Shell Programming and Scripting

I am posting this info only after having confirmed the fact that there are no prior posts avialable that serves my purpose .

In my Korn Shell script file , there are two shell variables which needs to be tested within a single if statement .

Sample variables :
$var1 and $var2

The check should be :
if $var1 is NULL(empty in Unix) OR $var2 == "Y" ,
then
do logic 1
else
do logic 2
fi

The way I wrote the if-else block is ;

Code:

 
if [ -z $var1 -o  $var2 == "Y" ]; then
     do logic 1
else
    do logic 2
fi

The problems I am facing are :
1. Even if for cases where $var1 is empty , logic 2 is executed
2. The script runs but with an error message like "argument expected"

I tested the two conditions individually , and they executed as expected , but when they are commbined with and -o operator , they yield improper result.

Can please somoeone help me ?

Thanks
Kumarjit Ghosh.

Change your condition to

Code:

if [ -z "$var1" -o "$var2" == "Y" ];

[ ] test needs = not ==

Code:

if [ -z "$var1" -o "$var2" = "Y" ]
then

Thanks a ton to all you guys , your help is truly appreciated .

But what difference does the double quotes around the shell variable make ?

Why without the double quotes the IF-ELSE conditions is not matching , but using them works wonders ?

Can you please guide me on the significance of double quoting shell variables and the way Unix interprets them while comparing?

All I know about double quote usage is that it is tolerant to variable expransions while single quote considers literal values of the constituent string.

Thanks
Kumarjit.

"$var1" is robust against

Code:

var1="two words"

and

Code:

var1="*"

Didnt understand ....
Request you to be more explicit .


Thanks in advance
Kumarjit

Code:

var1="two words"
if [ -z $var1 -o $var2 = "Y" ]; then
 echo "true"
else
 echo "false"
fi
bash: [: too many arguments
false
if [ -z "$var1" -o "$var2" = "Y" ]; then
 echo "true"
else
 echo "false"
fi
false