I am a beginner in bash programming. I want to obtain PIDs from processes, in order to use trap and kill to receive and send signals to a program in the same file.
In particular, I start the program opening a screen in this way:
screen -d -m "start program"
process_id=`/bin/ps -fu $USER| grep "program" | grep -v "grep" | awk '{print $2}'`
The variable process_id contains two PIDs, not one. If I run without a screen, I don't have this issue (anyway, I have to open the screen).
Does anyone have solutions to this problem?
Another question: If I write
screen -d -m "start program">log
the log file isn't printed. Any suggestions?
For your first question, pgrep(or process grep) is what you are looking for.
For instance, the following will return a list of PIDs of all bash processes running.
preg bash
And if you read the docs:
-signal
Defines the signal to send to each matched process. Either the numeric or the symbolic signal name can be used.
Second question, you could either use the -LogFile flag if your version of screen supports it. Or specify the log file in your .screenrc configuration file.
This has already been answered.
If you can't access the user's home directory where the configuration file .screenrc is usually put, you could change the $SCREENRC environment variable to explicitly set to an alternative path for it.