UNIX script type checking

[Runt888]

Is there anyway to do type checking in a UNIX script? I need to check if a command line argument to my script is an integer. I'm using a while loop to get each character, but I don't know how to check if the character is an integer. I tried checking if it was between 0 and 9, but I got an error. If there's a way to get the ascii value, then I could check that. Any ideas? Thanks.

Kelly

 

[pmcmicha]

Sure...

$ SOME_SCRIPT VAR1 VAR2 VAR3

$ SOME_SCRIPT TIME IS 4

SCRIPT:
#!/usr/bin/ksh -p

# TEST VAR3 for number only
CHKLET=`grep "[A-Za-z]" $3 > /dev/null 2>&1`
if [ $? -eq 0 ]
then
print "Variable 3 needs to be a number only...."
read PAUSE?"Press [Enter] to continue."
USAGE
fi

Something to that effect.

 

[Runt888]

Thanks for the reply. I never thought about using grep. I have one question: what is the ? for in the if statement? Thanks.

Kelly

 

[SamBones]

In the Korn shell, you can also force a variable to only allow integers. Try this...
[tt]
typeset -i INTEGER_VAR
[/tt]

 

[pmcmicha]

$?: This is a special request to the shell for a Return Code. So if the RC is not equal to 0 for any command (unless otherwise stated in a man page), then it failed.

So if the grep found any letter then it would be successful and exit with a status of 0, otherwise it exits with a status of 1.

 

[Runt888]

Ah...makes sense. Thanks for all the replies. SamBones, your solution sounds great, but I'm using the bash shell. Thanks though. And thank you pmcmicha for your solution.

Kelly

 

[Annihilannic]

Works for bash too. Annihilannic.