CODEWITHSUNDEEP
linuxbashshell
steviekm3
steviekm3

Reputation: 1053

How to have bash shell not execute rest of semicolon list of commands?

Suppose you do following in bash at command prompt:

cmd1;cmd2;cmd3

If cmd1 fails how do you get bash not to do cmd2.

Upvotes: 2

Views: 671

Answers (1)

jsp
jsp

Reputation: 1283

cmd1 && cmd2 && cmd3

Explanation

Execute cmd1. If it fails, cmd2 and cmd3 will not be executed.

Why? Because false logically ANDed with anything else is always equal to false, so if cmd1 returns false there is no need to evaluate cmd2 and cmd3. And by like reasoning, if cmd1 succeeds and cmd2 fails, don't execute cmd3.

Note

Just to make things a little more confusing, POSIX systems (like Linux and other UNIX variants) return 0 for success and non-zero for failure.

So, when I say failure above

  • false = non-zero = failure

  • true = zero = success

Why? Because the numerical value of the return code is used indicate different failure codes.

For example,

$ ls /root
ls: cannot open directory /root: Permission denied
$ echo $? 
2
$ asdf
asdf: command not found...
$ echo $?
127
$ ls /
bin  boot  data  dev  etc  home  lib  ...
$ echo $?
0

ls returns "1" for minor problems and "2" for more serious problems. The bash shell returns "127" to indicate "command not found", and ls / returns "0" to indicate success.

Upvotes: 9

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