2010-07-05 110 views
11

如何检测用户是否在iPhone 4或3G/3GS上运行应用程序?iPhone - 如何检测iPhone版本?

我需要检测硬件,而不是iOS版本。

感谢您的任何帮助。

+0

您可以拨打[currentDevice](http://developer.apple.com/iphone/library/documentation/uikit/reference/UIDevice_Class/Reference/UIDevice.html#//apple_ref/doc/uid/TP40006902 -CH3-SW10)在UIDevice上查看模型属性。 **编辑:**虽然...文档建议这不包括确切的型号。 – 2010-07-05 07:09:17

回答

11

随意使用这个类 - 我发现它here

使用

UIDeviceHardware *h=[[UIDeviceHardware alloc] init]; 
[self setDeviceModel:[h platformString]]; 
[h release]; 

UIDeviceHardware.h

// 
// UIDeviceHardware.h 
// 
// Used to determine EXACT version of device software is running on. 

#import <Foundation/Foundation.h> 

@interface UIDeviceHardware : NSObject 

- (NSString *) platform; 
- (NSString *) platformString; 

@end 

UIDeviceHardware.m

// 
// UIDeviceHardware.m 
// 
// Used to determine EXACT version of device software is running on. 

#import "UIDeviceHardware.h" 
#include <sys/types.h> 
#include <sys/sysctl.h> 

@implementation UIDeviceHardware 

- (NSString *) platform{ 
    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]; 
    free(machine); 
    return platform; 
} 

- (NSString *) platformString{ 
    NSString *platform = [self platform]; 
    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:@"iPod1,1"]) return @"iPod Touch 1G"; 
    if ([platform isEqualToString:@"iPod2,1"]) return @"iPod Touch 2G"; 
    if ([platform isEqualToString:@"i386"]) return @"iPhone Simulator"; 
    return platform; 
} 

@end 
+1

比我的回答要好得多,但有人需要在iPhone 4上运行此功能,并查看该机型的机器字符串。 我认为这可能是“iPhone3,1”。 – 2010-07-05 07:28:07

+0

+1好点。我现在太累了,但我会明天尝试并发布结果 - 除非有人打我 – SeniorShizzle 2010-07-05 07:38:18

+0

我只是使用hw.machine(和hw.model)。 iPhone1,1也被称为“原始iPhone”或“iPhone 2G”(非官方),Apple列出了“iPod Touch第二代”和“iPod Touch第三代”,其中一个在此处不存在(可能是iPod2,2), – 2010-07-05 12:01:34

6
#import <sys/utsname.h> 

+ (NSString*)deviceModelName { 

     struct utsname systemInfo; 

     uname(&systemInfo); 

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

     if([modelName isEqualToString:@"i386"]) { 
      modelName = @"iPhone Simulator"; 
     } 
     else if([modelName isEqualToString:@"iPhone1,1"]) { 
      modelName = @"iPhone"; 
     } 
     else if([modelName isEqualToString:@"iPhone1,2"]) { 
      modelName = @"iPhone 3G"; 
     } 
     else if([modelName isEqualToString:@"iPhone2,1"]) { 
      modelName = @"iPhone 3GS"; 
     } 
     else if([modelName isEqualToString:@"iPhone3,1"]) { 
      modelName = @"iPhone 4"; 
     } 
     else if([modelName isEqualToString:@"iPhone4,1"]) { 
      modelName = @"iPhone 4S"; 
     } 
     else if([modelName isEqualToString:@"iPod1,1"]) { 
      modelName = @"iPod 1st Gen"; 
     } 
     else if([modelName isEqualToString:@"iPod2,1"]) { 
      modelName = @"iPod 2nd Gen"; 
     } 
     else if([modelName isEqualToString:@"iPod3,1"]) { 
      modelName = @"iPod 3rd Gen"; 
     } 
     else if([modelName isEqualToString:@"iPad1,1"]) { 
      modelName = @"iPad"; 
     } 
     else if([modelName isEqualToString:@"iPad2,1"]) { 
      modelName = @"iPad 2(WiFi)"; 
     } 
     else if([modelName isEqualToString:@"iPad2,2"]) { 
      modelName = @"iPad 2(GSM)"; 
     } 
     else if([modelName isEqualToString:@"iPad2,3"]) { 
      modelName = @"iPad 2(CDMA)"; 
     } 

     return modelName; 
    } 
+0

您的代码缺少iPhone 5 – SpaceDog 2012-09-20 15:59:12

+0

这很简单...在最后一个'if else'之后,您可以添加其他{modelName = @“iPhone 5”;} – 2012-09-21 05:38:20

+1

如果没有'if'则不要使用'else' iPhone 5.任何未知设备都会错误地回退到称为iPhone 5. – kirb 2012-12-10 14:53:03

1

您可以使用此代码检测iPhone操作系统版本。 float version = [[[UIDevice currentDevice] systemVersion] floatValue];

if (version >= 3.0) { 
    // Only executes on version 3 or above. 
}