Integer expression expected error in shell script - bash

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 [ "$age" -le "7"] -o [ "$age" -ge " 65" ]
then
echo " You can walk in for free "
elif [ "$age" -gt "7"] -a [ "$age" -lt "65"]
then
echo " You have to pay for ticket "
fi
When I'm trying to open this script it asks me for my age and then it says
./bilet.sh: line 6: [: 7]: integer expression expected
./bilet.sh: line 9: [: missing `]'
I don't have any idea what I'm doing wrong. If someone could tell me how to fix it I would be thankful, sorry for my poor English I hope you guys can understand me.

Answer 1

You can use this syntax:
#!/bin/bash
echo " Write in your age: "
read age
if [[ "$age" -le 7 || "$age" -ge 65 ]] ; then
echo " You can walk in for free "
elif [[ "$age" -gt 7 && "$age" -lt 65 ]] ; then
echo " You have to pay for ticket "
fi

Answer 2

If you are using -o (or -a), it needs to be inside the brackets of the test command:
if [ "$age" -le "7" -o "$age" -ge " 65" ]
However, their use is deprecated, and you should use separate test commands joined by || (or &&) instead:
if [ "$age" -le "7" ] || [ "$age" -ge " 65" ]
Make sure the closing brackets are preceded with whitespace, as they are technically arguments to [, not simply syntax.
In bash and some other shells, you can use the superior [[ expression as shown in kamituel's answer. The above will work in any POSIX-compliant shell.

Answer 3

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/

Answer 4

./bilet.sh: line 6: [: 7]: integer expression expected
Be careful with " "
./bilet.sh: line 9: [: missing `]'
This is because you need to have space between brackets like:
if [ "$age" -le 7 ] -o [ "$age" -ge 65 ]
look: added space, and no " "

Answer 5

Try this:
If [ $a -lt 4 ] || [ $a -gt 64 ] ; then \n
Something something \n
elif [ $a -gt 4 ] || [ $a -lt 64 ] ; then \n
Something something \n
else \n
Yes it works for me :) \n

Answer 6

If you are just comparing numbers, I think there's no need to change syntax, just correct those lines, lines 6 and 9 brackets.
Line 6 before: if [ "$age" -le "7"] -o [ "$age" -ge " 65" ]
After: if [ "$age" -le "7" -o "$age" -ge "65" ]
Line 9 before: elif [ "$age" -gt "7"] -a [ "$age" -lt "65"]
After: elif [ "$age" -gt "7" -a "$age" -lt "65" ]