Using if elif fi in shell scripts

Question

I'm not sure how to do an if with multiple tests in shell. I'm having trouble writing this script:

echo "You have provided the following arguments $arg1 $arg2 $arg3"
if [ "$arg1" = "$arg2" && "$arg1" != "$arg3" ]
then
    echo "Two of the provided args are equal."
    exit 3
elif [ $arg1 = $arg2 && $arg1 = $arg3 ]
then
    echo "All of the specified args are equal"
    exit 0
else
    echo "All of the specified args are different"
    exit 4
fi

The problem is I get this error every time:

./compare.sh: [: missing `]' command not found

Quite a few commentators have suggested you use [[ rather than [ but that makes your script bash-specific. You'll have fewer maintenance and portability issues if you can stick to plain old Bourne shell (sh) syntax. — Adrian Pronk, Mar 01 '10 at 22:01

score 47 · Answer 1 · edited Apr 27 '17 at 11:57

Josh Lee's answer works, but you can use the "&&" operator for better readability like this:

echo "You have provided the following arguments $arg1 $arg2 $arg3"
if [ "$arg1" = "$arg2" ] && [ "$arg1" != "$arg3" ]
then 
    echo "Two of the provided args are equal."
    exit 3
elif [ $arg1 = $arg2 ] && [ $arg1 = $arg3 ]
then
    echo "All of the specified args are equal"
    exit 0
else
    echo "All of the specified args are different"
    exit 4 
fi

score 40 · Accepted Answer · answered Mar 01 '10 at 21:46

40

sh is interpreting the && as a shell operator. Change it to -a, that’s [’s conjunction operator:

[ "$arg1" = "$arg2" -a "$arg1" != "$arg3" ]

Also, you should always quote the variables, because [ gets confused when you leave off arguments.

answered Mar 01 '10 at 21:46

Josh Lee

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score 8 · Answer 3 · edited Apr 27 '17 at 11:58

8

I have a sample from your code. Try this:

echo "*Select Option:*"
echo "1 - script1"
echo "2 - script2"
echo "3 - script3 "
read option
echo "You have selected" $option"."
if [ $option="1" ]
then
    echo "1"
elif [ $option="2" ]
then
    echo "2"
    exit 0
elif [ $option="3" ]
then
    echo "3"
    exit 0
else
    echo "Please try again from given options only."
fi

This should work. :)

edited Apr 27 '17 at 11:58

Falko

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answered Jan 29 '13 at 07:49

Gaurav Kolarkar_InfoCepts

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Gawd this looks quirky to me. Why no `then` after `else`? It may actually not be quirky, but that's how it looks, compared to C. – Rolf Nov 10 '17 at 22:01
space is required after $option and before = to run it fine, else every time 1 – PiyusG Feb 09 '18 at 07:46

score 8 · Answer 4 · answered Mar 01 '10 at 21:46

8

Use double brackets...

if [[ expression ]]

answered Mar 01 '10 at 21:46

Henryk Konsek

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2

To note, this is a solution because the `[[` construct is built into the shell while `[` is another name for the `test` command and hence is subject to its syntax -- see `man test` – glenn jackman Mar 01 '10 at 23:54
5

Technically, `[` is a shell builtin, but `[[` is a shell keyword. That’s the difference. – Josh Lee Mar 02 '10 at 06:45
1

So see the results of the bash commands `type [`, `type [[`, `help [` and `help [[`. – Apostle Feb 10 '14 at 14:31
@JoshLee no, there is a file called `/usr/bin/[` – LtWorf Mar 03 '17 at 10:14

score 5 · Answer 5 · edited Jun 16 '20 at 03:48

5

Change [ to [[, and ] to ]].

edited Jun 16 '20 at 03:48

NAND

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answered Mar 01 '10 at 21:45

Ethan Post

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Better yet, change '[' to 'test' – William Pursell Mar 02 '10 at 14:11

score 1 · Answer 6 · edited Jun 16 '20 at 12:40

This is working for me,

 # cat checking.sh
 #!/bin/bash
 echo "You have provided the following arguments $arg1 $arg2 $arg3"
 if [ "$arg1" = "$arg2" ] && [ "$arg1" != "$arg3" ]
 then
     echo "Two of the provided args are equal."
     exit 3
 elif [ $arg1 == $arg2 ] && [ $arg1 = $arg3 ]
 then
     echo "All of the specified args are equal"
     exit 0
 else
     echo "All of the specified args are different"
     exit 4
 fi

 # ./checking.sh
 You have provided the following arguments
 All of the specified args are equal

You can add set -x in script to troubleshoot the errors.

Using if elif fi in shell scripts

6 Answers6

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