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Python phonenumber regex doesn't work well enough

I have this regex code that I am using in my code:

pattern = re.compile('\d{3,4}(\/?)(\d{6,6})')
m= pattern.match('0481/987421')
if m:
    print "yes"
else:
    print "no"

It's a regex that should work for phonenumbers like this: dddd / dddddddd so there are 3 or 4 digits first, then a forward slash or not, and then exactly 6 digits. It works great, for example 21/484135 doesn't work and other wrong things don't work either. But the problem with this regex is that when my first characters are correct and I type something random behind it, it still prints yes. I mean something like this: 0481/9874214879516874 I think because the regex matches the first 11 characters, it returns its matches, no matter what's behind.

How can I solve this problem?

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2 answers


You need to bind your expression. Add $

or \Z

at the end to make sure nothing follows. You can also add ^

to bind it at the beginning of the line, but this is not required when using with match()

.



pattern = re.compile(r"^\d{3,4}/?\d{6}\Z")
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I would suggest using phonenumbers instead of writing your own regex. Here's an example of parsing a Belgian phone number:

>>> x = phonenumbers.parse("0481/987421", "BE")
>>> x
PhoneNumber(country_code=32,
            national_number=481987421L,
            extension=None,
            italian_leading_zero=False,
            country_code_source=None,
            preferred_domestic_carrier_code=None)


It throws an exception on invalid phone numbers:

>>> x = phonenumbers.parse("0481/9874214879516874", "BE")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/local/lib/python2.7/dist-packages/phonenumbers/phonenumberutil.py", line 2038, in parse
    "The string supplied is too long to be a phone number.")
phonenumbers.phonenumberutil.NumberParseException: (4) The string supplied is too long to be a phone number.
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