LINUX CLASSES - PROGRAMMING

Can I Debug a Shell Script?

Debugging Shell Scripts

Sometimes shell scripts just don't work the way you think they should, or you get strange error messages when running your script. Just remember: The computer is always right. It's easy to omit a significant blank, quotation mark, or bracket, or to mistakenly use a single quotation mark when you should have used double quotation marks or a backtick. When you get unexpected behavior from a shell script and you're thinking "I just know this is coded right . . . the computer is wrong!"--remember: The computer is always right. But fortunately, there is help. By prefixing your shell invocation with bash -x, you can turn on tracing to see what's really happening. Let's say we have the following listarg script:

count=1
until [ "$*" = "" ]
do
echo "Arg $count : $1 "
shift
count=$count+1
done

At first glance, it looks golden, but for some reason the counter is not incrementing properly. Try running it with the command

% bash -x listarg abc def

and look at the trace output as the shell executes. The lines prefixed with a plus sign show the progress of the running script, and the lines without the plus sign are the script's normal output.

+ count=1
+ [ abc def = ]
+ echo Arg 1 : abc
Arg 1 : abc
+ shift
+ count=1+1
Hmmm . . .
+ [ def = ]
+ echo Arg 1+1 : def
Arg 1+1 : def
Not Good!
+ shift
+ count=1+1+1
+ [ = ]

Instead of printing Arg 2 : def, we got Arg 1+1 : def. But the trace output line reading count=1+1 nails the problem. You forgot to use the expr command, so the shell is treating this as a string concatenate instead of a mathematical calculation.

Note: You can always press Ctrl-C to stop a running shell script. This is handy if you accidentally create a script with an infinite loop (one that will never end by itself).

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Comments - most recent first
(Please feel free to answer questions posted by others!)

Boss     (26 Sep 2012, 07:05)
Bob,

I got a script which when run, calls few other scripts conditionally, when executed. I would want to know how to debug the execution. Its a big no. of files and they dont have 'set -xv' also in them. How will i get to know which scripts(file name) are being executed. Also is there a way wherein i get to see the commands being executed by the various shell scripts that are called(executed) without having to change each of those scripts to add a set or print command.
Prat     (19 Aug 2011, 09:42)
Hey Bob!!Many thanks for the great explanation of each and every topic.
Just wanted to suggest you one thing, if we get a lessons on database connectivity as well, would be really appreciated.
Ragz     (20 Jul 2011, 12:59)
Lolz..good one! Now I know how to bust some one :)
Bob Rankin     (20 Jul 2011, 12:56)
@Ragz - Look for the common elements in my answer, and his question. Then you'll discover the "secret" behind it. :-)
Ragz     (20 Jul 2011, 12:51)
Lol @ Bob! What ever your reply was to Amit..I'm sure you meant every bit of it..that was really hilarious!
BTW..what did it mean? Would it even make sense if I learn a new language?
Bob Rankin     (03 Feb 2011, 11:22)
@Amit - AIE YUM TAR CEUEO DAY ON A FUDD VMII I MESTY EO TTEI. :-)
Amit Singh     (02 Feb 2011, 23:13)
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John Ndambuki     (29 Jul 2010, 16:41)

+ An alternative solution:

Solution #3
-----------

count=1
until [ "$*" = "" ]
do echo "Arg $count : $1 "
shift
count=$(($count+1))
done
John Ndambuki     (29 Jul 2010, 04:41)
Just adding 2 more solutions for this:

Solution #1
-----------

count=1
until [ "$*" = "" ]
do echo "Arg $count : $1 "
shift
let count+=1
done

Solution #2
-----------

count=1
until [ "$*" = "" ]
do echo "Arg $count : $1 "
shift
let count=count+1
done
senthil     (28 Apr 2010, 04:31)
'''sh -x listarg abc def ''' Also Works , thanks
Bob Rankin     (07 Apr 2010, 13:29)
That doesn't work either. As I mentioned above, you need to use the expr command like this:
count=`expr $count + 1`
nolan     (06 Apr 2010, 17:38)
minor tweak needed...
line 6 needs a space on each side of the "+".

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