Question

Let’s say I have a bash script that executes three scripts in parallel

./script1 &
./script2 &
./script3 &

Now, let us say that ./script4 depends on script1, script2 and script3. How can I force it to wait for those, while still executing the three scripts in parallel?

Answer 1

You can use wait a built-in command available in Bash and in some other shells.
^{(see equivalent command WAITFOR on Windows)}

wait documentation

Wait for each specified process to complete and return its termination
status.

Syntax
      wait [n ...]
Key
   n   A process ID or a job specification

Each n can be a process ID or a job specification; if a job
specification is given, all processes in that job’s pipeline are
waited for.

If n is not given, all currently active child processes are waited
for, and the return status is zero.

If n specifies a non-existent process or job, the return status is
127. Otherwise, the return status is the exit status of the last process or job waited for.

Simple solution

Below wait waits indefinitely for all currently active child processes to be all ended (i.e. in this case the three scripts).

./script1 &
./script2 &
./script3 &
wait       # waits for all child processes
./script4

Store the PIDs in shell local variables

./script1 & pid1=$!
./script2 & pid2=$!
./script3 & pid3=$!
wait $pid1 $pid2 $pid3  # waits for 3 PIDs
./script4

Store the PIDs in temporary files

./script1 & echo $! >1.pid
./script2 & echo $! >2.pid
./script3 & echo $! >3.pid
wait $(<1.pid) $(<2.pid) $(<3.pid)
rm 1.pid 2.pid 3.pid            # clean up
./script4

This last solution pollutes the current directory with three files (1.pid, 2.pid and 3.pid). One of these file may be corrupted before wait call. Moreover these files could be left in the file-system in case of crash.