在iOS 5上通过命令行启动GUI应用程序（越狱） [英] Launch GUI app on iOS 5 through the command line (jailbreak)

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:

http://kramerapps.com/cydia/

此链接到回购网站：

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

更新：对于iOS 6.x， open 的当前版本似乎不工作。请参阅@ Nate对另一个问题的回答。

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： https：// github.com/conradev/Open 。
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.

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