从 iPhone 应用程序以编程方式拨打电话并在结束通话后返回应用程序

Question

我正在尝试从我的 iphone 应用程序发起呼叫，我是通过以下方式进行的..

-(IBAction) call:(id)sender
{

    UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"Call Besito" message:@"\n\n\n"
                                                   delegate:self cancelButtonTitle:@"Cancel"  otherButtonTitles:@"Submit", nil];


    [alert show];
    [alert release];
}

- (void)alertView:(UIAlertView *)alertView willDismissWithButtonIndex:(NSInteger)buttonIndex
{
    if (buttonIndex != [alertView cancelButtonIndex])
    {
      NSString *phone_number = @"0911234567"; // assing dynamically from your code
        [[UIApplication sharedApplication] openURL:[NSURL URLWithString:[NSString    stringWithFormat:@"tel:%@", phone_number]]]; 
         NSString *phone_number = @"09008934848";
        NSString *phoneStr = [[NSString alloc] initWithFormat:@"tel:%@",phone_number];
        NSURL *phoneURL = [[NSURL alloc] initWithString:phoneStr];
        [[UIApplication sharedApplication] openURL:phoneURL];
        [phoneURL release];
        [phoneStr release];
    }
}

通过上面的代码..我可以成功拨打电话..但是当我结束通话时，我无法返回我的应用程序

所以，我想知道如何实现，那..也请告诉我我们如何使用 webview 发起呼叫...

Answer 1

这是我的代码：

NSURL *url = [NSURL URLWithString:@"telprompt://123-4567-890"]; 
[[UIApplication  sharedApplication] openURL:url]; 

使用它以便在通话结束后返回应用程序。

Answer 2

UIWebView *callWebview = [[UIWebView alloc] init];
NSURL *telURL = [NSURL URLWithString:@"tel:+9196815*****"];
[callWebview loadRequest:[NSURLRequest requestWithURL:telURL]];

请试试这个，它一定会让你回到你的应用程序。

Answer 3

NSString *phoneNumber =  // dynamically assigned
NSString *phoneURLString = [NSString stringWithFormat:@"tel:%@", phoneNumber];
NSURL *phoneURL = [NSURL URLWithString:phoneURLString];
[[UIApplication sharedApplication] openURL:phoneURL];

Answer 4

以下代码将不会返回到您的应用[在拨打电话之前不会显示警报视图]

UIWebView *callWebview = [[UIWebView alloc] init];
NSURL *telURL = [NSURL URLWithString:[NSString stringWithFormat:@"tel://%@",phoneNumber]];
[callWebview loadRequest:[NSURLRequest requestWithURL:telURL]];

以下代码将返回到您的应用“alertview will show before make call”

UIWebView *callWebview = [[UIWebView alloc] init];
NSURL *telURL = [NSURL URLWithString:[NSString stringWithFormat:@"telprompt://%@", phoneNumber]];
[callWebview loadRequest:[NSURLRequest requestWithURL:telURL]];

Answer 5

你也可以使用 webview，这是我的代码：

NSURL *url = [NSURL URLWithString:@"tel://123-4567-890"];
UIButton *btn = [[UIButton alloc]initWithFrame:CGRectMake(50, 50, 150, 100)];
[self.view addSubview:btn];
UIWebView *webview = [[UIWebView alloc] initWithFrame:CGRectMake(50, 50, 150, 100)];
webview.alpha = 0.0;        
[webview loadRequest:[NSURLRequest requestWithURL:url]];
// Assume we are in a view controller and have access to self.view
[self.view insertSubview:webview belowSubview:btn];
[webview release];
[btn release];

Answer 6

结束通话后无法返回应用，因为它是应用。