Bash: integer expression expected

bashscriptingshell-script

I have a file out.csv I have to check if the name inputed by user exists in the file(comma separated) or not. I am taking name using read but while checking for equality I am getting error

    IFS=","
    while read tname tnum
        do
            if [ $tname -eq $name ]; then
                flag=1
                break
            fi
        done < out.csv
    echo "$ch"

Best Answer

You're getting this error since you are trying to compare string using equality operators intended for integers, -eq, -ne, -gt, and similar are integer functions.

To compare strings use = to compare for equality OR != to compare for non-equality.

Check this for more on comparison operators.

if [ $tname -eq $name ]; then

should be changed to:

if [ "$tname" = "$name" ]; then

(also remember to quote your variables).

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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

Bash script error: integer expression expected

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.

Best Answer

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