Shell Programming and Scripting

What does a command "(2>&1)" does?

I will not be running any scripts so I guess I can run the 2nd command as follow assuming that I have to kill XYZ process if running..

sudo -u user2 "ps -ef | grep XYZ | awk '{print $2}' | xargs -r kill" >> $LOG 2>&1

Yes that should be fine, just grep for [X]YZ instead.

The square brackets tell grep match any of the following single characters (a single X here). So this will find XYZ, but because the command line of the grep has the square brackets, grep wont match it's self and we avoid killing the grep command. eg:

Code:

# ps -ef | grep ntpd
    root  278664  209062   0   Oct 03      -  1:03 /usr/sbin/xntpd 
    root 1540282  659564   0 09:11:58 pts/29  0:00 grep ntpd

Code:

# ps -ef | grep [n]tpd
    root  278664  209062   0   Oct 03      -  1:03 /usr/sbin/xntpd

2>&1 means redirect standard error (the stream most error messages are printed on) to the same location as standard output. This will allow your log to capture both normal program output and any error messages on the error stream.

Is the following command will show up in a logs if [X]YZ is running? The actual ouput of ps -ef |grep XYZ ?

I want to kill it if running. I'm assuming that the commad "{print $2}" print the PID of the XYZ process and kill it.

Thanks,
Afi