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check if a string is numeric

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# 1  
Old 11-10-2011
check if a string is numeric
I checked all the previous threads related to this and tried this.
My input is all numbers or decimals greater than zero everytime.
I want to check the same in the korn shell script.
Just validate the string to be numeric.
This is what I am doing.

Code:
 
var="12345"
if  [[ "$var" !=  +([0-9]) -o  "$var"  != +([0-9]).+([0-9]) ]] ; then
echo "not numric" 
else
echo "it is numeric"
fi

I get a syntax error -o unexpected.
-o is used as a logical or operator right ?
Please let me know if am doing something wrong.
# 2  
Old 11-10-2011
|| is a boolean or
# 3  
Old 11-10-2011
How about this thread?
# 4  
Old 11-10-2011
Code:
$ var=123A24
$ echo $var | grep -q "^[0-9]*$" && echo "OK" || echo "Not OK"
Not OK
$ var=12345
$ echo $var | grep -q "^[0-9]*$" && echo "OK" || echo "Not OK"
OK

# 5  
Old 11-10-2011
Since you have the Korn shell you can do this:

Code:
[ ! -z "$VAR" -a -z "${VAR/[0-9]*/}" ] && echo "$VAR is all-numeric"

# 6  
Old 11-10-2011
Quote:
Originally Posted by Corona688
Since you have the Korn shell you can do this:

Code:
[ ! -z "$VAR" -a -z "${VAR/[0-9]*/}" ] && echo "$VAR is all-numeric"

I think the OP wanted to cover the floats as well as the ints.
# 7  
Old 11-10-2011
The problem with your expression is that the number may contain a + and/or a decimal point or it may not but you arent making that clear to ksh so how about a simpler regular expression like...
Code:
var="12345"

if  [[ "$var" = *([+-])*([0-9])*(.)*([0-9]) ]]; then
    echo "it is numeric"
else
    echo "not numeric" 
fi

This User Gave Thanks to shamrock For This Post:
megha2525
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