Question

-(void)application:(UIApplication *)application didReceiveRemoteNotification:(NSDictionary *)userInfo
{
     [[NSNotificationCenter defaultCenter] postNotificationName:@"MessageReceived" object:self];

    UIApplicationState state = [application applicationState];
    if (state == UIApplicationStateActive)
    {

//        NSString *cancelTitle = @"Close";
        NSString *showTitle = @"Show";
//        NSString *message = [[userInfo valueForKey:@"aps"] valueForKey:@"alert"];
        UIAlertView *alertView = [[UIAlertView alloc] initWithTitle:@"New Order"
                                                            message:@"You just received new order"
                                                           delegate:self
                                                  cancelButtonTitle:nil
                                                     otherButtonTitles:showTitle, nil];

        UILocalNotification *localNotifcation = [[UILocalNotification alloc] init];
        localNotifcation.userInfo = userInfo;
        localNotifcation.soundName = UILocalNotificationDefaultSoundName;
//        localNotifcation.alertBody = message;
        localNotifcation.fireDate = [NSDate date];
        [[UIApplication sharedApplication] scheduleLocalNotification:localNotifcation];

        [alertView show];

    }
    else {
        //Do stuff that you would do if the application was not active
    }
}

请给我一个解决方案，如果现在通知正在使用x-code 6.3，我怎样才能在应用程序运行时发出声音？

Answer 1

您可以手动添加声音文件，并在应用处于活动状态时播放。

if ([application applicationState] == UIApplicationStateActive) {
        NSLog(@"active");

      NSString *soundPath = [[NSBundle mainBundle] pathForResource:@"alarm2" ofType:@"wav"];
        NSURL *soundURL = [NSURL fileURLWithPath:soundPath];
        AudioServicesCreateSystemSoundID((__bridge CFURLRef)soundURL, &_mySound);
        AudioServicesPlaySystemSound(self.mySound);
    }

如需更多默认声音，请查看以下答案：Playing system sound without importing your own

希望有所帮助

Answer 2

添加以下框架

#include <AudioToolbox/AudioToolbox.h>

使用以下代码 -

- (void)application:(UIApplication *)application didReceiveRemoteNotification:(NSDictionary *)userInfo
{

    UIApplicationState state = [application applicationState];
    if (state == UIApplicationStateActive)
    {

        SystemSoundID soundID;
        CFBundleRef mainBundle = CFBundleGetMainBundle();
        CFURLRef ref = CFBundleCopyResourceURL(mainBundle, (CFStringRef)@"mySoundName.wav", NULL, NULL);
        AudioServicesCreateSystemSoundID(ref, &soundID);
        AudioServicesPlaySystemSound(soundID);
    }
    else {
        // Push Notification received in the background
    }
}

或您可以使用系统声音 -

AudioServicesPlaySystemSound(1007);

振动 -

AudioServicesPlaySystemSound(kSystemSoundID_Vibrate);

请参阅here所有声音列表。

Answer 3

试试这段代码。将didReceiveRemoteNotification的此代码放在AppDelegate。m

中

SystemSoundID soundID;
    CFBundleRef mainBundle = CFBundleGetMainBundle();
    CFURLRef ref = CFBundleCopyResourceURL(mainBundle, (CFStringRef)@"Voicemail.wav", NULL, NULL);
    AudioServicesCreateSystemSoundID(ref, &soundID);
    AudioServicesPlaySystemSound(soundID);

这段代码对我来说是完美的工作..

如果应用程序在ios中处于活动状态，如何在通知出现时发出声音？

3 个答案: