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I want to use the following regex with awk to validate phone numbers:

echo 012-3456-7890 | awk '/^\(?0[1-9]{2}\)?(| |-|.)[1-9][0-9]{3}( |-|.)[0-9]{4}$/ {print $0}'

But I am getting the following error:

awk: line 1: regular expression compile failed (missing operand)
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  • 1
    Are you using mawk? This works with GNU awk, but with mawk there are two problems: (| |-|.) has a leading | without a regex atom before it, and mawk doesn't support {n}.
    – muru
    Jun 5 '18 at 4:59
  • Yes I need a regex works with MAWK. The first pipe symbol is to match the no space condition. Is there any workaround for this problem?
    – sci9
    Jun 5 '18 at 5:22
  • You could handle the optional space/dash/period using [ -.]?, but this still leaves the {n} problem open, so I think you should switch from mawk to something else (gawk, Perl, Ruby, Python, Java, ....). Actually, grep -e or grep -P should work too. I don't see any compelling reason, why you are using a programming language, if you can simply grep for it. Jun 5 '18 at 5:33
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Since the ranges used here are of fixed length, you could simply write out the entire range [0-9]{3} => [0-9][0-9][0-9]. And instead of (| |-|.), ( |-|.)? - though I am confused: are you allowing any character (.), in addition to space and -? Then it could just be .? since space and - are matched by . anyway. If you're matching the literal period ., then you should use [- .]? instead (the leading - is to avoid interpretation as a character range). So:

^\(?0[1-9]{2}\)?(| |-|.)[1-9][0-9]{3}( |-|.)[0-9]{4}$

Becomes:

^\(?0[1-9][1-9]\)?[- .]?[1-9][0-9][0-9][0-9][- .][0-9][0-9][0-9][0-9]$

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