如何检查iPad是否是iPad Pro

Question

新的iPad专业版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？我没有升级到Xcode 7.1也没有设备,所以我还没有检查它.这项检查有效吗？

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

Answer 1

+(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 2

你可以用它

#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

正如HAS在其答案中所述,在您的代码中添加此扩展:

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
        }
    }
}

并检查

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

Answer 4

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)