如何获取 iPhone 设备类型？

Question

我想获取 iphone 设备类型，如果它是 xcode 4.0 中的 iphone 2 或 3G 或 4。

有什么办法可以得到吗？

如果是的话，请告诉我。

Answer 1

我发现了一些新的算法来获取设备类型和 iOS 版本。

#include <sys/sysctl.h>
#include <sys/utsname.h>


- (NSString *) platformString{

NSLog(@"[UIDevice currentDevice].model: %@",[UIDevice currentDevice].model);
NSLog(@"[UIDevice currentDevice].description: %@",[UIDevice currentDevice].description);
NSLog(@"[UIDevice currentDevice].localizedModel: %@",[UIDevice currentDevice].localizedModel);
NSLog(@"[UIDevice currentDevice].name: %@",[UIDevice currentDevice].name);
NSLog(@"[UIDevice currentDevice].systemVersion: %@",[UIDevice currentDevice].systemVersion);
NSLog(@"[UIDevice currentDevice].systemName: %@",[UIDevice currentDevice].systemName);
NSLog(@"[UIDevice currentDevice].batteryLevel: %f",[UIDevice currentDevice].batteryLevel);      
struct utsname systemInfo;
uname(&systemInfo);
NSLog(@"[NSString stringWithCString:systemInfo.machine encoding:NSUTF8StringEncoding]: %@",[NSString stringWithCString:systemInfo.machine encoding:NSUTF8StringEncoding]);
NSString *platform = [NSString stringWithCString:systemInfo.machine encoding:NSUTF8StringEncoding];

    if ([platform isEqualToString:@"iPhone1,1"])    return @"iPhone 1G";
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 CDMA";
if ([platform isEqualToString:@"iPhone3,3"])    return @"Verizon iPhone 4";
if ([platform isEqualToString:@"iPhone4,1"])    return @"iPhone 4S";
if ([platform isEqualToString:@"iPhone5,1"])    return @"iPhone 5 (GSM)";
if ([platform isEqualToString:@"iPhone5,2"])    return @"iPhone 5 (GSM+CDMA)";
if ([platform isEqualToString:@"iPhone5,3"])    return @"iPhone 5c (GSM)";
if ([platform isEqualToString:@"iPhone5,4"])    return @"iPhone 5c (GSM+CDMA)";
if ([platform isEqualToString:@"iPhone6,1"])    return @"iPhone 5s (GSM)";
if ([platform isEqualToString:@"iPhone6,2"])    return @"iPhone 5s (GSM+CDMA)";
if ([platform isEqualToString:@"iPhone7,2"])    return @"iPhone 6";
if ([platform isEqualToString:@"iPhone7,1"])    return @"iPhone 6 Plus";
if ([platform isEqualToString:@"iPod1,1"])      return @"iPod Touch 1G";
if ([platform isEqualToString:@"iPod2,1"])      return @"iPod Touch 2G";
if ([platform isEqualToString:@"iPod3,1"])      return @"iPod Touch 3G";
if ([platform isEqualToString:@"iPod4,1"])      return @"iPod Touch 4G";
if ([platform isEqualToString:@"iPod5,1"])      return @"iPod Touch 5G";
if ([platform isEqualToString:@"iPad1,1"])      return @"iPad";
if ([platform isEqualToString:@"iPad2,1"])      return @"iPad 2 (WiFi)";
if ([platform isEqualToString:@"iPad2,2"])      return @"iPad 2 (Cellular)";
if ([platform isEqualToString:@"iPad2,3"])      return @"iPad 2 (Cellular)";
if ([platform isEqualToString:@"iPad2,4"])      return @"iPad 2 (WiFi)";
if ([platform isEqualToString:@"iPad2,5"])      return @"iPad Mini (WiFi)";
if ([platform isEqualToString:@"iPad2,6"])      return @"iPad Mini (Cellular)";
if ([platform isEqualToString:@"iPad2,7"])      return @"iPad Mini (Cellular)";
if ([platform isEqualToString:@"iPad3,1"])      return @"iPad 3 (WiFi)";
if ([platform isEqualToString:@"iPad3,2"])      return @"iPad 3 (Cellular)";
if ([platform isEqualToString:@"iPad3,3"])      return @"iPad 3 (Cellular)";
if ([platform isEqualToString:@"iPad3,4"])      return @"iPad 4 (WiFi)";
if ([platform isEqualToString:@"iPad3,5"])      return @"iPad 4 (Cellular)";
if ([platform isEqualToString:@"iPad3,6"])      return @"iPad 4 (Cellular)";
if ([platform isEqualToString:@"iPad4,1"])      return @"iPad Air (WiFi)";
if ([platform isEqualToString:@"iPad4,2"])      return @"iPad Air (Cellular)";
if ([platform isEqualToString:@"i386"])         return @"Simulator";
if ([platform isEqualToString:@"x86_64"])       return @"Simulator";
return @"Unknown";
}

Answer 2

实际上，UIDevice 类没有方法platformString。如果使用未记录的方法，您的应用程序将被 Apple 拒绝。

[[UIDevice currentDevice] model] // e.g. "iPod touch"

会成功的。

Answer 3

Caleb 是对的，您不应该检查设备类型，而应该检查功能。例如，您可以检查设备是否支持多任务，如下所示：

UIDevice *device = [UIDevice currentDevice];
if ([device respondsToSelector:@selector(isMultitaskingSupported)] && [device isMultitaskingSupported]) {
    // ...code to be executed if multitasking is supported.
    // respondsToSelector: is very useful
}

如果必须，您可以检查 iOS 版本，但要知道这不能替代检查对象是否 respondsToSelector:。

#define IOS4_0 40000

// You'd probably want to put this in a convenient method
NSArray *versions = [[UIDevice currentDevice].systemVersion componentsSeparatedByString:@"."];
NSInteger major = [[versions objectAtIndex:0] intValue];
NSInteger minor = [[versions objectAtIndex:1] intValue];
NSInteger version = major * 10000 + minor * 100;

if (version >= IOS4_0) {
    // ...code to be executed for iOS4.0+
}

据我所知，没有记录的方法来检查设备型号。

Answer 4

我在我的项目中添加了一个静态方法来检查设备类型：iPad、iPhone4（或更低）和iPhone5，以处理不同的屏幕尺寸。

+ (DeviceType) currentDeviceType
{
    DeviceType device = DEVICE_TYPE_IPAD ;
    if( [[UIDevice currentDevice] userInterfaceIdiom] == UIUserInterfaceIdiomPhone )
    {
        if( [[UIScreen mainScreen] bounds].size.height >= 568 || [[UIScreen mainScreen] bounds].size.width >= 568 )
        {
            device = DEVICE_TYPE_IPHONE5 ;
        }
        else
        {
            device = DEVICE_TYPE_IPHONE4 ;
        }
    }

    return device ;
}

您还可以使用 UIUserInterfaceIdiomPad 类型。

据我所知，iPod 和 iPhone 的目的是平等对待。但如果您需要确定它是否是 iPod，请检查它是否可以打电话。

Answer 5

看看这个 https://github.com/erica/uidevice-extension/

[[UIDevice currentDevice] platformString] //@"iPhone 4"

Answer 6

这篇文章帮助我区分了 iPhone/iPad 类型：

您可以获取型号type (iPhone, iPad, iPod) 如上所示：

[[UIDevice currentDevice] model]

为了进一步指定，您可以导入并查找通过以下代码返回为 char 数组的特定模型类型：

struct utsname u;
uname(&u);
char *type = u.machine;

并将其转换为 NSString：

NSString *strType = [NSString stringWithFormat:@"%s", type];

Answer 7

通常的建议是不要担心您运行的设备类型，而是测试您需要的功能。如果您查看设备类型，您一定会做出不正确的假设。例如，在不同市场销售的每种型号的版本可能会有所不同，并且新设备可能会报告相同的类型但具有不同的功能。因此，测试功能而不是模型。

Answer 8

您可以定义一些扩展 @csb 对此线程的回答的宏。

#define IS_IPHONE4 ([[UIScreen mainScreen] bounds].size.width == 480 || [[UIScreen mainScreen] bounds].size.height == 480)
#define IS_IPHONE5 ([[UIScreen mainScreen] bounds].size.width == 568 || [[UIScreen mainScreen] bounds].size.height == 568)
#define IS_IPAD    ([[UIScreen mainScreen] bounds].size.width == 768 || [[UIScreen mainScreen] bounds].size.height == 768)