How to programmatically make cocoa application active

25
votes

I've got a background process that makes a transparent window appear when a hotkey is pressed:

[window makeKeyAndOrderFront:nil];
[[content animator] setAlphaValue:1.0]; // alpha was 0.0

... the window shows up fine & in front of the other windows (as I want it to), however until I manually click the window whatever application was active when the window appears remains active. I was expecting the 'makeKeyAndOrderFront' to make the application active as well, however adding a NSLog line to my -applicationWillBecomeActive shows it's not getting any active notification until the mouse click is performed.

Does anyone know how I can set my application active @ the same time I issue the -makeKeyAndOrderFront ? I need it active so that it can begin accepting keyboard input - any assistance needed :-)

3
objective-ccocoamacosnsapplication
Don't do this - Mac apps don't generally steal focus for good reason. – AlBlue
But in this case, the focus is expected to be in his app because the user has used the hotkey. – dreamlax
@AlBlue "generally". Lots of great apps do things that aren't "generally" done. – hmedia1

3 Answers

54
votes

Look at [[NSApplication sharedApplication] activateIgnoringOtherApps : YES];.

12
votes

For 10.6 or greater

[[NSRunningApplication currentApplication] activateWithOptions:(NSApplicationActivateAllWindows | NSApplicationActivateIgnoringOtherApps)];
6
votes

Swift 4.2+ of @Arvin answer

NSRunningApplication.current.activate(options: [.activateIgnoringOtherApps, .activateAllWindows])