以编程方式在iPhone上发出呼叫并在挂断后返回到同一个应用程序 [英] placing a call programmatically on iPhone and return to the same app after hangup

问题描述

我会撒谎在我的应用程序中拨打电话（我可以使用电话:)，但我想在用户挂机后返回我离开的应用程序。
这可能吗？

I would lie to place a call in my app (I can use the tel:) but I would like to return to my app where I left after the users hangs up. Is that possible?

推荐答案

我从Apple网站获得此代码并且运行正常：

I got this code from Apple site and it works perfectly:

-(IBAction) dialNumber:(id)sender{

NSString *aPhoneNo = [@"tel://" stringByAppendingString:[itsPhoneNoArray objectAtIndex:[sender tag]]] ; NSURL *url= [NSURL URLWithString:aPhoneNo];

NSString *osVersion = [[UIDevice currentDevice] systemVersion];

if ([osVersion floatValue] >= 3.1) { 
UIWebView *webview = [[UIWebView alloc] initWithFrame:[UIScreen mainScreen].applicationFrame]; 
[webview loadRequest:[NSURLRequest requestWithURL:url]]; 
webview.hidden = YES; 
// Assume we are in a view controller and have access to self.view 
[self.view addSubview:webview]; 
[webview release]; 
} else { 
// On 3.0 and below, dial as usual 
[[UIApplication sharedApplication] openURL: url];
}


}

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