ios - UIDevice currentDevice 模型可能的值

Question

返回的所有可能值是[[UIDevice currentDevice] model];多少？它没有 记录在案。

Answer 1

可能的值是iPod touch, iPhone, iPhone Simulator, iPad,iPad Simulator

如果您想知道哪些硬件iOS正在损坏，例如，iPhone3下面是该代码iPhone4iPhone5

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注意：下面的代码可能不包含所有设备的字符串，我和其他人在GitHub 上维护相同的代码，所以请从那里获取最新的代码

目标-C：GitHub/DeviceUtil

斯威夫特：GitHub/DeviceGuru

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

- (NSString*)hardwareDescription {
    NSString *hardware = [self hardwareString];
    if ([hardware isEqualToString:@"iPhone1,1"]) return @"iPhone 2G";
    if ([hardware isEqualToString:@"iPhone1,2"]) return @"iPhone 3G";
    if ([hardware isEqualToString:@"iPhone3,1"]) return @"iPhone 4";
    if ([hardware isEqualToString:@"iPhone4,1"]) return @"iPhone 4S";
    if ([hardware isEqualToString:@"iPhone5,1"]) return @"iPhone 5";
    if ([hardware isEqualToString:@"iPod1,1"]) return @"iPodTouch 1G";
    if ([hardware isEqualToString:@"iPod2,1"]) return @"iPodTouch 2G";
    if ([hardware isEqualToString:@"iPad1,1"]) return @"iPad";
    if ([hardware isEqualToString:@"iPad2,6"]) return @"iPad Mini";
    if ([hardware isEqualToString:@"iPad4,1"]) return @"iPad Air WIFI";
    //there are lots of other strings too, checkout the github repo
    //link is given at the top of this answer

    if ([hardware isEqualToString:@"i386"]) return @"Simulator";
    if ([hardware isEqualToString:@"x86_64"]) return @"Simulator";

    return nil;
}

- (NSString*)hardwareString {
    size_t size = 100;
    char *hw_machine = malloc(size);
    int name[] = {CTL_HW,HW_MACHINE};
    sysctl(name, 2, hw_machine, &size, NULL, 0);
    NSString *hardware = [NSString stringWithUTF8String:hw_machine];
    free(hw_machine);
    return hardware;
}

Answer 2

我刚刚在 iPod Touch、iPhone、Phone Retina、iPhone 5、iPad、iPad Retina 和 iPad Mini 上进行了测试。所以这是我的结论：

iPod touch
iPhone
iPad

在模拟器上 - 如果您是一名开发人员，开发有时在模拟器上根本不起作用的功能，这可能很有用 - 您将获得以下值：

iPhone Simulator
iPad Simulator

Answer 3

我相信解释的最佳答案（这里没有写的东西）是说值本身是一个字符串值。可能的答案是字符串，例如：“iPhone”、“iPad”等。

Answer 4

这些答案都不适用于新型号。这是一个枚举：

public enum DeviceType {
 case iPad(String?)
 case iPhone(String?)
 case simulator(String?)
 case appleTV(String?)
 case unknown
}

我写的扩展我认为在新型号出现时更简洁和更可扩展。

extension UIDevice {
    public static func getDevice() -> DeviceType {
        var info = utsname()
        uname(&info)
        let machineMirror = Mirror(reflecting: info.machine)
        let code = machineMirror.children.reduce("") { identifier, element in
            guard let value = element.value as? Int8, value != 0 else {
                return identifier
            }
            return identifier + String(UnicodeScalar(UInt8(value)))
        }

        if code.lowercased().range(of: "ipad") != nil {
            if let range = code.lowercased().range(of: "ipad") {
                var mutate = code
                mutate.removeSubrange(range)
                return .iPad(mutate)
            }else{
                return .iPad(nil)
            }
        }else if code.lowercased().range(of: "iphone") != nil {
            if let range = code.lowercased().range(of: "iphone") {
                var mutate = code
                mutate.removeSubrange(range)
                return .iPhone(mutate)
            }else{
                return .iPhone(nil)
            }
        }else if code.lowercased().range(of: "i386") != nil || code.lowercased().range(of: "x86_64") != nil{
            return .simulator(code)
        }else if code.lowercased().range(of: "appletv") != nil {
            if let range = code.lowercased().range(of: "appletv") {
                var mutate = code
                mutate.removeSubrange(range)
                return .appleTV(mutate)
            }else{
                return .appleTV(nil)
            }
        }else{
            return .unknown
        }
    }
}