Preventing a program from breaking my Bash script

Question

I'm using S3cmd in a bash script upon startup. If it returns an error code, the script is ready to do something. However, s3cmd seem to (sometimes) break it all when an error occurs, and outputs information on screen. It just exists my script.

How do I prevent a program from breaking my Bash script? If something is wrong, I just want the bash script to keep on doing the next thing in line.

EDIT: It seems this only happens with /etc/rc.local. If I runt the script as something else (/home/whateverscript) it does as I want it to.

Then you probably have relative paths messing up your execution. Also, as you said the script is run upon startup, you may have unset variables on which your code rely. — Rubens, Aug 04 '13 at 18:17
Yes, @VaughnCato, but it only seems to be the case when it's run as rc.local. — Paolo, Aug 04 '13 at 18:17
If you can show us how you are using the S3cmd inside you bash script, that would be great. Thanks! — SSaikia_JtheRocker, Aug 04 '13 at 18:47
@JtheRocker s3cmd sync --recursive --delete-removed --config="$HEMMAPPEN/.s3cfg" "$SOURCEFOLDER" "$TARGETFOLDER/" — Paolo, Aug 04 '13 at 23:09

score 0 · Answer 1 · answered Aug 05 '13 at 06:42

May be you can wrap your

s3cmd sync --recursive --delete-removed --config="$HEMMAPPEN/.s3cfg" "$SOURCEFOLDER" "$TARGETFOLDER/"

in the script as below,

outputText=$(s3cmd sync --recursive --delete-removed --config="$HEMMAPPEN/.s3cfg" "$SOURCEFOLDER" "$TARGETFOLDER/" 2>&1;echo ,$?)

This will redirect your stderr to stdout (2>&1) and the variable outputText would contain your desired output (in the form *stdout,exit_status*) of the command, if it's required later in the script context.

If you don't want stdout but only the status of the command in outputText, you can use the following

status=$(s3cmd sync --recursive --delete-removed --config="$HEMMAPPEN/.s3cfg" "$SOURCEFOLDER" "$TARGETFOLDER/" > /dev/null 2>&1;echo $?)

status variable would contain the status of teh command that was run.

I hope it makes sense. Please comment, if it hasn't solved your problem.

Preventing a program from breaking my Bash script

1 Answers1