Hiding internal work outputs by shell script

Question

I make a shell script that can update a tools. I want that when i run the

root@host:# sh script.sh

then the output will be

Its installing
Its done

But now my script look like

root@host:# sh script.sh
Its installing
Showing internat script working outputs
Its done

As i know stty -echo command hide stdin in terminal and stty echo enable stdin show in terminal. But how can i hide the stander output by shell script ?

Answer 1

Redirect both stdout and stderr to /dev/null:

> /dev/null 2>&1

OR in bash:

&> /dev/null

You can do it for all programs spawned by your script by using exec with the redirection at the beginning of your script.

exec > /dev/null 2>&1

Unless the programs you invoke in your script access the terminal directly (rare), this should cover you.

One note on redirections: The order matters. It needs to be > /dev/null 2>&1, NOT 2>&1 >/dev/null. I used to think the latter would work because > looks like an arrow which makes me think of pointers (as does the word "redirect"), and if I point stderr to stdout and then point stdout to /dev/null, then both should be pointing to /dev/null. That is not the case, though. File descriptors are not pointers and it is more helpful to think of > as sort of an assignment to a file descriptor rather than pointing. (Sort of, because technically, a fd is just a number and you need system functions like dup2 to open a different file into the same file descriptor; but I think assignment is a good high-level abstraction).