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Top Forums Shell Programming and Scripting if -z not working in SH shell Post 302144389 by rijeshpp on Wednesday 7th of November 2007 11:26:20 PM
Old 11-08-2007
if -z not working in SH shell

Hello all,

Any idea how to check whether a variable holding null value or not. if -z option works fine in bash, where as it is not working in sh.

bash-3.00$ sh
$ TEST=
$ if [ -z $TEST ] ; then
> echo "Null"
> else
> echo "Not null"
> fi
sh: test: 0403-004 Specify a parameter with this command.
Not null
$

The test fails. The same works in bash.

bash-3.00$ TEST=
bash-3.00$ if [ -z $TEST ] ; then
> echo "Null"
> else
> echo "Not null"
> fi
Null
bash-3.00$

I need this testing working in sh. Please help.

Thanks,
Rijesh.
 

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DEFINED(3)								 1								DEFINED(3)

defined - Checks whether a given named constant exists

SYNOPSIS
bool defined (string $name) DESCRIPTION
Checks whether the given constant exists and is defined. Note If you want to see if a variable exists, use isset(3) as defined(3) only applies to constants. If you want to see if a function exists, use function_exists(3). PARAMETERS
o $name - The constant name. RETURN VALUES
Returns TRUE if the named constant given by $name has been defined, FALSE otherwise. EXAMPLES
Example #1 Checking Constants <?php /* Note the use of quotes, this is important. This example is checking * if the string 'TEST' is the name of a constant named TEST */ if (defined('TEST')) { echo TEST; } ?> SEE ALSO
define(3), constant(3), get_defined_constants(3), function_exists(3), The section on Constants. PHP Documentation Group DEFINED(3)
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