Number of arguments in bash script

Question

I want to test the number of arguments passed to a Linux shell script. If the number of arguments is not 2 or 4, it should print something. Unfortunately it does not work. Can anyone explain what I am doing wrong?

#!/bin/bash
if [[ $# -ne 2 ]] || [[ $# -ne 4 ]];
then
    echo "here";
fi

possible duplicate of checking number of arguments bash script — Maximin
– Maximin, Commented Dec 6, 2014 at 18:42

Gilles Quénot · Accepted Answer · 2014-12-06 19:01:42Z

4

You should replace logical OR by logical AND, so :

#!/bin/bash

if [[ $# -ne 2 && $# -ne 4 ]]; then
   echo "here"
fi

In arithmetic form:

#!/bin/bash

if (($# != 2 && $# != 4)); then
   echo "here"
fi

As you can see, no need to use 2 [[ ]]

edited Dec 6, 2014 at 19:01

answered Dec 6, 2014 at 18:46

Gilles Quénot

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Comments

user3130569 · Accepted Answer · 2014-12-06 18:39:36Z

1

Logic.

if [[ $# -ne 2 ]] && [[ $# -ne 4 ]]; then
  echo "here"
fi

answered Dec 6, 2014 at 18:39

user3130569

2 Comments

user3130569 Over a year ago

It's -ne (not equal)

user3130569 Over a year ago

Also, the code posted by OP also has a problem: a value is always not 2 or not 4

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