How to properly debug shell scripts24 Feb 2019
I recently learned the proper way to debug shell scripts. It seems shell scripting has built-in features that can help developers find problems in their script. In this post I will be sharing how I debug scripts before and what is wrong with it while also showing the proper way to do it.
Table of Contents
- Printing commands upon execution.
- Running the entire script on debug mode.
- Terminating the entire script upon error.
Before, to debug a script I’m developing, I would do it like this.
echo "ls pictures" ls pictures
This prints a message before the command is executed. This worked out well for me before however this means I have to run echo for every command that I want to debug. What if I am debugging multiple consecutive lines of a script? Suppose I have a script called foo.sh with contents like these:
#!/bin/bash pwd whoami ls pictures uname
If I’m just interested to see if the first three commands, pwd, whoami and ls are being executed then with my previous way of debugging I would edit its contents to look like this.
#!/bin/bash echo "pwd" pwd echo "whoami" whoami echo "ls pictures" ls pictures uname
Thats a lot of echo and it looks tedious even though I’m just debugging the first three commands.
A better way to do this is to execute
before the commands I want to print. This command prints all the commands upon execution in the script until it is disabled by executing
The contents of foo.sh would look like this
#!/bin/bash set -x # Start printing commands upon execution. pwd whoami ls pictures set +x # Stop printing commands upon execution. uname
This is better because I only need to execute two commands to print all the commands being executed and I don’t have to execute echo which is another command that may potentially introduce some bugs upon execution of the script.
I can run the entire script on debug mode upon its execution from the command line if I pass the parameter “-x” to the script’s shebang which is always at the first line. Executing the command “set -x” is no longer necessary. The script foo.sh would look like this.
#!/bin/bash -x pwd whoami ls pictures uname
There are times that if a command the script is executing fails, I want it to terminate the script immediately, preventing the execution of subsequent commands.
Suppose that I run the script foo.sh and the pictures directory doesn’t exist. The command “ls pictures” will not execute properly and will result an error but will still execute the uname command. If I don’t want to execute uname if an error occurs, I will insert the line below after the ls command.
[ ! $? -eq 0 ] && exit 1
and the script to to look like this
#!/bin/bash set -x # Start printing commands upon execution. pwd whoami ls pictures [ ! $? -eq 0 ] && exit 1 uname set +x # Stop printing commands upon execution.
This is how I used to do it but this also has the same problem from the previous section. Do I have to add this statement after every command that should terminate the script upon error? That is a tedious task, fortunately I can just add another parameter “-e” on the set command to achieve the same result and the script would look like this.
#!/bin/bash set -xe # Start printing commands upon execution. pwd whoami ls pictures set +xe # Stop printing commands upon execution. uname
I can also add “-e” as a parameter for the script’s shebang if I want the script to exit upon error of any command like this:
#!/bin/bash -xe pwd whoami ls pictures uname
That’s it! These are the better and proper ways I learned to debug my scripts.