python re for validation string

Question

Below is code:

import re
string = "02:222222"
if re.match(r'[a-fA-F0-9]+[a-fA-F0-9]+:+[a-fA-F0-9]+[a-fA-F0-9]+$',string):
    print "pass"
else:
    print "fail"

above code is printing "pass"

my expected output should be "fail"

Below are the few examples:

string = 00:20
expected output: pass
string = 00:202
expected ouput: fail
string = 00:2z
expected output: fail
string = 000:2
expected ouput: fail

What exactly do you expect to match? Explain the structure of the matching input. Your example is too narrow. Is it two digits followed by colon followed by two digits? — DYZ
– DYZ, Commented Jul 18, 2017 at 5:53

Mustofa Rizwan · Accepted Answer · 2017-07-18 05:49:18Z

5

You can try this:

^[a-fA-F0-9]{2}:[a-fA-F0-9]{2}$

Demo

answered Jul 18, 2017 at 5:49

Mustofa Rizwan

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Laur Ivan · Accepted Answer · 2017-07-18 06:01:13Z

3

Note that + in regexp means "one or more". If you want an exact number of characters (like "XX:XX"), then you should remove the "+"-es:

r'[a-fA-F0-9][a-fA-F0-9]:[a-fA-F0-9][a-fA-F0-9]$'

or, better:

r'([a-fA-F0-9][a-fA-F0-9]):([a-fA-F0-9][a-fA-F0-9])$'

to get the $1 and $2 components of the regexp.

According to this, you can also use:

^[a-fA-F0-9]{2}:[a-fA-F0-9]{2}$

or even more compact:

^[^\W_]{2}:[^\W_]{2}$

edited Jul 18, 2017 at 6:01

answered Jul 18, 2017 at 5:55

Laur Ivan

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Comments

Ebrahim Poursadeqi · Accepted Answer · 2017-07-18 05:57:37Z

2

you can simply remove + between your regular express string

+ in regular expression means 1 or more of the preceding expression

answered Jul 18, 2017 at 5:57

Ebrahim Poursadeqi

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