executing shell command in background from script

问题:

how can I execute a shell command in the background from within a bash script, if the command is in a string?

For example:

#!/bin/bash
cmd="nohup mycommand";
other_cmd="nohup othercommand";

"$cmd &";
"$othercmd &";

this does not work -- how can I do this?

回答1:

Leave off the quotes

$cmd &
$othercmd &

eg:

nicholas@nick-win7 /tmp
$ cat test
#!/bin/bash

cmd="ls -la"

$cmd &


nicholas@nick-win7 /tmp
$ ./test

nicholas@nick-win7 /tmp
$ total 6
drwxrwxrwt+ 1 nicholas root    0 2010-09-10 20:44 .
drwxr-xr-x+ 1 nicholas root 4096 2010-09-10 14:40 ..
-rwxrwxrwx  1 nicholas None   35 2010-09-10 20:44 test
-rwxr-xr-x  1 nicholas None   41 2010-09-10 20:43 test~

回答2:

Building off of ngoozeff's answer, if you want to make a command run completely in the background (i.e., if you want to hide its output and prevent it from being killed when you close its Terminal window), you can do this instead:

cmd="google-chrome";
"${cmd}" &>/dev/null &disown;

& (the first one) detaches the command from stdin.
>/dev/null detaches the shell session from stdout and stderr.
&disown removes the command from the shell's job list.

You can also use &! instead of &disown; they're both the same command.

Also, when putting a command inside of a variable, it's more proper to use eval "${cmd}" rather than "${cmd}":

cmd="google-chrome";
eval "${cmd}" &>/dev/null &disown;

If you run this command directly in Terminal, it will show the PID of the process which the command starts. But inside of a shell script, no output will be shown.

Here's a function for it:

#!/bin/bash

# Run a command in the background.
_evalBg() {
    eval "$@" &>/dev/null &disown;
}

cmd="google-chrome";
_evalBg "${cmd}";

http://felixmilea.com/2014/12/running-bash-commands-background-properly/

回答3:

This works because the it's a static variable. You could do something much cooler like this:

filename="filename"
extension="txt"
for i in {1..20}; do
    eval "filename${i}=${filename}${i}.${extension}"
    touch filename${i}
    echo "this rox" > filename${i}
done

This code will create 20 files and dynamically set 20 variables. Of course you could use an array, but I'm just showing you the feature :). Note that you can use the variables $filename1, $filename2, $filename3... because they were created with evaluate command. In this case I'm just creating files, but you could use to create dynamically arguments to the commands, and then execute in background.

回答4:

For example you have a start program named run.sh to start it working at background do the following command line. ./run.sh &>/dev/null &