如何打开“共享菜单" OSX上的偏好?

问题描述

与Safari一样，尝试实现一个按钮，该按钮在单击时会打开系统偏好设置">扩展">共享菜单"窗格.

Much like Safari, trying to implement a button that when clicked opens System Preferences > Extensions > Share Menu pane.

我尝试过:

NSURL *URL = [NSURL URLWithString:@"x-apple.systempreferences:com.apple.preferences.extensions?Share_Menu"];
[[NSWorkspace sharedWorkspace] openURL:URL];

但是在新版本上似乎不起作用，有什么主意吗?

However it seems like that is not working on newer versions, any ideas?

推荐答案

您可以使用脚本桥来执行以下操作:

You can use Scripting Bridge to do something like this:

 SBSystemPreferencesApplication *systemPrefs =
[SBApplication applicationWithBundleIdentifier:@"com.apple.systempreferences"];

[systemPrefs activate];

SBElementArray *panes = [systemPrefs panes];
SBSystemPreferencesPane *notificationsPane = nil;

for (SBSystemPreferencesPane *pane in panes) {
    if ([[pane id] isEqualToString:@"com.apple.preferences.extensions"]) {
        notificationsPane = pane;
        break;
    }

}

[systemPrefs setCurrentPane:notificationsPane];

SBElementArray *anchors = [notificationsPane anchors];

for (SBSystemPreferencesAnchor *anchor in anchors) {
    if ([anchor.name isEqualToString:@"Extensions"]) {
        [anchor reveal];
    }
}

当然，您需要将ScriptingBridge框架添加到项目中，并添加Scripting Bridge头文件以获取系统首选项.您可以在Apple的开发人员文档中找到有关如何使用Scripting Bridge的更多详细信息.

Of course you need to add the ScriptingBridge framework to your project and a Scripting Bridge header file for system preferences. More details on how to use Scripting Bridge you can find in the developer documentation from Apple.

希望这会有所帮助

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