Error executing shell script

Question

I am trying to execute the following shell script

#!/bin/sh
echo "start"
if [ $# != 2 || $1 != "first" || $1 != "second" ]
then
    echo "Error"
fi
echo "done"

and I'm getting the following output: start ./autobuild.sh: line 3: [: missing `]' ./autobuild.sh: line 3: !=: command not found ./autobuild.sh: line 3: !=: command not found done

I have no idea how to resolve the errors. Even if I use -ne instead of !=, I get the same errors. Please help.

***`==`*** doesn't exist in dash if you're running Ubuntu. Use ***`=`*** instead. — Mmmh mmh, Sep 04 '13 at 07:21
I would suggest `!=` is not the actual syntax error : http://pic.dhe.ibm.com/infocenter/aix/v7r1/index.jsp?topic=%2Fcom.ibm.aix.baseadmn%2Fdoc%2Fbaseadmndita%2Fkorn_shell_conditional_exp.htm — Mmmh mmh, Sep 04 '13 at 07:28

score 2 · Accepted Answer · edited May 23 '17 at 11:56

Your syntax is incorrect. If you want multiple conditions in an if statement you need to have multiple [] blocks. Try:

if [ $# != 2 ] || [ $1 != "first" ] || [ $1 != "second" ]

But, it's better to use [[ (if your shell supports it) as it is safer to use. I would go for:

if [[ $# -ne 2 || $1 != "first" || $1 != "second" ]]

See this question on brackets: Is [[ ]] preferable over [ ] in bash scripts?

score 1 · Answer 2 · answered Sep 04 '13 at 07:31

1

While OR ing the conditions should be seperate as follows :

#!/bin/sh
echo "start"
if [ $# != 2]  || [ $1 != "first" ] || [ $1 != "second" ]
then
    echo "Error"
fi
echo "done"

answered Sep 04 '13 at 07:31

Shuhail Kadavath

448
3
13

score 0 · Answer 3 · answered Sep 04 '13 at 07:25

0

Try substituting -NE for !=.

if-usage samples

answered Sep 04 '13 at 07:25

vikingsteve

38,481
23
112
156

score 0 · Answer 4 · answered Sep 04 '13 at 07:29

0

This should work:

#!/bin/sh
echo "start"
if [ $# -ne 2 -o $1 -ne "first" -o $1 -ne "second" ]
then
    echo "Error"
fi
echo "done"

answered Sep 04 '13 at 07:29

Vijay

65,327
90
227
319

Error executing shell script

4 Answers4