Check if string is neither empty nor space in shell script

bash, freebsd, shell

Solution

You need a space on either side of the `!=`. Change your code to:

str="Hello World"
str2=" "
str3=""

if [ ! -z "$str" -a "$str" != " " ]; then
        echo "Str is not null or space"
fi

if [ ! -z "$str2" -a "$str2" != " " ]; then
        echo "Str2 is not null or space"
fi

if [ ! -z "$str3" -a "$str3" != " " ]; then
        echo "Str3 is not null or space"
fi

Problem

I am trying to run the following shell script which is supposed to check if a string is neither space nor empty. However, I am getting the same output for all the 3 mentioned strings. I have tried using the "[[" syntax as well but to no avail. Here is my code: ``` str="Hello World" str2=" " str3="" if [ ! -z "$str" -a "$str"!=" " ]; then echo "Str is not null or space" fi if [ ! -z "$str2" -a "$str2"!=" " ]; then echo "Str2 is not null or space" fi if [ ! -z "$str3" -a "$str3"!=" " ]; then echo "Str3 is not null or space" fi ``` I am getting the following output: ``` # ./checkCond.sh Str is not null or space Str2 is not null or space ```

Original source