Question

在我的应用中，我想拨打所选号码。为此，我有以下代码

 +(void)makeCallToSelectedContact:(NSString*)phoneNo{

    NSMutableString *phoneNumber = [NSMutableString stringWithString:phoneNo];

    [phoneNumber replaceOccurrencesOfString:@" " 
                                 withString:@"" 
                                    options:NSLiteralSearch 
                                      range:NSMakeRange(0, [phoneNumber length])];
    [phoneNumber replaceOccurrencesOfString:@"(" 
                                 withString:@"" 
                                    options:NSLiteralSearch 
                                      range:NSMakeRange(0, [phoneNumber length])];
    [phoneNumber replaceOccurrencesOfString:@")" 
                                 withString:@"" 
                                    options:NSLiteralSearch 
                                      range:NSMakeRange(0, [phoneNumber length])];

    NSLog(@"phoneNumber => %@",phoneNumber);
    if([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:[NSString stringWithFormat:@"tel:+%@",phoneNumber]]]) {
        [[UIApplication sharedApplication] openURL:[NSURL URLWithString:[NSString stringWithFormat:@"tel:+%@",phoneNumber]]];
    }
    else {
        NSLog(@"Unable to open");
        [self showAlertWithTitle:@"Alert" andMessage:@"This Device Doesn't Support Call Functionality"]; 
    }
}

让我们说如果我的电话号码不是+91 1234567890，那么它会正确拨打号码。但如果我的号码没有+＆amp;国家代码如1234567890然后它将其转换为+12 34567890，这是错误的电话号码。这是我的控制台日志

    2012-04-05 19:13:37.960 Search[22453:11003] phoneNumber => +911234567890

2012-04-05 19:13:47.928 Search[22453:11003] phoneNumber => 1234567890

我缺少什么？任何形式的帮助都表示赞赏。

Answer 1

您在创建网址时添加“+” - 查看：

if([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:[NSString stringWithFormat:@"tel:+%@",phoneNumber]]]) {
        [[UIApplication sharedApplication] openURL:[NSURL URLWithString:[NSString stringWithFormat:@"tel:+%@",phoneNumber]]];
}

telUR在openURL中无法正常工作

1 个答案: