nohup:ignoring input and appending output to 'nohup.out'

Question

I want to start my server through nohup.php but the command is not running and displays following error

nohup:ignoring input and appending output to 'nohup.out'

I am using ssh through putty, this is what i am doing

nohup php server1.php

score 95 · Accepted Answer · edited Jun 12 '16 at 01:03

95

That's not an error - it's normal behavior. It just informs you that once started the in/output is removed from your console.

To avoid the message you need to start it like

nohup php server1.php </dev/null &>/dev/null &

edited Jun 12 '16 at 01:03

jww

answered Jul 10 '14 at 18:06

lpaseen

1

Also see [Difference between 2>&-, 2>/dev/null, |&, &>/dev/null and >/dev/null 2>&1](http://unix.stackexchange.com/q/70963) on Unix and Linux stack exchange. – jww Jun 12 '16 at 01:10

score 0 · Answer 2 · answered Apr 07 '21 at 07:14

0

If you want to see the output of the executed command,You need to follow this step, execute the command $tailf nohup.out

Please note : Nohup command creates a file with name nohup.out in same location where you have executed the nohup command .

answered Apr 07 '21 at 07:14

Sudhir Gaurav

is `tail` instead of `tailf`. – LianSheng Mar 21 '23 at 15:15
1

tailf will print out the last 10 lines of the given file and then wait for this file to grow We can use tail -f command to show no_of_lines of the given file and then wait for this file to grow Read more at: https://www.commandlinux.com/man-page/man1/tailf.1.html – Sudhir Gaurav Apr 03 '23 at 12:21
It turns out that there is really a `tailf` command, but my knowledge is too little, thank you for adding it! – LianSheng Apr 03 '23 at 14:47
use tail -100f , it will work for you . – Sudhir Gaurav Apr 06 '23 at 18:16

2 Answers2