通话后加载应用

Question

I am currently trying to make a speed dial app for various numbers the user might enter. I am loading the phone all like so after clicking a UITableViewCell

- (IBAction)Dialer:(id)sender{
    NSURL *url = [ [ NSURL alloc ] initWithString: @"tel:09-410-7078" ];
    [[UIApplication sharedApplication] openURL:url];
}

that loads up the phone dialer and dials the number.. I'm woundering if after the phone call has ended is its possible to load the app from where it exited for the phone call.. or if there is a better way to do what I am trying to do?

Answer 1

From iOS 5.0 onwards, we return to the app after the call ends. Find the below code.

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"telprompt:18004912200"]]; 

Answer 2

Actually you should keep in mind that an application can be unloaded at any moment. But when your application is being unloaded your app delegate gets some messages ( applicationWillTerminate , applicationWillResignActive , applicationDidEnterBackground ). You should read this post . In those methods you should save some parameters of your application (current page, settings, etc.) and use them when your application is launched or become active again.

If you want to start your application manually you shouldn't do so. User will be surprised by such behavior.

Answer 3

I searched very long time and 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];
}


}