通过命令行在 iOS 5 上启动 GUI 应用程序（越狱）
我相信您过去可以通过命令行（通过 SSH）在越狱的 iOS 设备上启动 GUI 应用程序，方法是执行如下命令：
launch com.apple.Calculator
但这在我的 iOS 5 设备上不起作用（未找到启动）。
我也尝试过：
launchctl start com.apple.Calculator
但这也给了我一个错误（没有这样的过程）。
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).
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评论（1）
那些
launch/launchctl命令对我来说也不起作用。有效的方法是从 Cydia 安装命令行实用程序open并执行注意
calculator中的小写 c，这是我的计算器应用程序的包标识符。这是 Cydia 内容的开发者网站：
http://kramerapps.com/cydia/
此链接指向存储库站点：
http://moreinfo.thebigboss.org/moreinfo/depiction.php?file=openData
更新：对于 iOS 6.x，当前版本的
open似乎没有去工作。请参阅 @Nate 对下面评论中链接的另一个问题的回答。更新 2：Cydia 中的
open包已更新，现在可在 iOS 6 上使用。更新 3：以下是该包的源代码：https://github.com/conradev/Open。
如果查看
open.m文件，您可以看到SpringBoardServices私有框架中的函数SBSLaunchApplicationWithIdentifier才是实际打开应用程序的函数。Those
launch/launchctlcommands didn't work for me either. What did work was to install the command-line utiltyopenfrom Cydia and just executeNotice 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
opendoesn't seem to work. See @Nate's answer to another question linked below in the comments.Update 2: The
openpackage 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.mfile, you can see that the functionSBSLaunchApplicationWithIdentifierfrom theSpringBoardServicesprivate framework is what actually opens the app.