multiple if condition throws error in shell script

Question

I'm using below if condition in shell script

   if [! -z "$failedTC"] || [-n "$failedTC"];
   then
   echo "ITS not empaty AND NOT NULL"
   else
   echo "ITS empty or NULL"
   fi

The above code I'm checking the $failedTC shouldn't be null and empty. But the if condition throws below error message in jenkins. I used the above script in jenkins Execute Shell window.

/tmp/jenkins371454092166709762.sh: line 23: [!: command not found /tmp/jenkins371454092166709762.sh: line 23: [-n: command not found

Any leads....

Add a shebang and then paste your script there: http://www.shellcheck.net/ — Cyrus, Nov 01 '18 at 18:50

EJK · Accepted Answer · 2018-11-01T18:23:21.363

4

You need to add a space after the opening brackets and before the closing brackets in your conditionals:

Change

if [! -z "$failedTC"] || [-n "$failedTC"];

to

if [ ! -z "$failedTC" ] || [ -n "$failedTC" ];

edited Nov 01 '18 at 18:23

answered Nov 01 '18 at 18:15

EJK

12,090
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score 2 · Answer 2 · answered Nov 01 '18 at 18:16

2

The error is that, it needs a space before and after every square bracket[].

Try this:

if [ ! -z "$failedTC" ] || [ -n "$failedTC" ];
   then
   echo "ITS not empaty AND NOT NULL"
   else
   echo "ITS empty or NULL"
fi

answered Nov 01 '18 at 18:16

Mayank Porwal

31,737
7
30
50

score 2 · Answer 3 · answered Nov 01 '18 at 18:17

2

"[" and "]" are commands, it should be wrapped around spaces " [ ". that is the best practice. try this

  if [ ! -z "$failedTC" ] || [ -n "$failedTC" ];
   then
   echo "ITS not empaty AND NOT NULL"
   else
   echo "ITS empty or NULL"
   fi

answered Nov 01 '18 at 18:17

stack0114106

8,220
3
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multiple if condition throws error in shell script

3 Answers3