How do I check if the argument passed to a script is an integer?

I am writting a script that will take to integers and want to be able to check before I go on.


I am using bourne shell.

Thanks in advance

echo $string | grep "^[0-9]+$"
echo $?

Sorry but that does not work.

I have this script that is working except for negative integers or when someone enters a + infront of the integer.

Why isn't this accepting negative integers or integer with a + sign?


#!/bin/sh
checkinput()
{
case $1 in [0-9]|[0-9]*[0-9][\.]) return 0 ;; esac
return 1
}

exitusage()
{
echo "Usage: Proj2 Integer1 Integer2"
exit
}

[ $# -ne 2 ] && exitusage
checkinput $1 || exitusage && A=$1
checkinput $2 || exitusage && B=$2

Thanks in advance

replace your checkinput() func with this...

checkinput() 
{ 

echo $1 | grep '^(\+|-)?[0-9]+$'

return $?

}

I certanly apreciate all your help. but this:
checkinput()
{

echo $1 | grep '^(\+|-)?[0-9]+$'

return $?

}

doesn't work either. I'll keep tryng.

Thanks a million.

it is always a good idea to test things on command line before putting in a script.. which I hope you are already doing..

you can do few tests echoing the return code...

echo -3243 | grep '^(\+|-)?[0-9]+$'
echo $?

echo -abc | grep '^(\+|-)?[0-9]+$'
echo $?

let me know if these tests work on your unix system...

Cheers!
Vishnu.

I think you need to use -E with grep to invoke extended regular expressions. But also, if the argument passes the test, it will go to stdout. The -q option will fix that. So try this:

checkinput() 
{ 

echo $1 | grep -Eq '^(\+|-)?[0-9]+$'

return $?

}