Checking if string contains integer

haz · #1 (**permalink**) 09-10-2006

G'day guys, first post so be gentle.

I need help with some code to work out if a variable (string) contains any integers. The valid variable (string) must contain only letters.

Also need to be able to work out if a variable contains only integers.

Any help greatly appreciated.

vino · #2 (**permalink**) 09-10-2006

Welcome to the forums. We have a search feature which you can use to see if similiar problems have encountered before.

See - ksh : find value type

ghostdog74 · #3 (**permalink**) 09-10-2006

you can try this?

Code:

if ! [ $var -ge 0 -o $var -lt 0 ]; then 
    echo "string"
....

haz · #4 (**permalink**) 09-10-2006

Sorry, should have specified it is a Bourne shell script that I am writing.

This is what I've tried writing, where am i going wrong?

Code:

namenum=$1

if [ "$namenum" -eq *[a-zA-Z]* ] ; then
      echo "$namenum is all alphabetic"
else
      echo "not all alphabetic"
fi

tayyabq8 · #5 (**permalink**) 09-10-2006

Try:

Code:

#! /bin/sh

namenum=$1

echo $namenum | grep [0-9] > /dev/null
if [ $? -eq 0 ] ; then
        echo "non-alpha"
else
        echo "alpha"
fi

haz · #6 (**permalink**) 09-10-2006

Thanks, that did it perfectly. I changed it so it would check the other way round using -ne instead of -eq.

Andrek · #7 (**permalink**) 09-11-2006

try "tr"
Do a man on tr

basiclly

# echo "AbCd67Fg8" | tr -d "[:alpha:]"
# 678

So test for null variables

ie

#!/bin/sh
var=AbCd67g8
if [ -z `echo $var | tr -d "[:alpha:]"` ]
then
echo "ONLY Alpha chars here!" # You had only A-z
else
echo "Sorry you have other stuff" # You have soming else besides A-z?
fi

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