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
我喜欢埃里卡·萨顿的作品。她包括苹果电视设备和其他你可能想不到的设备。
https://github.com/erica/uidevice-extension/blob/master/UIDevice-Hardware.h
其他回答
NSString *deviceType = [UIDevice currentDevice].model;
- (BOOL)deviceiPhoneOriPod
{
NSString *deviceType = [UIDevice currentDevice].model;
if([deviceType rangeOfString:@"iPhone"].location!=NSNotFound)
return YES;
else
return NO;
}
我知道答案已经被勾选了,但为了将来的参考,你可以总是使用设备屏幕大小来找出它是哪台设备:
if ([[UIDevice currentDevice] userInterfaceIdiom] == UIUserInterfaceIdiomPhone) {
CGSize result = [[UIScreen mainScreen] bounds].size;
if (result.height == 480) {
// 3.5 inch display - iPhone 4S and below
NSLog(@"Device is an iPhone 4S or below");
}
else if (result.height == 568) {
// 4 inch display - iPhone 5
NSLog(@"Device is an iPhone 5/S/C");
}
else if (result.height == 667) {
// 4.7 inch display - iPhone 6
NSLog(@"Device is an iPhone 6");
}
else if (result.height == 736) {
// 5.5 inch display - iPhone 6 Plus
NSLog(@"Device is an iPhone 6 Plus");
}
}
else if ([[UIDevice currentDevice] userInterfaceIdiom] == UIUserInterfaceIdiomPad) {
// iPad 9.7 or 7.9 inch display.
NSLog(@"Device is an iPad.");
}
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;
}
添加到Arash的代码,我不关心我的应用程序使用什么模型,我只想知道什么样的设备,所以,我可以测试如下:
if (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPad)
{
NSLog(@"I'm definitely an iPad");
} else {
NSString *deviceType = [UIDevice currentDevice].model;
if([deviceType rangeOfString:@"iPhone"].location!=NSNotFound)
{
NSLog(@"I must be an iPhone");
} else {
NSLog(@"I think I'm an iPod");
}
}
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