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I'm familiar with

explorer /select,/path/to/something

To open the folder that contains a file, with that file highlighted, in windows explorer. Anybody know of a one liner (or one liner applescript) to do the same for OS X with the Finder? Google search didn't seem to reveal much in the way of one liners.

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up vote 8 down vote accepted

The open command normally acts as if you double-clicked a file, but it has a -R flag to reveal the argument in finder. Therefore, you're looking for:

open -R /path/to/something

For further information, consult the open man page.

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Finder's AppleScript dictionary has a reveal command:

tell app "Finder" to reveal POSIX file "/private/etc"

But it doesn't bring Finder to the front or use your default view options for the created window.

This should additionally do both of those:

tell application "Finder"
    reopen
    activate
    set selection to {}
    set target of window 1 to (POSIX file "/private/etc")
end tell
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I have modified the code found in this article, it's not tested, but it's still something.

property the_path : "/Library/Scripts/file.ext"

--No more edits.

set the_folder to (POSIX file the_path) as alias
tell application "Finder"
if window 1 exists then
set target of window 1 to the_folder
else
reveal the_folder
end if
end tell
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