我需要在我的应用程序中以按钮点击的方式进行编程调用。
为此,我找到了以下代码。
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"tel:1-800-555-1212"]];
这个在 iPhone SDK 3.0 和 iPhone 2.0 上是否可行?
可以请任何人帮忙吗?
先谢谢了。
我需要在我的应用程序中以按钮点击的方式进行编程调用。
为此,我找到了以下代码。
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"tel:1-800-555-1212"]];
这个在 iPhone SDK 3.0 和 iPhone 2.0 上是否可行?
可以请任何人帮忙吗?
先谢谢了。
将电话号码保存在一个单独的字符串中。
NSString *phoneNumber = @"1-800-555-1212"; // dynamically assigned
NSString *phoneURLString = [NSString stringWithFormat:@"tel:%@", phoneNumber];
NSURL *phoneURL = [NSURL URLWithString:phoneURLString];
[[UIApplication sharedApplication] openURL:phoneURL];
NSLog(@"Phone calling...");
UIDevice *device = [UIDevice currentDevice];
NSString *cellNameStr = [NSString stringWithFormat:@"%@",self.tableCellNames[indexPath.row]];
if ([[device model] isEqualToString:@"iPhone"] ) {
NSString *phoneNumber = [@"tel://" stringByAppendingString:cellNameStr];
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:phoneNumber]];
} else {
UIAlertView *warning =[[UIAlertView alloc] initWithTitle:@"Note" message:@"Your device doesn't support this feature." delegate:nil cancelButtonTitle:@"OK" otherButtonTitles:nil];
[warning show];
}
// VKJ
#import <CoreTelephony/CTTelephonyNetworkInfo.h>
#import <CoreTelephony/CTCarrier.h>
if ([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:@"tel://"]]) {
// Check if iOS Device supports phone calls
CTTelephonyNetworkInfo *netInfo = [[CTTelephonyNetworkInfo alloc] init];
CTCarrier *carrier = [netInfo subscriberCellularProvider];
NSString *mnc = [carrier mobileNetworkCode];
// User will get an alert error when they will try to make a phone call in airplane mode.
if (([mnc length] == 0)) {
// Device cannot place a call at this time. SIM might be removed.
} else {
// iOS Device is capable for making calls
}
} else {
// iOS Device is not capable for making calls
}
if ( ! [[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:@"sms:"]]) {
// iOS Device is not capable to send SMS messages.
}
别忘了添加CoreTelephony框架
if ([UIApplication instancesRespondToSelector:@selector(canOpenURL:)]) { NSURL *aURL = [NSURL URLWithString:@"tel:1234567890"]; if ([[UIApplication sharedApplication] canOpenURL:aURL]) { [[UIApplication sharedApplication] openURL:aURL]; } }
- joshpaul