通过命令行在iOS 5上启动GUI应用程序（越狱）

Question

I believe you used to be able to launch GUI apps on jailbroken iOS devices via the command line (over SSH) by executing a command like this:

launch com.apple.Calculator

but that is not working on my iOS 5 device ( launch not found ).

I also tried:

launchctl start com.apple.Calculator

but that also gives me an error ( no such process ).

Answer 1

Those launch / launchctl commands didn't work for me either. What did work was to install the command-line utilty open from Cydia and just execute

open com.apple.calculator

Notice the lowercase c in calculator, that was the bundle identifier for my calculator app.

Here's the developer's website for Cydia stuff:

http://kramerapps.com/cydia/

This links to the repo site:

http://moreinfo.thebigboss.org/moreinfo/depiction.php?file=openData

Update: For iOS 6.x, this current version of open doesn't seem to work. See @Nate's answer to another question linked below in the comments.

Update 2: The open package in Cydia has been updated and now works with iOS 6.

Update 3: Here is the source for the package: https://github.com/conradev/Open . If you look at the open.m file, you can see that the function SBSLaunchApplicationWithIdentifier from the SpringBoardServices private framework is what actually opens the app.