用于从其包标识符获取程序路径的Shell脚本

Question

In a shell script I need to know the path to a program from its bundle identifier.

The last answer of this SO question almost answers my question:

Eg for getting the path to XCode the idea is to launch this AppleScript one liner:

osascript -e "POSIX path of (path to application id \"com.apple.dt.Xcode\")"

This displays the path, but it also launches the Xcode program.

Does anybody have an idea how to display the path without launching the program?

Any other method than using AppleScript is of course welcome.

Answer 1

You can use the Finder to get the path:

osascript -e 'tell application "Finder" to POSIX path of ((application file id "com.apple.dt.Xcode") as alias)'

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Or, you can use a method of the NSWorkspace class ( macOS 10.6+ ):

/usr/bin/python -c 'from AppKit import NSWorkspace; print NSWorkspace.sharedWorkspace().URLForApplicationWithBundleIdentifier_("com.apple.dt.Xcode").path()'