Bash script: too many arguments in [ test ]

Question

I have the bash script below:

#!/bin/bash
#
[ $# -eq 1 -a $1 = "--help" -o $# -eq 0 ] && {
  echo Help will come here
}

When I run it:

$ ./script 
./script: line 3: [: too many arguments
$ ./script --help
Help will come here

As you can see, when I don't pass parameters ( $# -eq 0 ) it fails with "too many arguments". So, I tested it directly in terminal:

$ a=1;b=2;c=3
$ [ $a -eq 1 -a $b -eq 2 -o $c -eq 3 ] && echo ok
ok
$ [ $a -eq 0 -a $b -eq 2 -o $c -eq 3 ] && echo ok
ok
$ [ $a -eq 0 -a $b -eq 0 -o $c -eq 3 ] && echo ok
ok
$ [ $a -eq 0 -a $b -eq 0 -o $c -eq 0 ] && echo ok
$ [ $a -eq 0 -a $b -eq 2 -o $c -eq 0 ] && echo ok
$ [ $a -eq 1 -a $b -eq 2 -o $c -eq 0 ] && echo ok
ok

So, if it works perfectly in terminal why doesn't it work passing parameters?

Thanks,

Use more quotes: [ $# -eq 1 -a "$1" = "--help" -o $# -eq 0 ] (see the quotes around the variable expansion "$1"?). — gniourf_gniourf
– gniourf_gniourf, Commented Apr 9, 2017 at 17:57
@gniourf_gniourf: Why not just use $# -eq 0 -o "$*" = "--help"? — l'L'l
– l'L'l, Commented Apr 9, 2017 at 18:49
@l'L'l: my point was only about the quotes. There are better answers below ;). — gniourf_gniourf
– gniourf_gniourf, Commented Apr 9, 2017 at 19:33
@gniourf_gniourf, thank you, you were right about the quotes, I totally forgot them, and I think you were the first to mention it. If you had put as an answer I would've ticked yours as the right one. — Adriano_Pinaffo
– Adriano_Pinaffo, Commented Apr 11, 2017 at 2:39

Fred · Accepted Answer · 2017-04-09 18:07:58Z

3

Express your condition like this :

[ $# -eq 1 ] && [ "$1" = "--help" ] || [ $# -eq 0 ]

Actually, [ is a command, and the following elements in the commands are subject to word splitting. If an argument is empty (or contains whitespace and is unquoted), you can run into surprises. Using -a and -o is deprecated.

Please note that, if you want to use the && logical operator (instead of an if statement) before your echo statement, you will need to enclose the above inside braces, else the operator precedence (coupled with lazy evaluation) may yield incorrect results.

{ [ $# -eq 1 ] && [ "$1" = "--help" ] || [ $# -eq 0 ] ; } && { echo...

If you do not mind using Bash-specific syntax, you could also write :

[[ $# -eq 1 && $1 = "--help" || $# -eq 0 ]]

Note that in this case, double-quoting $1 is not required, because the [[ ]] construct is special shell syntax, not a command, and what is inside is not subject to word splitting. Because there is a single test, you do not need to enclose it inside braces before your && { echo....

edited Apr 9, 2017 at 18:07

answered Apr 9, 2017 at 17:55

Fred

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7 Comments

l'L'l Over a year ago

Re: "[[ ]] construct is special shell syntax, not a command, and what is inside is not subject to word splitting", is perhaps not entirely correct. github.com/koalaman/shellcheck/wiki/SC2048. In shellcheck.net using an unquoted $@ or $* prompts this warning while used inside [[ ]], so this may be an exception.

Fred Over a year ago

@l'L'l From the Bash manual : "Word splitting and filename expansion are not performed on the words between the [[ and ]]"

l'L'l Over a year ago

It's interesting that the warning comes up on shellcheck.net; I wonder if that should be reported.

Fred Over a year ago

Still from the manual, in the section about $@ : "When the expansion occurs within double quotes, each parameter expands to a separate word.". This invaluable behavior, also available for array expansions, can be a little counter-intuitive at first, because double quotes usually enclose a string that will be considered a single word. It should be understood as "expand all arguments, but make sure each argument stays a single word by not submitting it to word splitting".

l'L'l Over a year ago

In shellcheck when trying to use [[ $@ ]] it comes back with ^-- SC2199: Arrays implicitly concatenate in [[ ]]. Use a loop (or explicit * instead of @)., and then while using [[ $* ]] it returns ^-- SC2048: Use "$@" (with quotes) to prevent whitespace problems., so it is a bit confusing.

|

l'L'l · Accepted Answer · 2017-04-09 18:36:22Z

1

Your entire expression can be simplified to:

function help () {
    printf "%s\n" "help is on it's way."
}

[[ $# -eq 0 || "$*" = "--help" ]] && help ; echo "done." && exit 0 ;

This checks if the total sum of arguments is zero, or the argument(s) equals "--help". If either of those two things are true then it proceeds to the help function, otherwise echo "done" and exit.

answered Apr 9, 2017 at 18:36

l'L'l

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3 Comments

Adriano_Pinaffo Over a year ago

This works beautifully @l'L'l. I don't quite understand the difference between [] and [[]]. From what I know [] is a short for the command test.

l'L'l Over a year ago

@Adriano_epifas: In short, [[ is bash's improvement to the [ notation. Here's a detailed comparison: mywiki.wooledge.org/BashFAQ/031. Examples: stackoverflow.com/a/3427931/499581

l'L'l Over a year ago

You're welcome! [[ and [ are actually "test commands" not really "notation" as mentioned... ;_;

VIPIN KUMAR · Accepted Answer · 2017-04-09 18:16:29Z

When you are executing the script without parameter, you are getting the error because your condition is matching with blank character, see below -

$sh -x kk.sh 
+ '[' 0 -eq 1 -a = --help -o 0 -eq 0 ']'
kk.sh: line 3: [: too many arguments

As you can see that there is no value to match.

When you will execute below command in your terminal -

$[ $a -eq 1 -a $b -eq 2 -o $c -eq 3 ] && echo ok
-bash: [: too many arguments
$a=1;b=2;c=3
$[ $a -eq 1 -a $b -eq 2 -o $c -eq 3 ] && echo ok
ok ####it is printing this value bcoz you have set the variable to match, it doesn't matter condition is wrong or right but there is something to match.

To resolve this issue you can use one if condition in the beginning to assign a dummy value if there is no value.

Shubhraj · Accepted Answer · 2017-04-09 18:01:44Z

-1

Try this :-

[ $# -eq 1 ] || [ $1 = "--help" ] || [ $# -eq 0 ]

This way it will automatically display if --help is provided or 1 is typed.

I think $# is creating the problem as both there is $# for first and second conditions

answered Apr 9, 2017 at 18:01

Shubhraj

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4 Comments

Fred Over a year ago

$1 should be quoted, because an empty value or one with whitespace would cause the test to fail due to an incorrect number of arguments after word splitting.

Fred Over a year ago

Reuse of $# in two conditions is not the problem here, it is actually required to get the behavior I assume is expected by the OP.

Fred Over a year ago

The logical operator between the first two tests should be &&.

Adriano_Pinaffo Over a year ago

what @Fred said is right. the problems was that I didn't quote $1. Thank you

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