syntax error near unexpected token 'elif'

Question

I get the error on the 7th line. Any ideas? I checked for spaces and there are none.

#!/bin/bash
if test $# -eq 0
then
    echo "No arguments"
elif test $# -eq 1
    echo "$1"
elif test $# -eq 2
    echo "$1 $2"
else
    echo "More than 2 arguments"
fi

Are your quotes actually “ and ”? Can you make sure they are " throughout? — doneal24, Feb 22 '19 at 16:52
@KyanHamad: That isn't what Doug was getting at. The quotes you pasted here are curly quotes generally created in programs like microsoft word and will not work in code. You need to ensure you have straight quotes, regardless if it is a double or single quote. — jesse_b, Feb 22 '19 at 17:00
@Jesse_b I was using straight quotes. The issue I found out was that I needed to use then statements after the elif as well as quoting the $#. Thanks for clarifying! — Kyan Hamad, Feb 22 '19 at 17:10

Jeff Schaller · Accepted Answer · 2019-02-22T17:12:34.090

3

The syntax for if/elif/else/fi requires a then after each "elif":

#!/bin/bash
if test "$#" -eq 0
then
    printf 'No arguments\n'
elif test "$#" -eq 1
then
    printf '%s\n' "$1"
elif test "$#" -eq 2
then
    printf '%s %s\n' "$1" "$2"
else
    printf 'More than 2 arguments\n'
fi

I've also replaced your echos with printfs and quoted the instances of $#.

edited Feb 22 '19 at 17:12

answered Feb 22 '19 at 16:55

Jeff Schaller

67,283
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score 2 · Answer 2 · answered Feb 22 '19 at 16:55

2

You need a then after elif.

#!/bin/bash
if test $# -eq 0
then
    echo "No arguments"
elif test $# -eq 1
then
    echo "$1"
elif test $# -eq 2
then
    echo "$1 $2"
else
    echo "More than 2 arguments"
fi

answered Feb 22 '19 at 16:55

Freddy

25,565

syntax error near unexpected token 'elif'

2 Answers2