我刚刚升级到Xcode 4.5 GM,发现你现在可以将“4英寸视网膜”大小应用到故事板中的视图控制器上。
现在,如果我想创建一个同时在iPhone 4和5上运行的应用程序,当然我必须构建每个窗口两次,但我还必须检测用户的iPhone屏幕是3.5英寸还是4英寸,然后应用视图。
我该怎么做呢?
我刚刚升级到Xcode 4.5 GM,发现你现在可以将“4英寸视网膜”大小应用到故事板中的视图控制器上。
现在,如果我想创建一个同时在iPhone 4和5上运行的应用程序,当然我必须构建每个窗口两次,但我还必须检测用户的iPhone屏幕是3.5英寸还是4英寸,然后应用视图。
我该怎么做呢?
当前回答
这是我们的代码,在ios7/ios8上测试通过,适用于iphone4、iphone5、ipad、iphone6、iphone6p,无论在设备上还是模拟器上:
#define IS_IPAD (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPad)
#define IS_IPHONE (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPhone) // iPhone and iPod touch style UI
#define IS_IPHONE_5_IOS7 (IS_IPHONE && [[UIScreen mainScreen] bounds].size.height == 568.0f)
#define IS_IPHONE_6_IOS7 (IS_IPHONE && [[UIScreen mainScreen] bounds].size.height == 667.0f)
#define IS_IPHONE_6P_IOS7 (IS_IPHONE && [[UIScreen mainScreen] bounds].size.height == 736.0f)
#define IS_IPHONE_4_AND_OLDER_IOS7 (IS_IPHONE && [[UIScreen mainScreen] bounds].size.height < 568.0f)
#define IS_IPHONE_5_IOS8 (IS_IPHONE && ([[UIScreen mainScreen] nativeBounds].size.height/[[UIScreen mainScreen] nativeScale]) == 568.0f)
#define IS_IPHONE_6_IOS8 (IS_IPHONE && ([[UIScreen mainScreen] nativeBounds].size.height/[[UIScreen mainScreen] nativeScale]) == 667.0f)
#define IS_IPHONE_6P_IOS8 (IS_IPHONE && ([[UIScreen mainScreen] nativeBounds].size.height/[[UIScreen mainScreen] nativeScale]) == 736.0f)
#define IS_IPHONE_4_AND_OLDER_IOS8 (IS_IPHONE && ([[UIScreen mainScreen] nativeBounds].size.height/[[UIScreen mainScreen] nativeScale]) < 568.0f)
#define IS_IPHONE_5 ( ( [ [ UIScreen mainScreen ] respondsToSelector: @selector( nativeBounds ) ] ) ? IS_IPHONE_5_IOS8 : IS_IPHONE_5_IOS7 )
#define IS_IPHONE_6 ( ( [ [ UIScreen mainScreen ] respondsToSelector: @selector( nativeBounds ) ] ) ? IS_IPHONE_6_IOS8 : IS_IPHONE_6_IOS7 )
#define IS_IPHONE_6P ( ( [ [ UIScreen mainScreen ] respondsToSelector: @selector( nativeBounds ) ] ) ? IS_IPHONE_6P_IOS8 : IS_IPHONE_6P_IOS7 )
#define IS_IPHONE_4_AND_OLDER ( ( [ [ UIScreen mainScreen ] respondsToSelector: @selector( nativeBounds ) ] ) ? IS_IPHONE_4_AND_OLDER_IOS8 : IS_IPHONE_4_AND_OLDER_IOS7 )
其他回答
这个问题已经得到了上百次的回答,但这个解决方案对我来说是最好的,并且在引入新设备时帮助解决了这个问题,而我没有定义一个大小。
Swift 5助手:
extension UIScreen {
func phoneSizeInInches() -> CGFloat {
switch (self.nativeBounds.size.height) {
case 960, 480:
return 3.5 //iPhone 4
case 1136:
return 4 //iPhone 5
case 1334:
return 4.7 //iPhone 6
case 2208:
return 5.5 //iPhone 6 Plus
case 2436:
return 5.8 //iPhone X
case 1792:
return 6.1 //iPhone XR
case 2688:
return 6.5 //iPhone XS Max
default:
let scale = self.scale
let ppi = scale * 163
let width = self.bounds.size.width * scale
let height = self.bounds.size.height * scale
let horizontal = width / ppi, vertical = height / ppi
let diagonal = sqrt(pow(horizontal, 2) + pow(vertical, 2))
return diagonal
}
}
}
这是因为记住手机的英寸大小很容易,比如“5.5英寸”或“4.7英寸”,但很难记住准确的像素大小。
if UIScreen.main.phoneSizeInInches() == 4 {
//do something with only 4 inch iPhones
}
这也给了你这样做的机会:
if UIScreen.main.phoneSizeInInches() < 5.5 {
//do something on all iPhones smaller than the plus
}
默认值:尝试使用屏幕大小和比例来尝试计算对角线英寸。这是为了防止出现一些新的设备大小,它将尽力确定和代码,如最后一个例子,应该仍然工作。
CGFloat height = [UIScreen mainScreen].bounds.size.height;
NSLog(@"screen soze is %f",height);
if (height>550) {
// 4" screen-do some thing
}
else if (height<500) {
// 3.5 " screen- do some thing
}
用于检测所有版本的iPhone和iPad设备。
#define IS_IPAD (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPad)
#define IS_IPHONE (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPhone)
#define IS_IPHONE_5 (IS_IPHONE && [[UIScreen mainScreen] bounds].size.height == 568.0)
#define IS_IPHONE_6 (IS_IPHONE && [[UIScreen mainScreen] bounds].size.height == 667.0)
#define IS_IPHONE_6_PLUS (IS_IPHONE && [[UIScreen mainScreen] bounds].size.height == 736.0)
#define IS_RETINA ([[UIScreen mainScreen] scale] == 2.0)
在Swift 3中,你可以使用我的简单类KRDeviceType。
https://github.com/ulian-onua/KRDeviceType
它有很好的文档,并支持运算符==,>=,<=。
例如,要检测设备是否有iPhone 6/6s/7的边界,你可以使用next比较:
if KRDeviceType() == .iPhone6 {
// Perform appropiate operations
}
要检测设备是否有iPhone 5/5S/SE或更早(iPhone 4s)的边界,您可以使用下一个比较:
if KRDeviceType() <= .iPhone5 { //iPhone 5/5s/SE of iPhone 4s
// Perform appropiate operations (for example, set up constraints for those old devices)
}
我用hfossli的答案翻译给了Swift
let IS_IPAD = UIDevice.currentDevice().userInterfaceIdiom == .Pad
let IS_IPHONE = UIDevice.currentDevice().userInterfaceIdiom == .Phone
let IS_RETINA = UIScreen.mainScreen().scale >= 2.0
let SCREEN_WIDTH = UIScreen.mainScreen().bounds.size.width
let SCREEN_HEIGHT = UIScreen.mainScreen().bounds.size.height
let SCREEN_MAX_LENGTH = max(SCREEN_WIDTH, SCREEN_HEIGHT)
let SCREEN_MIN_LENGTH = min(SCREEN_WIDTH, SCREEN_HEIGHT)
let IS_IPHONE_4_OR_LESS = (IS_IPHONE && SCREEN_MAX_LENGTH < 568.0)
let IS_IPHONE_5 = (IS_IPHONE && SCREEN_MAX_LENGTH == 568.0)
let IS_IPHONE_6 = (IS_IPHONE && SCREEN_MAX_LENGTH == 667.0)
let IS_IPHONE_6P = (IS_IPHONE && SCREEN_MAX_LENGTH == 736.0)