awk - how to compare part of the string?

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# 1  
Old 10-11-2013
IBM awk - how to compare part of the string?

Need help for awk..
file will have comma separated numbers, I need check digits before 10 numbers eg ( 001)1234567890
Basically want to check country code of a mobile number.

abc,def,data, data,0011234567890, data,data

Script should be checking country code with 001, I will pass country code in a variable.

Last edited by radoulov; 10-11-2013 at 09:47 AM..
# 2  
Old 10-11-2013
awk -F, -v c=001 'substr($5,0,3)==c' file

This will print the line if the first three numbers of the 5th field are 001.
This User Gave Thanks to Subbeh For This Post:
# 3  
Old 10-11-2013
But how do I find 3 digits before 10 digits from right ? Number could be greater or less than 10... if greater it might have leading multiple zeros.. So need to compare from the right side of the string.

eg: number 1231234567890
country (123)1234567890

eg: number 00000000121234567890
country 012

eg number 12345
country : nothing will be there to compare as number is not 10 digit

Please help !

Last edited by vegasluxor; 10-11-2013 at 10:05 AM..
# 4  
Old 10-11-2013
try this:
awk -F, -v c=001 'substr($5,length($5)-12,3)==c' file

# 5  
Old 10-14-2013
Thanks a lot ! It worked !!! Smilie
# 6  
Old 10-14-2013
even shorter :
awk -F, -v c=001 '$5~c"..........$"' file

Note that this code does not check wheter the 10 characters that are on the right of c are digit or not , but neither does the code provided by Subbeh
# 7  
Old 10-14-2013
$ cat file
abc,def,data, data,0011234567890, data,data
abc,def,data, data,00112345678901, data,data
abc,def,data, data,001123456789, data,data
abc,def,data, data,001123456789x, data,data
abc,def,data, data,0021234567890, data,data
$ awk -F, -v c=001 '$5~"^"c"[[:digit:]]{10}$"' file
abc,def,data, data,0011234567890, data,data

(GNU awk on cygwin)

Last edited by CarloM; 10-14-2013 at 02:40 PM.. Reason: better test data
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