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I want to capture the name of an image file created with scrot. I can capture the filename like this:

$ image=$(scrot -s -e 'echo $n')
$ echo $image
2018-09-07-140419_185x76_scrot.png

I need to do that via su:

$ image=$(su someuser -c "scrot -s -e 'echo $n'")

However, image is empty.

I think this syntax is correct - this works:

$ greeting=$(su someuser -c "echo 'hello, world'")
$ echo $greeting
hello, world

I am not sure why this isn't working with scrot. I presume it something to do with how scrot launches its subshell - I guess it doesn't propagate its output?

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    Are you sure scrot is not simply failing because an appropriate DISPLAY is not set in the su environment and/or someuser doesn't have authority to access it? Commented Sep 7, 2018 at 13:30
  • No, the scrot part works just fine and the file is created successfully (did have to set DISPLAY etc for that to work however, so a good question to ask!) Commented Sep 7, 2018 at 13:54
  • btw, it would be a better idea to run scrot via "ssh -X someuser@localhost scrot ..." (with public key auth, so you don't have to type in a password) instead of having to give 'someuser' access to your display. Commented Sep 7, 2018 at 14:16
  • It's root issuing the su so no password is required. The user switch is needed to execute in the context of the user that the desktop belongs to otherwise scrot and other things I do later don't work. Commented Sep 7, 2018 at 15:41

1 Answer 1

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The "$n" inside the -c command with be expanded by the outer shell, not by scrot. You should escape it:

$ image=$(su someuser -c "scrot -s -e 'echo \$n'")
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  • escaping, yes! I tried that unsuccessfully but your answer made be look at it again. Because of exactly what I am doing, I needed a double-escape: scrot -s -e 'echo \\$n'. But escaping is the reason it wasn' working, so - accepted answer. Commented Sep 7, 2018 at 15:39

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