Bash script error: integer expression expected

bashcommand-substitutionshell-script

I have a problem with a bash script on raspberry pi:

x='gpio -g read 22'

if [ $x -ge 1 ]
then
gpio -g write 23 1
fi

The error is integer expression expected. Why?

Best Answer

That's because you are checking whether the string gpio -g read 22 is greater than 1. Since gpio -g read 22 is not a number, you get that error.

You don't explain what you are trying to do but I'm guessing you want to compare the output of the gpio command. To do that, you need to enclose the command in $() or backticks (``):

x=$(gpio -g read 22)

if [ "$x" -ge 1 ]
then
   gpio -g write 23 1
fi

Or, more simply:

[ "$(gpio -g read 22)" -ge 1 ] && gpio -g write 23 1

The assignment foo='command' doesn't run command. The variable foo takes the value of the string command and not its output.

Related Solutions

Bash: integer expression expected, using read/test

Actually you can set some attributes on variables using the declare (or the old typeset) builtin. declare -i var1 var2 will set integer attribute on those variables. After that assignments which attempt to set non-integer values to those variables will raise error.

But your problem is with the syntax. When using a variable's value you have to prefix its name with $:

if [ "$var1" -lt "$var2" ]; then
    echo "$var1 is lt $var2"
else
    echo "$var2 is lt $var1"
fi

Exceptions are the arithmetic evaluations, where is no need for the $:

if ((var1<var2)); then
    echo "$var1 is lt $var2"
else
    echo "$var2 is lt $var1"
fi

As a word of warning, inside [..] always double quote your variables to avoid word expansion messing up your expression's syntax. (I mean, you will have problems with unset variables, variables containing empty string and variables containing IFS characters.) Or you can use the newer and better [[..]] instead which handles such cases correctly:

if [[ $var1 -lt $var2 ]]; then
    echo "$var1 is lt $var2"
else
    echo "$var2 is lt $var1"
fi

Bash – integer expression expected bash

You need to use a $ to expand a variable in bash, unless it's inside of (( )). You are comparing the literal string "num" to the number 0. Use one of the following:

while [ "$num" -ge 0 ] # POSIX
while (( num >= 0 )); # bash

Best Answer

Related Solutions

Bash: integer expression expected, using read/test

Bash – integer expression expected bash

Related Question