[Linux 常見問題] Automatic exit from bash shell script on error

Source From Here
Question:
I've been writing some shell script and I would find it useful if there was the ability to halt the execution of said shell script if any of the commands failed. See below shell script for an example:
- test.sh

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1. #!/bin/sh  
2. #set -e  
3.   
4. doTesting()  
5. {  
6.     touch "Done"  
7. }  
8.   
9. chkResult()  
10. {  
11.     [ -e "Done" ] && touch "Done2"  
12. }  
13.   
14. rm -f Done*  
15. rm -f Error*  
16. echo "Running Testing..."  
17.   
18. doTesting  
19. chkResult  
20.   
21. touch "Done3"  
22. echo "Bye"  

If I execute this shell script and the result will be:

$ ./test.sh
Running Testing...
Bye
$ ls Done*
Done Done2 Done3

If we modify the test.sh on doTesting() to touch "Error" and execute it:

$ ./test.sh
Running Testing...
Bye
$ ls
Done3 Error test.sh

Here we don't expect Done3 to be generated while Done doesn't exist. Is there a global setting to make the script exit if one of the commands fails?

How-To
Use the set -e builtin:

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1. #!/bin/bash  
2. set -e  
3. # Any subsequent(*) commands which fail will cause the shell script to exit immediately  

Alternatively, you can pass -e on the command line:

$ bash -e test.sh
Running Testing...
$ ls
Error test.sh

Note.

The shell does not exit if the command that fails is part of the command list immediately following a while or until keyword, part of the test in an ifstatement, part of a && or || list, or if the command's return value is being inverted via !