如何生成脚本的n个线程，每个线程都有自己的进程ID？

Question

I'd like to start n threads of a script, each with their own process id.

I currently do this via cronjob like so:

* * * * *    php /path/to/script.php >> /log/script.log 2>&1
* * * * *    php /path/to/script.php >> /log/script.log 2>&1
* * * * *    php /path/to/script.php >> /log/script.log 2>&1

Each of these three threads all log to the same script.log, which pairs output with its pid.

How can I do the same without copy/paste from a script?

Would the following spawn each of these with a different pid (accessible from php's getmypid())? Or would they all share the same script-launcher.sh pid?

#!/bin/bash
# Let's call this `script-launcher.sh`
# Launch 3 threads at once with `script-launcher.sh 3`

N=${1-0}
for i in {1..$N}
do
   php /path/to/script.php >> /log/script.log 2>&1
done

Answer 1

Whenever you span a new process, the new process will gain a new pid. So in this case, each time your shell script spans an instance of php, each of those copies of php will have their own pid.

The {1..$N} syntax will not work, though, so you will need to change your script to

N=${1-0}
for i in $(seq 1 $N)
do
   php /path/to/script.php >> script.log 2>&1
done

Then, if you call your script as script-launcher.sh 42, you'll get 42 instances of PHP running.

To have your php script run in the background (asynchronously), instruct bash to so with &:

   php /path/to/script.php >> script.log 2>&1 &