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I need to create a script that returns the default app to be used for a given file extension (including the path). I had a look at the file command which can return the mime type and xdg-open which would open a file. But what I'd like to get is

myscript doc



myscript fun


How do I do that?

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2 Answers 2

up vote 1 down vote accepted

I grabbed a Linux expert in the office and we found a reasonable solution:

CURMINE=$(xdg-mime query filetype $SAMPLENAME)
CURDSK=$(xdg-mime query default $CURMINE)

if [ -f /.local/share/applications/$CURDSK ]; then
elif [ -f /usr/local/share/applications/$CURDSK ]; then
elif [ -f  /usr/share/applications/$CURDSK ]; then
    echo "Sorry no executable found for $1"
    exit 1

WHATTODO=$(grep "^Exec" $TRUEDSK | head -1)

Once we figured that there are only 3 locations for the desktop files it was not hard anymore.

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The command can be something like:

xdg-mime query default `xdg-mime query filetype example.odp` 


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Hi Samik, thx for the reply. I tried the command on Ubuntu 12.04 and ran into 2 problems: a) if example.odp doesn't exist it just throws an error and secondly (the bigger issue) I just got application/vnd.oasis.opendocument.presentation in return. Also the desktop file doesn't execute, I need to have something I can call with a set of parameters thereafter – stwissel Jul 5 '12 at 0:44
Quick update: I can get to the desktop file using xdg-mime query filetype example.odp || xdg-mime query default but then still the challenge is how to get to the executable including the path – stwissel Jul 5 '12 at 0:49

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