是否有方法确定正在运行应用程序的设备。如果可能的话,我想区分iPhone和iPod Touch。


当前回答

Dutchie432和Brian Robbins提供了很好的解决方案。但还有一款型号没有推出,那就是Verizon的iPhone 4。这是缺失的一行。

if ([platform isEqualToString:@"iPhone3,2"])    return @"iPhone 4"; //Verizon

其他回答

NSString *deviceType = [[UIDevice currentDevice] systemName];

我可以保证,上述建议将适用于iOS 7及以上版本。我相信iOS 6也能运行。但我不确定。

要识别iPhone 4S,只需检查以下内容:

var isIphone4S: Bool {

    let width = UIScreen.main.bounds.size.width
    let height = UIScreen.main.bounds.size.height
    let proportions = width > height ? width / height : height / width

    return proportions == 1.5 && UIDevice.current.model == "iPhone"
}

我在我的应用程序中使用这个。截至2012年12月。

- (NSString *) platformString {
    // Gets a string with the device model
    size_t size;  
    sysctlbyname("hw.machine", NULL, &size, NULL, 0);  
    char *machine = malloc(size);  
    sysctlbyname("hw.machine", machine, &size, NULL, 0);  
    NSString *platform = [NSString stringWithCString:machine encoding:NSUTF8StringEncoding];  
    free(machine); 

    if ([platform isEqualToString:@"iPhone1,1"])    return @"iPhone 2G";
    if ([platform isEqualToString:@"iPhone1,2"])    return @"iPhone 3G";
    if ([platform isEqualToString:@"iPhone2,1"])    return @"iPhone 3GS";
    if ([platform isEqualToString:@"iPhone3,1"])    return @"iPhone 4";
    if ([platform isEqualToString:@"iPhone3,2"])    return @"iPhone 4";
    if ([platform isEqualToString:@"iPhone3,3"])    return @"iPhone 4 (CDMA)";    
    if ([platform isEqualToString:@"iPhone4,1"])    return @"iPhone 4S";
    if ([platform isEqualToString:@"iPhone5,1"])    return @"iPhone 5";
    if ([platform isEqualToString:@"iPhone5,2"])    return @"iPhone 5 (GSM+CDMA)";

    if ([platform isEqualToString:@"iPod1,1"])      return @"iPod Touch (1 Gen)";
    if ([platform isEqualToString:@"iPod2,1"])      return @"iPod Touch (2 Gen)";
    if ([platform isEqualToString:@"iPod3,1"])      return @"iPod Touch (3 Gen)";
    if ([platform isEqualToString:@"iPod4,1"])      return @"iPod Touch (4 Gen)";
    if ([platform isEqualToString:@"iPod5,1"])      return @"iPod Touch (5 Gen)";

    if ([platform isEqualToString:@"iPad1,1"])      return @"iPad";
    if ([platform isEqualToString:@"iPad1,2"])      return @"iPad 3G";
    if ([platform isEqualToString:@"iPad2,1"])      return @"iPad 2 (WiFi)";
    if ([platform isEqualToString:@"iPad2,2"])      return @"iPad 2";
    if ([platform isEqualToString:@"iPad2,3"])      return @"iPad 2 (CDMA)";
    if ([platform isEqualToString:@"iPad2,4"])      return @"iPad 2";
    if ([platform isEqualToString:@"iPad2,5"])      return @"iPad Mini (WiFi)";
    if ([platform isEqualToString:@"iPad2,6"])      return @"iPad Mini";
    if ([platform isEqualToString:@"iPad2,7"])      return @"iPad Mini (GSM+CDMA)";
    if ([platform isEqualToString:@"iPad3,1"])      return @"iPad 3 (WiFi)";
    if ([platform isEqualToString:@"iPad3,2"])      return @"iPad 3 (GSM+CDMA)";
    if ([platform isEqualToString:@"iPad3,3"])      return @"iPad 3";
    if ([platform isEqualToString:@"iPad3,4"])      return @"iPad 4 (WiFi)";
    if ([platform isEqualToString:@"iPad3,5"])      return @"iPad 4";
    if ([platform isEqualToString:@"iPad3,6"])      return @"iPad 4 (GSM+CDMA)";

    if ([platform isEqualToString:@"i386"])         return @"Simulator";
    if ([platform isEqualToString:@"x86_64"])       return @"Simulator";

    return platform;
}  
- (BOOL)deviceiPhoneOriPod
  {
    NSString *deviceType = [UIDevice currentDevice].model;
    if([deviceType rangeOfString:@"iPhone"].location!=NSNotFound)
      return YES;
    else
      return NO;
  }

你可以像这样使用UIDevice类:

NSString *deviceType = [UIDevice currentDevice].model;

if([deviceType isEqualToString:@"iPhone"])
    // it's an iPhone