如何检查 iPad 是否为 iPad Pro

Question

新的 iPad pro 具有不同的尺寸和分辨率。如果我根据屏幕宽度检查是否正确？我还没有升级到 Xcode 7.1，我也没有设备，所以我还不能检查它。这个检查有用吗？

if([UIScreen mainScreen].bounds.size.width>1024)
    {
        // iPad is an iPad Pro
    }

Answer 1

按照以下步骤检查

if([[[UIDevice currentDevice] name] isEqualToString:@"iPad Pro"])
{
  //  do your stuff
}

Answer 2

你可以用这个

#define IS_IPAD (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPad)
#define SCREEN_WIDTH ([[UIScreen mainScreen] bounds].size.width)
#define SCREEN_HEIGHT ([[UIScreen mainScreen] bounds].size.height)
#define IS_IPAD_PRO_1366 (IS_IPAD && MAX(SCREEN_WIDTH,SCREEN_HEIGHT) == 1366.0)
#define IS_IPAD_PRO_1024 (IS_IPAD && MAX(SCREEN_WIDTH,SCREEN_HEIGHT) == 1024.0)

然后

 if (IS_IPAD_PRO_1366) {
    NSLog(@"It is ipad pro 1366");
  }

Answer 3

试试这个库：https://github.com/fahrulazmi/UIDeviceHardware

那么你的代码应该是：

NSString *platform = [UIDeviceHardware platformString];
if ([platform isEqualToString:@"iPad6,7"] || [platform isEqualToString:@"iPad6,8"]) {
     // iPad is an iPad Pro
}

或者这个更强大的库：https://github.com/InderKumarRathore/DeviceUtil

该解决方案不适用于模拟器。我想检查模拟器的设备类型，看来你必须检查屏幕大小。

Answer 4

+(BOOL) isIpad_1024
{

    if ([UIScreen mainScreen].bounds.size.height == 1024) {
        return  YES;
    }
    return NO;
}

+(BOOL) isIpadPro_1366
{

    if ([UIScreen mainScreen].bounds.size.height == 1366) {
        return  YES;
    }
    return NO;
}

Answer 5

如 HAS 在 their answer here 中所述，在您的代码中添加此扩展程序：

public extension UIDevice {
    var modelName: String {
        var systemInfo = utsname()
        uname(&systemInfo)
        let machineMirror = Mirror(reflecting: systemInfo.machine)
        let identifier = machineMirror.children.reduce("") { identifier, element in
            guard let value = element.value as? Int8 where value != 0 else { return identifier }
            return identifier + String(UnicodeScalar(UInt8(value)))
        }

        switch identifier {
        case "iPod5,1":                                 return "iPod Touch 5"
        case "iPod7,1":                                 return "iPod Touch 6"
        case "iPhone3,1", "iPhone3,2", "iPhone3,3":     return "iPhone 4"
        case "iPhone4,1":                               return "iPhone 4s"
        case "iPhone5,1", "iPhone5,2":                  return "iPhone 5"
        case "iPhone5,3", "iPhone5,4":                  return "iPhone 5c"
        case "iPhone6,1", "iPhone6,2":                  return "iPhone 5s"
        case "iPhone7,2":                               return "iPhone 6"
        case "iPhone7,1":                               return "iPhone 6 Plus"
        case "iPhone8,1":                               return "iPhone 6s"
        case "iPhone8,2":                               return "iPhone 6s Plus"
        case "iPad2,1", "iPad2,2", "iPad2,3", "iPad2,4":return "iPad 2"
        case "iPad3,1", "iPad3,2", "iPad3,3":           return "iPad 3"
        case "iPad3,4", "iPad3,5", "iPad3,6":           return "iPad 4"
        case "iPad4,1", "iPad4,2", "iPad4,3":           return "iPad Air"
        case "iPad5,1", "iPad5,3", "iPad5,4":           return "iPad Air 2"
        case "iPad2,5", "iPad2,6", "iPad2,7":           return "iPad Mini"
        case "iPad4,4", "iPad4,5", "iPad4,6":           return "iPad Mini 2"
        case "iPad4,7", "iPad4,8", "iPad4,9":           return "iPad Mini 3"
        case "iPad5,1", "iPad5,2":                      return "iPad Mini 4"
        case "iPad6,7", "iPad6,8":                      return "iPad Pro"
        case "i386", "x86_64":                          return "Simulator"
        default:                                        return identifier
        }
    }
}

并进行检查

if(UIDevice.currentDevice().modelName == "iPad Pro"){//Your code}

Answer 6

您可以使用此代码：

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

- (BOOL) isIpadPro{
    size_t size;
    sysctlbyname("hw.machine", NULL, &size, NULL, 0);
    char *machine = malloc(size);
    sysctlbyname("hw.machine", machine, &size, NULL, 0);
    NSString *platform = [NSString stringWithUTF8String:machine];
    free(machine);

    if ([platform isEqualToString:@"iPad6,8"])
        return YES;

    return NO;
}

Answer 7

你们复杂的答案是在开玩笑吗？

if([UIScreen mainScreen].bounds.size.width >= 1024) {
    // iPad pro (or hypothetical/future huge-screened iOS device)
} else {
    // not iPad pro
}

如果你只是做一个 >= 符号而不是 > 符号，它会非常有效。

（好吧，我知道我不应该对你详尽、具体的答案不屑一顾。当然，有时候特定设备比屏幕尺寸更重要。但为了快速、明显的答案......！）

Answer 8

iPad Pro 有一个错误，导致它目前的 webview 用户代理错误。用户代理看起来像这样：

Mozilla/5.0 (iPhone; CPU iPhone OS9_1 喜欢 Mac OS X) AppleWebKit/601.1.46(KHTML，像 Gecko)Mobile/13B143

我认为我们可以使用此错误在兼容模式下检测 iPad Pro for apps 运行。

-(BOOL)isiPadPro;
{
    UIWebView* webView = [[UIWebView alloc] initWithFrame:CGRectZero];
    NSString* userAgent = [webView stringByEvaluatingJavaScriptFromString:@"navigator.userAgent"];
    return [userAgent containsString:@"iPhone"] && ([[UIDevice currentDevice] userInterfaceIdiom] == UIUserInterfaceIdiomPad);
}

Answer 9

到目前为止，这个宏似乎没有任何问题。

#define IS_IPAD_PRO (MAX([[UIScreen mainScreen]bounds].size.width,[[UIScreen mainScreen] bounds].size.height) > 1024)

Answer 10

这个宏在横向和纵向都有效：

#define IS_IPAD_PRO_12_INCH (([UIScreen mainScreen].bounds.size.width == 1366 && [UIScreen mainScreen].bounds.size.height == 1024) || ([UIScreen mainScreen].bounds.size.width == 1024 && [UIScreen mainScreen].bounds.size.height == 1366))

Answer 11

当我在 Xcode 中的模拟器中进行测试时，这些解决方案中有 8 none 个有效。

诀窍是寻找 "nativeBounds" 尺寸高度，否则您将在模拟器中继续获得 1024 作为高度

#define iPadPro12 (UIDevice.currentDevice.userInterfaceIdiom == UIUserInterfaceIdiomPad && UIScreen.mainScreen.nativeBounds.size.height > 1024)

if (iPadPro12)
{
  //its ipad Pro 12.9 inch screen
}

Answer 12

SWIFT

这是所有 swift 喜欢的公认答案。

let isIpadPro:Bool = max(UIScreen.main.bounds.size.width, UIScreen.main.bounds.size.height) > 1024

Answer 13

我的全套设备检测。

#define IS_IPHONE (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPhone)
#define IS_RETINA ([[UIScreen mainScreen] scale] >= 2.0)

#define SCREEN_WIDTH ([[UIScreen mainScreen] bounds].size.width)
#define SCREEN_HEIGHT ([[UIScreen mainScreen] bounds].size.height)
#define SCREEN_MAX_LENGTH (MAX(SCREEN_WIDTH, SCREEN_HEIGHT))
#define SCREEN_MIN_LENGTH (MIN(SCREEN_WIDTH, SCREEN_HEIGHT))

#define IS_IPHONE_4_OR_LESS (IS_IPHONE && SCREEN_MAX_LENGTH < 568.0)
#define IS_IPHONE_5 (IS_IPHONE && SCREEN_MAX_LENGTH == 568.0)
#define IS_IPHONE_6 (IS_IPHONE && SCREEN_MAX_LENGTH == 667.0)
#define IS_IPHONE_6P (IS_IPHONE && SCREEN_MAX_LENGTH == 736.0)
#define IS_IPHONE_X (IS_IPHONE && SCREEN_MAX_LENGTH == 812.0)

#define IS_IPAD_PRO_97 (IS_IPAD && SCREEN_MAX_LENGTH == 1024.0)
#define IS_IPAD_PRO_105 (IS_IPAD && SCREEN_MAX_LENGTH == 1112.0)
#define IS_IPAD_PRO_129 (IS_IPAD && SCREEN_MAX_LENGTH == 1366.0)

Answer 14

您可以在 userAgent

中使用正则表达式来检测 iPad

var isIPadPro = /Macintosh/.test(navigator.userAgent) && 'ontouchend' in document;

如何检查 iPad 是否为 iPad Pro

How to check if iPad is iPad Pro

objective-c

ipad

ios

swift

xcode7