Question

我正在开发一个使用PJSIP Library的VOIP_App，它使用C-Language编写，根据情况自动调用该库中编写的大多数方法。

有一个名为on_incoming_call的方法自动调用并且用户收到调用，我想添加一些用户交互来接收调用，我需要创建一些callBack，应该在这个方法中调用，并且方法定义应写在Objective-C中。

以下是代码段：

/* Callback called by the library upon receiving incoming call */
static void on_incoming_call(pjsua_acc_id acc_id, pjsua_call_id call_id,
                         pjsip_rx_data *rdata)
{
    pjsua_call_info ci;

    PJ_UNUSED_ARG(acc_id);
    PJ_UNUSED_ARG(rdata);

    pjsua_call_get_info(call_id, &ci);



    PJ_LOG(3,(THIS_FILE, "....\n\n\n Incoming call from %.*s!!  \n\n\n",
          (int)ci.remote_info.slen,
          ci.remote_info.ptr));

    /* Automatically answer incoming calls with 200/OK */
    pjsua_call_answer(call_id, 200, NULL, NULL);
}

Answer 1

您可以通过在PJSIP库中的回调中发布通知来实现您的目标

static void on_incoming_call(pjsua_acc_id acc_id, pjsua_call_id call_id,
                     pjsip_rx_data *rdata){
   pjsua_call_info ci;
   NSString *callerNumber = [NSString stringWithUTF8String:ci.remote_info.ptr];

   [[NSNotificationCenter defaultCenter] postNotificationName:
     @"IncomingCall" object:callerNumber];

pjsua_call_answer(call_id, 200, NULL, NULL); //Comment out this line and instead call this in your UIViewController to accept the call on click of some button or whatever UI action you desire 

}

现在您可以在AppDelegate中查看此通知

- (void)applicationDidFinishLaunching:(UIApplication *)application
{
 [[NSNotificationCenter defaultCenter] addObserver:self selector:@selector(launchIncomingVC:) name:@"IncomingCall" object:nil];
}

然后该函数显示您的UIViewController接受或拒绝来电

-(void)launchIncomingVC:(NSNotification *) notif
{
    NSLog(@"LAUNCH INCOMING CALL VC");

    YourViewController *rvc = [[YourViewController alloc] init];
    UIStoryboard *storyboard = [UIStoryboard storyboardWithName:@"YourStoryBoardName" bundle:nil];
    rvc = [storyboard instantiateViewControllerWithIdentifier:@"YourViewControllerIdentifier"];

    rvc.paramCallerNumber = noti.object; //pass the caller number to show it on the incoming view

        dispatch_async(dispatch_get_main_queue(), ^{

            [self.window makeKeyAndVisible];

            UIViewController *topRootViewController = [UIApplication sharedApplication].keyWindow.rootViewController;
            while (topRootViewController.presentedViewController)
            {
                topRootViewController = topRootViewController.presentedViewController;
            }

            [topRootViewController presentViewController:incomingVC animated:YES completion:nil];
        });

}

如何在iOS中创建应该由库调用的callBack

1 个答案: