Run another command when previous command completes [duplicate]

Solution 1:

&& means to run it if the previous command was successful. In Unix that generally means exit status 0.

$ ls barfoofba && echo "This will not be echo'd"
ls: cannot access 'barfoofba': No such file or directory
$ ls bar && echo "This will be echo'd"  
This will be echo'd

In the first instance, ls did not succeed, so it exited with a non-zero exit status, and bash did not run the second command.

In the second example, I ls'd a existing file, so ls exited with 0 as exit status, and the command was executed.

If you want to run commands unconditionally, e.g. not dependent on the result of the first, you may separate them with ; like this

 command1; command2; command3 

and so forth.

Thus you may do

ls -lR ; sleep 4 ; gnome-terminal

In addition to && and ; you have || which is the opposite of &&: Only run the command if the previous command failed with a non-zero exit status. If the first command succeeds, the next will not be executed. If it fails, the next will be executed.

So in short:

• &&: Run if preceding command exited with 0
• ;: Run unconditionally
• ||: Run if preceding command exited with a non-zero exit status.
• &: Run both commands in paralell, the first in background and second in foreground.

Solution 2:

The command ls -lR exited with an exit-status different than zero, so the following commands have not been executed. Most probably ls was unable to open a subdirectory. It didn't happen in your first command because you didn't use the -R-option.

From man bash:

        command1 && command2
       command2  is  executed if, and only if, command1 returns an exit status
       of zero.

From man ls:

Exit status:
       0      if OK,
       1      if minor problems (e.g., cannot access subdirectory)
       2      if serious trouble (e.g., cannot access command-line argument).