我怎样才能获得iphone设备类型？

Question

我想获得iphone设备类型,如果它是iphone 2或3G或4 xcode 4.0.

有没有办法得到它？

如果是,请告诉我.

Answer 1

我找到了一些新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班级没有方法platformString.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"

会做的.

Answer 3

Caleb是对的,你不应该检查设备类型,而是检查功能.例如,您可以检查设备是否支持多任务处理,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,但是知道这不能代替检查对象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模型.