\"([\w\d\\\\]+)\",",test)
    print(m.group(1))

The error i get is:

AttributeError: 'NoneType' object has no attribute 'group'

Thanks for your help.

Avinash Raj · Accepted Answer · 2015-08-15 15:41:11Z

2

It seems like unicode chars present in your input string (\u002f). There isn't an actual backslash character present in the input. And also I think you want to get the characters present inside the double quotes which exists next to "string":. So use a negated character class like [^"]+ to match any character (including unicodes) but not of double quotes, zero or more times.

>>> st=',"string":"x\u002fg06HaX0M\u002fuTJh7Q",Bla'
>>> re.match(u',"string":"([^"]+)",',st).group(1)
'x/g06HaX0M/uTJh7Q'
>>> re.match(r',"string":"([^"]+)",',st).group(1)
'x/g06HaX0M/uTJh7Q'

edited Aug 15, 2015 at 15:41

answered Aug 15, 2015 at 15:33

Avinash Raj

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3 Comments

Neet33 Over a year ago

Why there isn't actual backslash?

Avinash Raj Over a year ago

\u002f is a Unicode representation of a single character.

Neet33 Over a year ago

p=''     for line in text.splitlines():         if 'string' in line:             for part in line.split(','):                 if 'string' in part:                     p=part     m=re.match(u'.*"string":"([^"]+)"',p).group(1)

Any idea why this code doesn't decode the unicode characters? print(m) gives string with \u002f

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