iOS 判断设备型号的三种方式
方法一:
//***宏定义
#define iPhone5 ([UIScreen instancesRespondToSelector:@selector(currentMode)] ? CGSizeEqualToSize(CGSizeMake(640, 1136), [[UIScreen mainScreen] currentMode].size) : NO)
方法二:
还有做适配,代码中不要写320和480什么的初始化坐标,要用屏幕高度和宽度初始化
#define Screen_height [[UIScreen mainScreen] bounds].size.height
#define Screen_width [[UIScreen mainScreen] bounds].size.width
方法三:
+ (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;
}
附件一种判断设备类型的方式(用于区分iphone、ipod、ipad):
if(UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPhone)
{
NSLog(@"这是iphone设备");
}
郑重声明:本站内容如果来自互联网及其他传播媒体,其版权均属原媒体及文章作者所有。转载目的在于传递更多信息及用于网络分享,并不代表本站赞同其观点和对其真实性负责,也不构成任何其他建议。