是否有方法确定正在运行应用程序的设备。如果可能的话，我想区分iPhone和iPod Touch。
当前回答
我在我的应用程序中使用这个。截至2012年12月。
- (NSString *) platformString {
// Gets a string with the device model
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 encoding:NSUTF8StringEncoding];
free(machine);
if ([platform isEqualToString:@"iPhone1,1"]) return @"iPhone 2G";
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";
if ([platform isEqualToString:@"iPhone3,3"]) return @"iPhone 4 (CDMA)";
if ([platform isEqualToString:@"iPhone4,1"]) return @"iPhone 4S";
if ([platform isEqualToString:@"iPhone5,1"]) return @"iPhone 5";
if ([platform isEqualToString:@"iPhone5,2"]) return @"iPhone 5 (GSM+CDMA)";
if ([platform isEqualToString:@"iPod1,1"]) return @"iPod Touch (1 Gen)";
if ([platform isEqualToString:@"iPod2,1"]) return @"iPod Touch (2 Gen)";
if ([platform isEqualToString:@"iPod3,1"]) return @"iPod Touch (3 Gen)";
if ([platform isEqualToString:@"iPod4,1"]) return @"iPod Touch (4 Gen)";
if ([platform isEqualToString:@"iPod5,1"]) return @"iPod Touch (5 Gen)";
if ([platform isEqualToString:@"iPad1,1"]) return @"iPad";
if ([platform isEqualToString:@"iPad1,2"]) return @"iPad 3G";
if ([platform isEqualToString:@"iPad2,1"]) return @"iPad 2 (WiFi)";
if ([platform isEqualToString:@"iPad2,2"]) return @"iPad 2";
if ([platform isEqualToString:@"iPad2,3"]) return @"iPad 2 (CDMA)";
if ([platform isEqualToString:@"iPad2,4"]) return @"iPad 2";
if ([platform isEqualToString:@"iPad2,5"]) return @"iPad Mini (WiFi)";
if ([platform isEqualToString:@"iPad2,6"]) return @"iPad Mini";
if ([platform isEqualToString:@"iPad2,7"]) return @"iPad Mini (GSM+CDMA)";
if ([platform isEqualToString:@"iPad3,1"]) return @"iPad 3 (WiFi)";
if ([platform isEqualToString:@"iPad3,2"]) return @"iPad 3 (GSM+CDMA)";
if ([platform isEqualToString:@"iPad3,3"]) return @"iPad 3";
if ([platform isEqualToString:@"iPad3,4"]) return @"iPad 4 (WiFi)";
if ([platform isEqualToString:@"iPad3,5"]) return @"iPad 4";
if ([platform isEqualToString:@"iPad3,6"]) return @"iPad 4 (GSM+CDMA)";
if ([platform isEqualToString:@"i386"]) return @"Simulator";
if ([platform isEqualToString:@"x86_64"]) return @"Simulator";
return platform;
}
其他回答
下面提到的代码片段应该有帮助:
if ([[UIDevice currentDevice] userInterfaceIdiom] == UIUserInterfaceIdiomPhone) {
// iPhone device
}
else if ([[UIDevice currentDevice] userInterfaceIdiom] == UIUserInterfaceIdiomPad) {
// iPad device
}
else {
// Other device i.e. iPod
}
更有用
#include <sys/types.h>
#include <sys/sysctl.h>
@interface UIDevice(Hardware)
- (NSString *) platform;
- (BOOL)hasRetinaDisplay;
- (BOOL)hasMultitasking;
- (BOOL)hasCamera;
@end
@implementation UIDevice(Hardware)
- (NSString *) platform{
int mib[2];
size_t len;
char *machine;
mib[0] = CTL_HW;
mib[1] = HW_MACHINE;
sysctl(mib, 2, NULL, &len, NULL, 0);
machine = malloc(len);
sysctl(mib, 2, machine, &len, NULL, 0);
NSString *platform = [NSString stringWithCString:machine encoding:NSASCIIStringEncoding];
free(machine);
return platform;
}
- (BOOL)hasRetinaDisplay {
NSString *platform = [self platform];
BOOL ret = YES;
if ([platform isEqualToString:@"iPhone1,1"]) {
ret = NO;
}
else
if ([platform isEqualToString:@"iPhone1,2"]) ret = NO;
else
if ([platform isEqualToString:@"iPhone2,1"]) ret = NO;
else
if ([platform isEqualToString:@"iPod1,1"]) ret = NO;
else
if ([platform isEqualToString:@"iPod2,1"]) ret = NO;
else
if ([platform isEqualToString:@"iPod3,1"]) ret = NO;
return ret;
}
- (BOOL)hasMultitasking {
if ([self respondsToSelector:@selector(isMultitaskingSupported)]) {
return [self isMultitaskingSupported];
}
return NO;
}
- (BOOL)hasCamera {
BOOL ret = NO;
// check camera availability
return ret;
}
@end
您可以读取属性
NSLog(@"platform %@, retita %@, multitasking %@", [[UIDevice currentDevice] platform], [[UIDevice currentDevice] hasRetinaDisplay] ? @"YES" : @"NO" , [[UIDevice currentDevice] hasMultitasking] ? @"YES" : @"NO");
以下是新模型的小更新:
- (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:@"iPhone3,1"]) return @"iPhone 4";
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:@"i386"]) return @"iPhone Simulator";
return platform;
}
我进一步将大的“isEqualToString”块转换为设备类型、生成和逗号后的其他限定符(我称之为子生成)的位掩码分类。它被包装在一个类中，带有一个单例调用SGPlatform，避免了大量重复的字符串操作。代码可用https://github.com/danloughney/spookyGroup
这个类允许你做这样的事情:
if ([SGPlatform iPad] && [SGPlatform generation] > 3) {
// set for high performance
}
and
switch ([SGPlatform deviceMask]) {
case DEVICE_IPHONE:
break;
case DEVICE_IPAD:
break;
case DEVICE_IPAD_MINI:
break;
}
设备的分类在platformBits方法中。这个方法对于本文的读者来说应该非常熟悉。我根据设备的类型和年代对其进行了分类，因为我最感兴趣的是整体性能，但源代码可以调整以提供您感兴趣的任何分类，如视网膜屏幕、网络功能等。
基于以上非常好的答案，以下是我想到的。这与@Rodrigo的回答非常相似，但解决了@Oliver对该回答的评论所关注的问题。这还提供了在输出字符串中包含模型字符串的选项。
+ (NSString *) deviceModel {
size_t size;
sysctlbyname("hw.machine", NULL, &size, NULL, 0);
char *model = malloc(size);
sysctlbyname("hw.machine", model, &size, NULL, 0);
NSString *deviceModel = [NSString stringWithCString:model encoding:NSUTF8StringEncoding];
free(model);
return deviceModel;
}
+ (NSString *) deviceName {
NSString *deviceModel = [DeviceGateway deviceModel];
if ([deviceModel isEqual:@"i386"]) return @"Simulator"; //iPhone Simulator
if ([deviceModel isEqual:@"iPhone1,1"]) return @"iPhone1G"; //iPhone 1G
if ([deviceModel isEqual:@"iPhone1,2"]) return @"iPhone3G"; //iPhone 3G
if ([deviceModel isEqual:@"iPhone2,1"]) return @"iPhone3GS"; //iPhone 3GS
if ([deviceModel isEqual:@"iPhone3,1"]) return @"iPhone4"; //iPhone 4 - AT&T
if ([deviceModel isEqual:@"iPhone3,2"]) return @"iPhone4"; //iPhone 4 - Other carrier
if ([deviceModel isEqual:@"iPhone3,3"]) return @"iPhone4"; //iPhone 4 - Other carrier
if ([deviceModel isEqual:@"iPhone4,1"]) return @"iPhone4S"; //iPhone 4S
if ([deviceModel isEqual:@"iPod1,1"]) return @"iPod1stGen"; //iPod Touch 1G
if ([deviceModel isEqual:@"iPod2,1"]) return @"iPod2ndGen"; //iPod Touch 2G
if ([deviceModel isEqual:@"iPod3,1"]) return @"iPod3rdGen"; //iPod Touch 3G
if ([deviceModel isEqual:@"iPod4,1"]) return @"iPod4thGen"; //iPod Touch 4G
if ([deviceModel isEqual:@"iPad1,1"]) return @"iPadWiFi"; //iPad Wifi
if ([deviceModel isEqual:@"iPad1,2"]) return @"iPad3G"; //iPad 3G
if ([deviceModel isEqual:@"iPad2,1"]) return @"iPad2"; //iPad 2 (WiFi)
if ([deviceModel isEqual:@"iPad2,2"]) return @"iPad2"; //iPad 2 (GSM)
if ([deviceModel isEqual:@"iPad2,3"]) return @"iPad2"; //iPad 2 (CDMA)
NSString *aux = [[deviceModel componentsSeparatedByString:@","] objectAtIndex:0];
//If a newer version exists
if ([aux rangeOfString:@"iPhone"].location != NSNotFound) {
int version = [[aux stringByReplacingOccurrencesOfString:@"iPhone" withString:@""] intValue];
if (version == 3) return @"iPhone4";
if (version == 4) return @"iPhone4s";
return @"Newer iPhone";
}
if ([aux rangeOfString:@"iPod"].location != NSNotFound) {
int version = [[aux stringByReplacingOccurrencesOfString:@"iPod" withString:@""] intValue];
if (version == 4) return @"iPod4thGen";
return @"Newer iPod";
}
if ([aux rangeOfString:@"iPad"].location != NSNotFound) {
int version = [[aux stringByReplacingOccurrencesOfString:@"iPad" withString:@""] intValue];
if (version == 1) return @"iPad3G";
if (version == 2) return @"iPad2";
return @"Newer iPad";
}
//If none was found, send the original string
return deviceModel;
}
+ (NSString *) deviceNameWithDeviceModel:(BOOL)shouldIncludeDeviceModel {
if (shouldIncludeDeviceModel) {
return [NSString stringWithFormat:@"%@ (%@)", [DeviceGateway deviceName], [DeviceGateway deviceModel]];
}
return [DeviceGateway deviceName];
}
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