Call a function in bash script with & [duplicate] - bash

This question already has answers here:
What does "program &" mean on the command line?
(2 answers)
Closed 5 years ago.
Recently I am learning how to write bash script.
I noticed that in other's code, there is a line calling a function like this:
restartApp ${array[app]} $num &
I know the first two are variables that passing to the function, but I really don't understand what is the last symbol & here?

If you end the line with & the process starts in the background. This means you can continue to use the shell and do not have to wait on the output.

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Trying to run multiple processes in bash only executes one time [duplicate]

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I have the following in my bash file:
echo "Spawning $1 processes"
for i in {1..$1}
do
(go run loadgen.go $2 &)
echo "done."
done
However, I can only seem to get my go file to execute once. I know that they're started in the background, but each of my go files should append to the same log file (I can reproduce this by running my bash script multiple times). Am I doing something wrong to get this to iterate multiple times?

Ways to execute bash script without "./"? [duplicate]

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How do I run a shell script without using "sh" or "bash" commands?
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I have a bash script called climb.sh. When I execute it I write
./climb.sh 1
while inside the directory in which the script is located. However, I want to do the same thing wherever I am, and across all shell sessions by simply calling
climb 1
Also, climb.sh takes an numeric argument and calls "cd ../" that many times. In order for the program to work, it has to run alongside the current process, not within some child process.
How to achieve all this?
Thanks

Find out what the (shell) script was invoked with [duplicate]

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How do I parse command line arguments in Bash?
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How to get exact command line string from shell?
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Suppose my script.sh could take a number of options and arguments. What is the best way to find out what the script was invoked with (form inside the script)?
For eg., someone called it with script.sh --foo_option bar_arg
Is there a way to echo that exact command they typed from inside the script?
I've tried echo !! which does not work inside a script.

bash script, execute command line and keep going the next for loop [duplicate]

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Loop background job
(3 answers)
Closed 4 years ago.
Is it possible to run a command in a for loop without waiting for that command ended, while keeping going to the next iteration?
Because I have to send multi-files at the same time asap via many ssh connections, therefore I couldn't wait until the command ended one by one.
Maybe is it related to something like 'xterm' or 'gnome-terminal'?
Yes, you can execute the command in background by adding & at its end.
So the syntax looks like programName [arguments] & (at least for bourne compatible shells)

Regarding operation flow in a bash script [duplicate]

This question already has answers here:
bash script order of execution
(2 answers)
Closed 7 years ago.
I was just wondering, in a bash script, if I have a command on one line to say rsync a large file (~10GB) then the next command on the next line is meant to rename then move that same file, will the script know to wait for the rsync to complete before attempting the rename and move?
Is there a flag or something I can put on each line to make it wait before executing the next command?
Sorry if this seems like a total noob question, but alas I am a noob!
Thanks in advance for any info!
All commands are executed in sequence, so, a command will wait for the previous command to complete. Only when you add a & to the end of a command will it run in the background, and thus not wait for completion before executing the next.

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