CODEWITHSUNDEEP
shell
sateayam
sateayam

Reputation: 1089

Shell script elif

I am new in shell script, trying to catch the return value of a program, and do something with it.

I have this script below

#!/bin/sh


if [ $# !=2 ] ; then
        echo "Usage : param1 param2 "
        exit 1;
elif [ $# -eq 2 ]; then
        ./callprogram
        $out = $?
        echo "$out"
fi

if [ $out==0 ]; then
    echo "out ok"
fi

It keeps getting me error of 

"[: 11: 0: unexpected operator

out ok

I have no clue why line 11 is wrong. if I remove "fi", it will promt that it needs "fi". Can anyone help with this matter?

Thank you

Upvotes: 1

Views: 4206

Answers (3)

Michał Šrajer
Michał Šrajer

Reputation: 31182

Change:

if [ $out==0 ]; then

to:

if [ $out = 0 ]; then

add spaces, and change '==' to '='. Note, that bash, executed as a bash accepts ==. But if you run is as a sh it will say "unexpected operator".

Why:

The [ is a command (or symlink to test binary, depending on your OS and shell). It expects $out and == and 0 and ] to be separate command arguments. If you miss the space around them, you have one argument $out==0.

BTW:

It's safer to always enquote the variables like that:

if [ "$var" ......

instead of 

if [ $var

because when variable is empty, then you can get another error because of wrong number of arguments (no argument instead of empty string).

Upvotes: 2

William Pursell
William Pursell

Reputation: 212464

You have several problems. The one that is giving you the error is that you need a space after != on

if [ $# != 2 ]

(although -ne would be better than !=). It appears that you are calling the script with 11 arguments, and then calling [ with the arguments 11 !=2, and it does not know what to do with !=2 because you meant != 2 but forgot the space. Also, you want 

out=$?

on the assignment (no $ on the LHS) and

if [ $out = 0 ]

on the comparison. (Spaces around the operator, which is '=' instead of '=='. '==' will work on many shells, but '=' works in more shells.)

But your script would be better written without the explicit reference to $?

#!/bin/sh

if test $# != 2; then
        echo "Usage: $0 param1 param2 " >&2  # Errors go to stderr, not stdout
        exit 1;
fi
# you know $# is 2 here. No need to check

if ./callprogram; then
        echo "out ok"
fi

Upvotes: 1

dogbane
dogbane

Reputation: 274758

You need a space after the [ and you need to use -eq (equals) or -ne (not equals) to compare numbers in your if-statement.

To assign a variable use out=$?, not $out = $?. There should be no spaces on either side of the = sign.

Try this:

if [ $# -ne 2 ] ; then
        echo "Usage : param1 param2 "
        exit 1
elif [ $# -eq 2 ]; then
        ./callprogram
        out=$?
        echo "$out"
fi

if [ $out -eq 0 ]; then
    echo "out ok"
fi

Upvotes: 3

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