如何檢查iPad是否爲iPad Pro

Question

新款iPad Pro具有不同的尺寸和分辨率。如果我根據屏幕寬度檢查它是否正確？我沒有升級到Xcode 7.1，也沒有設備，所以我不能檢查它。這個檢查工作嗎？

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

Answer 1

到目前爲止，這個宏似乎沒有任何問題。

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

Answer 2

按照下面的步驟檢查

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

Answer 3

你可以使用這個

#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) 

然後

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

Answer 4

試試這個庫：https://github.com/fahrulazmi/UIDeviceHardware

那麼你的代碼應該是：

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

還是這個更強大的庫：https://github.com/InderKumarRathore/DeviceUtil

該解決方案不能爲仿真工作。我想檢查模擬器的設備類型，看起來像你必須檢查屏幕大小。

Answer 5

+(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 6

如前所述由their answer here了，在代碼中添加此擴展：

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 
     } 
    } 
} 

而對於檢查

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

Answer 7

您可以使用此代碼：

#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 8

你們是在開玩笑嗎？你的複雜答案？

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

如果你只是做了>>符號而不是>符號，它將工作得非常好。

（好的，我知道我不應該對你的徹底的具體答案不屑一顧，當然有時候特定的設備比屏幕尺寸更重要，但對於快速明顯的答案......）

Answer 9

iPad Pro存在一個bug，使得它目前在webview中有一個錯誤的useragent。 用戶代理看起來喜歡這樣的：

的Mozilla/5.0（iPhone; CPU iPhone OS9_1，例如Mac OS X）爲AppleWebKit/601.1.46（KHTML，例如Gecko）手機/ 13B143

我認爲，我們可以用這個用於檢測運行在兼容模式下的應用程序的iPad Pro的缺陷。

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

Answer 10

在橫向和縱向此宏的工作方式：

#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

當我在模擬器中進行測試的Xcode 8這些解決方案的工作。

訣竅是尋找「nativeBounds」大小HIGHT否則，你將繼續得到1024作爲高度仿真

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

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