My nestedif on unix shell scripting using sh has a syntax error and i can't seem to fix it

Question

I'm coding a nestedif program as shown below but there is a syntax error that says "Syntax error: "then" unexpected (expecting "fi")

#!bin/sh

choose=0
echo "Do you want to choose a color?"

read choose

if [ $choose == "Y" ] ; then

echo "1. Blue"

echo "2. Green"

echo -n "Select your choice [1 or 2]? "

read choice

 if [ $choice == "1" ] ; then

        echo "you chose blue"
 else [ $choice == "2" ] ; then

        echo "you chose green"
 fi

else     

 echo " invalid choice"

fi

I expect it to be if i choose 1 the output will be blue then if i choose 2 the output will be green else invalid choice

Whenever you have a shell script error, a good first step is to cut and paste your code into shellcheck.net and correct the errors (important) and warnings (might be important) that it identifies. If you have trouble understanding its messages, then come here and ask. — John1024
– John1024, Commented May 9, 2019 at 1:40
Some suggestions. (1) Recognize "y" as well as "Y", and have the prompt specify what input is expected. (2) Put double quotes around your variable references to avoid errors when the input is empty: if [ "$choose" == "Y" ] ; then .... (3) Indent your code consistently. — Keith Thompson
– Keith Thompson, Commented May 9, 2019 at 2:13
When the user has more choices, consider case "{choice}" in ... esac. — Walter A
– Walter A, Commented May 9, 2019 at 8:17
"Invalid choice" is an error message, and should be written to stderr. echo 'invalid choice' >&2 — William Pursell
– William Pursell, Commented May 17, 2019 at 20:54

John1024 · Accepted Answer · 2019-05-17 22:41:33Z

2

Replace:

else [ $choice == "2" ] ; then

with:

elif [ "$choice" = "2" ] ; then

Notes:

else is used for the final clause, one without conditions, in an if statement. Use elif if you want to apply a test, such as, in this case, [ "$choice" = "2" ].
Always put shell variables in double-quotes unless you explicitly want the shell to apply expansions such as word splitting and pathname expansion.
== inside [...] is not POSIX. Bash accepts it but other shells, such as dash, don't. For greater portability, use =.

edited May 17, 2019 at 22:41

answered May 9, 2019 at 1:42

John1024

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6 Comments

qaqu Over a year ago

Worked but it said another error which is " ./try.sh: 6: [: y: unexpected operator

John1024 Over a year ago

Did you try using shellcheck.net? It will save you a lot of time.

Keith Thompson Over a year ago

@qaqu: When I copy-and-paste the script from your question and make this change, it works. Also, line 6 is read choose; there's no "y" on that line. You must have been using a script that differs from the one you posted.

William Pursell Over a year ago

Better yet, use else [ "$choice" = "2" ]; . == is a non-standard operator

John1024 Over a year ago

@WilliamPursell Good observation. Answer updated to mention that == inside [...] is not supported by the POSIX standard.

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