Shell Programming and Scripting

when i run a select query in database, it will return the objid based on the condtion. If there is no objid tht meets the condition, I will not get any value returned and the variable will hold nothing... so now how to chk tht the variable is blank or not. I tried the below piece of code:

if [ -z $objid ]; then
echo "No objid returned"
else
<do this>
.
.
.
fi

but the above code throws the following error:
test: argument expected.

Or else please let me know how to check whether the variable holds the numerical or not, this will also serve my purpose.

Please help


Thanks,
Harish

That is supposed to work.

Try this.

Replace

Code:

if [ -z $objid ]; then

with

Code:

if [ x${objid} = x ] ; then

Another way :

Code:

if [ -z "$objid" ]; then

Jean-Pierre.

Not sure about your shell, but...

The form "[ ]" is essentially calling the external "test" command, and thus everything is evaluated by the shell first. That's why you get "test: argument expected" when $objid is NULL, since the expression evaluates to "[ -z ]".

When using "[ ]", you really need to quote all values/variables. But, you can use the super-secret double bracket:

if [[ -z $myvar ]]; then.....

WIth the double-brackets, the test is a shell builtin. So nothing needs to be quoted, since it is not evaluated prior to being tested. Plus, you will save dozens of milliseconds by avoiding an external program.

The [[ is a korn shell special.

Oh. Isn't everybody using ksh?

It does work with other ksh-ish shells (zsh, bash, etc.)

Thanks all... its working!!!!!

The quotes/secret backets solved my issue... Thanks gus2000 for explaining the concept