Integer expression expected error in shell script

Question

I\'m a newbie to shell scripts so I have a question. What Im doing wrong in this code?

#!/bin/bash
echo \" Write in your age: \"
read age
if [ \"$age\" -le \         


        

Answer 1

This error can also happen if the variable you are comparing has hidden characters that are not numbers/digits.

For example, if you are retrieving an integer from a third-party script, you must ensure that the returned string does not contain hidden characters, like "\n" or "\r".

For example:

#!/bin/bash

# Simulate an invalid number string returned
# from a script, which is "1234\n"
a='1234
'

if [ "$a" -gt 1233 ] ; then
    echo "number is bigger"
else
    echo "number is smaller"
fi

This will result in a script error : integer expression expected because $a contains a non-digit newline character "\n". You have to remove this character using the instructions here: How to remove carriage return from a string in Bash

So use something like this:

#!/bin/bash

# Simulate an invalid number string returned
# from a script, which is "1234\n"
a='1234
'

# Remove all new line, carriage return, tab characters
# from the string, to allow integer comparison
a="${a//[$'\t\r\n ']}"

if [ "$a" -gt 1233 ] ; then
    echo "number is bigger"
else
    echo "number is smaller"
fi

You can also use set -xv to debug your bash script and reveal these hidden characters. See https://www.linuxquestions.org/questions/linux-newbie-8/bash-script-error-integer-expression-expected-934465/