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How can I spawn a process after a delay in a shell script? I want a command to start 60 seconds after the script starts, but I want to keep running the rest of the script without waiting 60 seconds first. Here's the idea:
#!/bin/sh # Echo A 60 seconds later, but without blocking the rest of the script sleep 60 && echo "A" echo "B" echo "C" The output should be
B C ... 60 seconds later A I need to be able to do this all in one script. Ie. no creating a second script that is called from the first shell script.
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/bin/sleep is Linux or Unix command to delay for a specified amount of time. You can suspend the calling shell script for a specified time. For example, pause for 10 seconds or stop execution for 2 mintues. In other words, the sleep command pauses the execution on the next shell command for a given time.
The bash wait command is a Shell command that waits for background running processes to complete and returns the exit status. Unlike the sleep command, which waits for a specified time, the wait command waits for all or specific background tasks to finish.
wait is typically used in shell scripts that spawn child processes that execute in parallel. To illustrate how the command works, create the following script: #!/bin/bash sleep 30 & process_id=$! echo "PID: $process_id" wait $process_id echo "Exit status: $?"
Shell loops are slow and bash's are the slowest. Shells aren't meant to do heavy work in loops. Shells are meant to launch a few external, optimized processes on batches of data.
& starts a background job, so
sleep 60 && echo "A" & Can't try it right now but
(sleep 60 && echo "A")&
should do the job
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