Running shell script command after executing an application

Question

I have written a shell script to execute a series of commands. One of the commands in the shell script is to launch an application. However, I do not know how to continue running the shell script after I have launched the application.

For example:

...
cp somedir/somefile .
./application
rm -rf somefile

Once I launched the application with "./application" I am no longer able to continue running the "rm -rf somefile" command, but I really need to remove the file from the directory.

Anyone have any ideas how to compete running the "rm -rf" command after launching the application?

Thanks

Answer 1

As pointed out by others, you can background the application (man bash 'job control', eg).

Also, you can use the wait builtin to explicitely await the background jobs later:

 ./application &
 echo doing some more work

 wait # wait for background jobs to complete 
 echo application has finished

You should really read the man pages and bash help for more details, as always:

• http://unixhelp.ed.ac.uk/CGI/man-cgi?sh
• http://www.gnu.org/s/bash/manual/bash.html#Job-Control-Builtins

Answer 2

Start the application in the background, this way the shell is not going to wait for it to terminate and will execute the consequent commands right after starting the application:

./application &

In the meantime, you can check the background jobs by using the jobs command and wait on them via wait and their ID. For example:

$ sleep 100 &
[1] 2098
$ jobs
[1]+  Running                 sleep 100 &
$ wait %1

Answer 3

你需要使用'＆'在后台启动命令，甚至可能是nohup。

nohup ./application > log.out 2>&1 

Answer 4

将已启动的流程置于后台：

./application &