使用iOS SDK和完整的Cocoa Touch / Objective-C代码确定用户的设备 [英] Determine user's device using iOS SDK and full Cocoa Touch / Objective-C code

问题描述

我根据不同的来源编写了以下UIDevice类别。我已经升级了 platformCode 方法，所以它的水平低于可以看到的水平。

I have written the following UIDevice category based on different sources. I've upgraded the platformCode method so it's less low-level than can be seen.

这很有效，但 platformCode 方法是低级别的。你知道这种呼叫是否可以用Cocoa Touch代码替换？以下是相关代码：

This works perfectly, but the platformCode method is low level. Do you know if this kind of call can be replaced with Cocoa Touch code? Here's the relevant code:

UIDevice_enhanced.h

@interface UIDevice (Enhanced)

typedef enum {
    kUnknownPlatform = 0,
    kiPhone1G,
    kiPhone3G,
    kiPhone3GS,
    kiPhone4,
    kiPhone4Verizon,
    kiPhone4S,
    kiPodTouch1G,
    kiPodTouch2G,
    kiPodTouch3G,
    kiPodTouch4G,
    kiPad,
    kiPad2Wifi,
    kiPad2GSM,
    kiPad2CMDA,
    kSimulator
} PlatformType;

- (NSString *) platformName;
- (PlatformType) platform;

@end

UIDevice_enhanced.m

#import "UIDevice_enhanced.h"
#include <sys/utsname.h>

@interface UIDevice (Enhanced)
- (NSString *) platformCode;
@end 


@implementation UIDevice (Enhanced)

// Utility method (private)
- (NSString*) platformCode {
    struct utsname systemInfo;
    uname(&systemInfo);
    NSString* platform =  [NSString stringWithCString:systemInfo.machine encoding:NSUTF8StringEncoding];

    return platform;
}

// Public method to use
- (NSString*) platformName {
    NSString* platform = [self platformCode];

    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 @"Verizon iPhone 4";
    if ([platform isEqualToString:@"iPhone4,1"])    return @"iPhone 4S";
    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:@"iPad1,1"])      return @"iPad";
    if ([platform isEqualToString:@"iPad2,1"])      return @"iPad 2 (WiFi)";
    if ([platform isEqualToString:@"iPad2,2"])      return @"iPad 2 (GSM)";
    if ([platform isEqualToString:@"iPad2,3"])      return @"iPad 2 (CDMA)";
    if ([platform isEqualToString:@"i386"])         return @"Simulator";

    return platform;
}

// Public method to use
- (PlatformType) platform {
    NSString *platform = [self platformCode];
    if ([platform isEqualToString:@"iPhone1,1"])    return kiPhone1G;
    if ([platform isEqualToString:@"iPhone1,2"])    return kiPhone3G;
    if ([platform isEqualToString:@"iPhone2,1"])    return kiPhone3GS;
    if ([platform isEqualToString:@"iPhone3,1"])    return kiPhone4;
    if ([platform isEqualToString:@"iPhone3,2"])    return kiPhone4Verizon;
    if ([platform isEqualToString:@"iPhone4,1"])    return kiPhone4S;
    if ([platform isEqualToString:@"iPod1,1"])      return kiPodTouch1G;
    if ([platform isEqualToString:@"iPod2,1"])      return kiPodTouch2G;
    if ([platform isEqualToString:@"iPod3,1"])      return kiPodTouch3G;
    if ([platform isEqualToString:@"iPod4,1"])      return kiPodTouch4G;
    if ([platform isEqualToString:@"iPad1,1"])      return kiPad;
    if ([platform isEqualToString:@"iPad2,1"])      return kiPad2Wifi;
    if ([platform isEqualToString:@"iPad2,2"])      return kiPad2GSM;
    if ([platform isEqualToString:@"iPad2,3"])      return kiPad2CMDA;
    if ([platform isEqualToString:@"i386"])         return kSimulator;

    return kUnknownPlatform;
}

@end

推荐答案

这可以被视为目标-c方式：

This may be considered an "objective-c" way of doing it:

// Utility method (private)
- (NSString *)platformCode {
    // This may or not be necessary 
    // Im not sure if you can have a device thats not currentDevice can you?
    // if ([self isEqual:[UIDevice currentDevice]]) {

    NSString* platform =  [[self.systemName copy] autorelease];
    return platform;

    // Could probably shorten to just
    // return [[self.systemName copy] autorelease];

    // or - return [NSString stringWithString:self.systemName];
}

这是 utsname <的对象版本/ code> machine （来自此行： NSString * platform = [NSString stringWithCString：systemInfo.machine encoding：NSUTF8StringEncoding]; ）。

utsname ：

< sys / utsname.h> header定义结构utsname，其中包括
至少以下成员：

The <sys/utsname.h> header defines structure utsname, which includes at least the following members:

char sysname []   此操作系统实现的名称

char nodename [] 依赖于实现的
通信网络中此节点的名称

char release []    $当前版本级别b $ b此实现

char版本[]   此
版本的当前版本级别

char machine []   运行
系统的硬件类型名称

char sysname[]   name of this implementation of the operating system
char nodename[] name of this node within an implementation-dependent communications network
char release[]   current release level of this implementation
char version[]   current version level of this release
char machine[]   name of the hardware type on which the system is running

UIDevice 类参考：

systemName 接收器代表的设备
上运行的操作系统的名称。 （只读）

@property（非原子，只读，保留）NSString * system

---

但是，由于systemName只返回@iPhone OS，要获取实际的设备型号，必须使用c代码。这是另一种方法：

---

But, since systemName only returns @"iPhone OS", to get the actual device model number, you have to use c code. Here's another way to do it:

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

- (NSString *)machine {
     size_t size;

    // Set 'oldp' parameter to NULL to get the size of the data
    // returned so we can allocate appropriate amount of space
    sysctlbyname("hw.machine", NULL, &size, NULL, 0); 

    // Allocate the space to store name
    char *name = malloc(size);

    // Get the platform name
    sysctlbyname("hw.machine", name, &size, NULL, 0);

    // Place name into a string
    NSString *machine = [NSString stringWithCString:name];

    // Done with this
    free(name);

    return machine;
}

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