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I've got a bash script that rsync's two directories. Sometimes there's a change, odds are there's nothing.

I want to run a command only if rsync actually made a change (add/update a file). Otherwise I want to just skip it. Is there a return response I should be looking at?

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up vote 7 down vote accepted

If you use the -i option (and don't use the -v option), rsync will only print lines to STDOUT for any changes that were made. Depending on your script, this could look like

if [ -n "$(rsync -i /dir1 /dir2)" ]; then
  run_command;
fi
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That seems to do exactly what I want... thanks. Didn't think of using the -i flag like that. – raccettura Jun 30 '10 at 16:51
    
To clarify, if you want to run the script only if a change was made, then == should be !=. – hamx0r Mar 3 '15 at 0:18
    
Well as it is, this code is wrong from what I have seen. As for that code you get it Rsync ran FINE or HAS ERROR running, but this code doesn't check if there was a change. I don´t know if other version of Rsync could that, but I´m using RSYNC in Raspbmc (an old implementation we have) and OSMC. This is sad as RSYNC seems to be a wonderful tool. – eBrain Nov 9 '15 at 15:38

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