How do I start a command from terminal so that terminal is not the parent?

Question

Let's take example of a command "example-command".

1. I open terminal
2. I write example-command in terminal, and example-command executes.
3. Now if I close terminal, example-command gets killed too.
4. I now try with "example-command &", but the same behaviour.

How do I execute a command so that when I close the terminal, the command doesn't get terminated?

Answer 1

example-command &; disown {pid}

or just

disown

Answer 2

There are two ways, identical in result.

1. Use nohup when you start your program. E.g., nohup example-command. You can background and work with it normally; it will simply continue running after you've quit.
2. Alternatively, as @alamar noted, if you use bash as your shell, you can us the disown command. Unfortunately, as far as I know, disown is bash-specific; if you use another shell, such tcsh, you may be restricted to the nohup form above.

Answer 3

nohup example-command

Answer 4

You can also use the 'at' or 'batch' commands and give it the current time.

Answer 5

disown is a bash builtin. You could create a wrapper shellscript for your command such as

#!/bin/bash
$1 &
P=`which $1`
disown `pidof ${P}`

Not the most robust script (by any means) but may help get you going. For example:

$./launch_script.sh myProgram

You can also do this in the source of the program if you are editing it.

Answer 6

You could also consider using the screen command.

Answer 7

Hello,

Please search for similar questions first.

Besides the ways listed above, you can do:

setsid command_name

For example:

setsid xclock

Thanks

Answer 8

Run: example-command

Press: Control-Z

Run: bg