InsanelyMac Forum > initiating a call on iphone

Full Version: initiating a call on iphone

InsanelyMac Forum > Apple World > Gadgets and Handheld Devices > iPhone

Mr. JZ

Sep 19 2008, 11:22 AM

Hi,

i am developing an application that needs to initiate a call, i am trying

self.number = @"18003998651";

UIApplication *app = [UIApplication sharedApplication]; NSString *num = [NSString stringWithFormat:@"tel://%@", self.number]; NSString *fixNum = [num stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSURL *url = [NSURL URLWithString:fixNum]; [app openURL:url];

i have also tried it with "tel:%@" instead of @"tel://%@"

but it shows alert : url unsupported, this url wasnt loaded

could you tell me what is the problem

Secondly how can i connect my xcode to my iphone device, so that application runs directly into iphone device

thanks,

JSn1™

Sep 20 2008, 03:38 AM

In the SDK, there is a call button in which you just have to set the phone number.
You can locate it on library.

Mr. JZ

Sep 22 2008, 01:05 PM

hi,

could you please elaborate more on that. By the way which sdk version you are talking about, i have 2.0

thanks

MoDs

Sep 29 2008, 04:48 AM

You can initiate a call like this:

CODE

-(NSString *)phoneNumberIdentifier:(NSString *)telephoneNumber
{
    NSString *tempString = [[NSString stringWithFormat:@"%@%@", @"tel://", telephoneNumber] retain];
    return tempString;
}

-(BOOL)callNumber:(NSString *)number
{
    if (![number isEqualToString:@""])
        return FALSE;
    else {
        [[UIApplication sharedApplication] openURL: [NSURL URLWithString:[self phoneNumberIdentifier:number]]];
    }

    return TRUE;
}

then just call

CODE

-(BOOL)callNumber:(NSString *)number

PS: This will work on the real device, but in the simulator, it won't. It's just a simulator, after all!

Pacon!

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