I don't set any values for $pass_tc11; so it is returning null while echoing. How to compare it in if clause?
Here is my code. I don't want "Hi" to be printed...
-bash-3.00$ echo $pass_tc11
-bash-3.00$ if [ "pass_tc11" != "" ]; then
> echo "hi"
> fi
hi
-bash-3.00$
First of all, note you are not using the variable correctly:
if [ "pass_tc11" != "" ]; then
# ^
# missing $
Anyway, to check if a variable is empty or not you can use -z --> the string is empty:
if [ ! -z "$pass_tc11" ]; then
echo "hi, I am not empty"
fi
or -n --> the length is non-zero:
if [ -n "$pass_tc11" ]; then
echo "hi, I am not empty"
fi
From man test:
-z STRING
the length of STRING is zero
-n STRING
the length of STRING is nonzero
$ [ ! -z "$var" ] && echo "yes"
$
$ var=""
$ [ ! -z "$var" ] && echo "yes"
$
$ var="a"
$ [ ! -z "$var" ] && echo "yes"
yes
$ var="a"
$ [ -n "$var" ] && echo "yes"
yes
fedorqui has a working solution but there is another way to do the same thing.
Chock if a variable is set
#!/bin/bash
amIEmpty='Hello'
# This will be true if the variable has a value
if [ $amIEmpty ]; then
echo 'No, I am not!';
fi
Or to verify that a variable is empty
#!/bin/bash
amIEmpty=''
# This will be true if the variable is empty
if [ ! $amIEmpty ]; then
echo 'Yes I am!';
fi
tldp.org has good documentation about if in bash:
http://tldp.org/LDP/Bash-Beginners-Guide/html/sect_07_01.html
Source: Stackoverflow.com