cocoa - 如何从 Mac 上的 cocoa 应用程序内部启动和终止应用程序

Question

我的 Cocoa 应用程序需要启动和终止其他应用程序。请让我知道任何可以执行以下操作的示例代码:

1. 从 Cocoa 代码内部启动应用程序
2. 从 Cocoa 代码内部终止应用程序

Answer 1

正如前面提到的，在 NSWorkspace 类的帮助下启动其他应用程序非常容易，例如:

- (BOOL)launchApplicationWithPath:(NSString *)path
{
    // As recommended for OS X >= 10.6.
    if ([[NSWorkspace sharedWorkspace] respondsToSelector:@selector(launchApplicationAtURL:options:configuration:error:)])
        return nil != [[NSWorkspace sharedWorkspace] launchApplicationAtURL:[NSURL fileURLWithPath:path isDirectory:NO] options:NSWorkspaceLaunchDefault configuration:nil error:NULL];

    // For older systems.
    return [[NSWorkspace sharedWorkspace] launchApplication:path];
}

您必须做更多的工作才能退出另一个应用程序，特别是如果目标是 10.6 之前的版本，但这并不太难。这是一个例子:

- (BOOL)terminateApplicationWithBundleID:(NSString *)bundleID
{
    // For OS X >= 10.6 NSWorkspace has the nifty runningApplications-method.
    if ([[NSWorkspace sharedWorkspace] respondsToSelector:@selector(runningApplications)])
        for (NSRunningApplication *app in [[NSWorkspace sharedWorkspace] runningApplications])
            if ([bundleID isEqualToString:[app bundleIdentifier]])
                return [app terminate];

    // If that didn‘t work then try using the apple event method, also works for OS X < 10.6.

    AppleEvent event = {typeNull, nil};
    const char *bundleIDString = [bundleID UTF8String];

    OSStatus result = AEBuildAppleEvent(kCoreEventClass, kAEQuitApplication, typeApplicationBundleID, bundleIDString, strlen(bundleIDString), kAutoGenerateReturnID, kAnyTransactionID, &event, NULL, "");

    if (result == noErr) {
        result = AESendMessage(&event, NULL, kAEAlwaysInteract|kAENoReply, kAEDefaultTimeout);
        AEDisposeDesc(&event);
    }
    return result == noErr;
}