Quick regex question about alphabetic string

Shell Programming and Scripting
1

Hi guys,

Pretty new to regex, and i know im doing something wrong here. I'm trying to get a regex command that restricts a string to be 8 characters long, and the first character cannot be 0. Here's what i have so far...

echo "01234" | grep "^[0][0-9]{8}*$"

Thanks very much!

-Crawf

EDIT: Sorry about the wrong title, after two long hours of trying another regex command to work, about 30 seconds after posting managed to figure out an answer by myself...Sorry!

2

no need regular expression. KISS

[[ ${#num} -ne 8 ]] && echo "not equal 8 characters" && exit
case $num in
 0* ) echo "0 as first character""  &&     exit ;;
esac
3

Aha! That works as well!

Thanks mate!

-Crawf

4

If you are using ksh93, the following is one way of testing the numeric string is of length 8 and does not have a leading zero

#!/bin/ksh93

num="12345678"
req_len=8

if [[ ${#num}-$num == $req_len-${num/?('0')({7,8}(\d))*/\2} ]]
then
   print "OK - $num matches criteria"
else
   print "Sorry - $num contains either a leading zero or is not of length ${req_len}!"
fi
5

Can that be done in the Bourne shell? Thats what i'm working in...

Looks just what i need though! Thanks!

-Crawf

6

$echo "01234567"|awk 'BEGIN{FS=null}{if(length==8 && $1!='0') print"Valid"}'
$echo "91234567"|awk 'BEGIN{FS=null}{if(length==8 && $1!='0') print"Valid"}'
Valid

7

Thanks Abhishek Ghose! That works well too!

Thanks for the support guys!

-Crawf

8 
case $num in [1-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]) echo fine;;
  *) echo not good at all, play again;;
esac