问题描述
我相信你以前可以通过命令行（通过SSH）通过执行这样的命令，在越狱的iOS设备上启动GUI应用程序：
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
但这不适用于我的iOS 5设备（ launch not found ）。
but that is not working on my iOS 5 device (launch not found).
我也试过：
launchctl start com.apple.Calculator
但这也给我一个错误（没有这样的过程）。
but that also gives me an error (no such process).
推荐答案
那些启动 / launchctl 命令也没有为我工作。什么工作是安装命令行utilty 打开从Cydia并只需执行
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
注意
这里是开发者的Cydia的网站：
Here's the developer's website for Cydia stuff:
此链接到回购网站：
更新：对于iOS 6.x， open 的当前版本似乎不工作。请参阅@ Nate对另一个的回答。
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.
更新2：Cydia中的打开包已更新，现在可与iOS 6配合使用。
Update 2: The open package in Cydia has been updated and now works with iOS 6.
更新3：以下是软件包的来源：。
如果你查看 open.m 文件，你可以看到 SBSLaunchApplicationWithIdentifier code> SpringBoardServices 私人框架是什么打开应用程序。
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.
这篇关于在iOS 5上通过命令行启动GUI应用程序（越狱）的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！