判断iphone设备型号

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判断设备是否是 iphone5

2013-03-26 0 个评论作者：qingsi00

我要投稿

1.在程序中涉及到全屏的宽高的时候，初始化的时候，代码中不要写320和480

#define Screen_height [[UIScreen mainScreen] bounds].size.height

#define Screen_width [[UIScreen mainScreen] bounds].size.width

2.使用宏定义方法判断

#define iPhone5 ([UIScreen instancesRespondToSelector:@selector(currentMode)] ? CGSizeEqualToSize(CGSizeMake(640, 1136), [[UIScreen mainScreen] currentMode].size) : NO)

if（iPhone5）{

/*代码操作*/

}else{

/*代码操作*/

}

在网上又找到另外一种方法：

+ (NSString*)deviceString

{

// 需要#import "sys/utsname.h"

struct utsname systemInfo;

uname(&systemInfo);

NSString *deviceString = [NSString stringWithCString:systemInfo.machine encoding:NSUTF8StringEncoding];

if ([deviceString isEqualToString:@"iPhone1,1"]) return @"iPhone 1G";

if ([deviceString isEqualToString:@"iPhone1,2"]) return @"iPhone 3G";

if ([deviceString isEqualToString:@"iPhone2,1"]) return @"iPhone 3GS";

if ([deviceString isEqualToString:@"iPhone3,1"]) return @"iPhone 4";

if ([deviceString isEqualToString:@"iPhone4,1"]) return @"iPhone 4S";

if ([deviceString isEqualToString:@"iPhone5,2"]) return @"iPhone 5";

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

if ([deviceString isEqualToString:@"iPod1,1"]) return @"iPod Touch 1G";

if ([deviceString isEqualToString:@"iPod2,1"]) return @"iPod Touch 2G";

if ([deviceString isEqualToString:@"iPod3,1"]) return @"iPod Touch 3G";

if ([deviceString isEqualToString:@"iPod4,1"]) return @"iPod Touch 4G";

if ([deviceString isEqualToString:@"iPad1,1"]) return @"iPad";

if ([deviceString isEqualToString:@"iPad2,1"]) return @"iPad 2 (WiFi)";

if ([deviceString isEqualToString:@"iPad2,2"]) return @"iPad 2 (GSM)";

if ([deviceString isEqualToString:@"iPad2,3"]) return @"iPad 2 (CDMA)";

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

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

NSLog(@"NOTE: Unknown device type: %@", deviceString);

return deviceString;

}

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