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Simple regular expression

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# 1  
Old 05-29-2012
Simple regular expression
Hello,

I don't understand why my regular expression doesn't work.
simply I want to check a phone number of 8 digits doesn't start with 0

Code:
regex1='^([1-9]{1})([0-9]{7})$'

 if [ "`expr \"$1\" : \"$regex1\"`" != "0" ]
        then
         ------
fi

for example this number should be valid 12345678
and this 01234567 or 1234567 not

Thanks.
Last edited by joeyg; 05-29-2012 at 02:44 PM.. Reason: Wrap commands and data with CodeTags
# 2  
Old 05-29-2012
May be ..
Code:
$ regex1='^[1-9]\{1\}[0-9]\{7\}$'
$ x=0122
$ if [ "`expr "$x" : "$regex1"`" -ne 0 ]; then echo yes ;else echo no; fi
no
$ x=12121212
$ if [ "`expr "$x" : "$regex1"`" -ne 0 ]; then echo yes ;else echo no; fi
yes
$

# 3  
Old 05-29-2012
It worked with me by just adding \

Thanks.
# 4  
Old 05-30-2012
Quote:
Originally Posted by eng_asa
It worked with me by just adding \

It would, but remember that, with "!=" you are comparing as string.
"-ne" is for integer comparison with is more accurate in this case since the return value from "expr" would always be an integer.
# 5  
Old 05-30-2012
do we need of the specify the character limit as {1} ?
the character class with a range [1-9] by default matches for a single character within the range right ? in that case no need of specifying
Code:
[1-9]{1}

instead we can go with

Code:
^[1-9][0-9]{7}

I am recently learning regex.. so please correct me if i am wrong
# 6  
Old 05-30-2012
That is correct chidori, but there would need to be a dollar sign at the end, otherwise longer numbers would be matched too.
And in a basic regular expression, which is what expr uses, that would be:
Code:
^[1-9][0-9]\{7\}$

# 7  
Old 05-30-2012
yeah. Got it... Thanks
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