Shell Programming and Scripting

Hello Folks,
I am working on filtering a file having some special characters.
Let's say for an example a file contains person name and phone number based on positions. First 5 characters name and next 10 characters phone number. My task is to , if there is any special character in phone number (position 6-15), then that record should be removed from the file

Code:

Input File
 
RAMUN1237891111
SUPER9019199911
GOODS9*9&889000
YESIR234^789909

Desired output

Code:

RAMUN1237891111
SUPER9019199911

PLease help me with the awk script to accomplish this

Try:

Code:

awk '/^......*[^[:digit:]]/ {next}1' input

try this..

Code:

 awk '/^[a-z0-9A-Z]+$/' file

Code:

grep -iE '^[a-z]{5}[0-9]{10}' file

Thanks for your replies..
All the above command works. But I want to check for special characters at specific position. In the above sample input, if special character is present in first 5 position, the record should not be filtered.

Appreciate your help

Hi

Code:

awk 'substr($0,6) ~ /^[a-z0-9A-Z]+$/' file

Guru.

I think you'll find that the script I provided allows the 1st five characters to be any value (except the null byte and the newline character) and will discard any line if anything other than decimal digits follow that (no matter how many more characters are on the line).

The script provided by elixir_sinari requires the 1st five characters to be uppercase or lowercase letters, the next 10 characters to be decimal digits (anything after those 1st fifteen characters will be accepted and printed).

The script provided by pamu accept and print any line that contains one or more alphanumeric characters as long as no other characters are present on the line.

If you really want to allow any character in the 1st five positions and exactly 10 decimal digits after that, a combination of the above suggestions:

Code:

grep -E '^.{5}[0-9]{10}$' file

should work.