Shell 脚本一个接一个地运行

Question

I have 5 shell scripts. Each has a java command. Previous jobs output is input to the next job.

I created a superScript.sh

//mail - to inform beginning
sh script1.sh;
sh script2.sh;
sh script3.sh;
sh script4.sh;
sh script5.sh;
//mail to inform end

Sample script1.sh

cd toBaseDirectory;
java -cp /path/to/application.jar main.class parameter

But all the jobs are started at the same time. How can I make this sequential?

Answer 1

Try to run javas like this

java -cp /path/to/application.jar main.class parameter & wait

Answer 2

If want to run the second command only if the first exited successfully. To do so, join them with &&

command1 && command2

simple Example

~$ cat abc.sh 
#!/usr/bin/env bash
echo "hi"

~$ cat pqr.sh 
#!/usr/bin/env bash
echo "batman say :  $1"

If abc.sh execute successfully then only execute pqr.sh

~$ retval=$(./abc.sh) && result=$(./pqr.sh "$retval") && echo "$result"
batman say :  hi

you can also try similar approach with your java command execution using shell script

Note: To execute Shell scripts as command sequentially, waiting for the first to finish before the next one starts. You can use; command1; command2

Answer 3

wait waits for a process to finish

You can also choose to run the second command only if the first exited successfully. To do so, join them with &&

cmd1 && cmd2

but In repetitive cases like this I recommended using a simple loop in the body of your script:

for n in {1..5} ; do sh script${n}.sh ; done

The loop not only to run them in order but is easier to tweak and reuse when needed, with using brace expansion