macos - 调用 TransformProcessType() 时，应用程序菜单不显示

Question

如果您像这样调用 TransformProcessType() :

ProcessSerialNumber psn = { 0, kCurrentProcess }; 
(void) TransformProcessType(&psn, kProcessTransformToForegroundApplication);

然后，除非您在应用程序中足够早地调用它(例如，在 applicationWillFinishLaunching 中)，否则 cocoa 应用程序菜单不会显示。

Answer 1

我向 Apple 寻求帮助，他们为我提供了很好的帮助。引用:

The reason why the menu bar isn't show when you call TransformProcessType is that your app is already the active app (that is, [[NSRunningApplication currentApplication] isActive] returns YES) and the menu bar for an app is shown when the app is activated

这是他们的解决方法:

- (void)transformStep1 {
    for (NSRunningApplication * app in [NSRunningApplication runningApplicationsWithBundleIdentifier:@"com.apple.finder"]) {
        [app activateWithOptions:NSApplicationActivateIgnoringOtherApps];
        break;
    }
    [self performSelector:@selector(transformStep2) withObject:nil afterDelay:0.1];
}

- (void)transformStep2
{
    ProcessSerialNumber psn = { 0, kCurrentProcess }; 
    (void) TransformProcessType(&psn, kProcessTransformToForegroundApplication);

    [self performSelector:@selector(transformStep3) withObject:nil afterDelay:0.1];
}

- (void)transformStep3
{
    [[NSRunningApplication currentApplication] activateWithOptions:NSApplicationActivateIgnoringOtherApps];
}