ios 检测视网膜显示屏
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Detect Retina Display
提问by Pierre Valade
Does iOS SDK provides an easy way to check if the currentDevice has an high-resolution display (retina) ?
iOS SDK 是否提供了一种简单的方法来检查 currentDevice 是否具有高分辨率显示器（视网膜）？
The best way I've found to do it now is :
我现在找到的最好的方法是：
if ([[UIScreen mainScreen] respondsToSelector:@selector(scale)] == YES && [[UIScreen mainScreen] scale] == 2.00) {
// RETINA DISPLAY
}
回答by sickp
In order to detect the Retina display reliably on all iOS devices, you need to check if the device is running iOS4+ and if the [UIScreen mainScreen].scaleproperty is equal to 2.0. You CANNOT assume a device is running iOS4+ if the scaleproperty exists, as the iPad 3.2 also contains this property.
为了在所有 iOS 设备上可靠地检测 Retina 显示，您需要检查设备是否运行 iOS4+ 以及该[UIScreen mainScreen].scale属性是否等于 2.0。如果该scale属性存在，您不能假设设备正在运行 iOS4+ ，因为 iPad 3.2 也包含此属性。
On an iPad running iOS3.2, scale will return 1.0 in 1x mode, and 2.0 in 2x mode -- even though we know that device does not contain a Retina display. Apple changed this behavior in iOS4.2 for the iPad: it returns 1.0 in both 1x and 2x modes. You can test this yourself in the simulator.
在运行 iOS3.2 的 iPad 上，缩放将在 1x 模式下返回 1.0，在 2x 模式下返回 2.0——即使我们知道该设备不包含 Retina 显示屏。Apple 在 iOS4.2 中为 iPad 更改了这种行为：它在 1x 和 2x 模式下都返回 1.0。您可以在模拟器中自行测试。
I test for the -displayLinkWithTarget:selector:method on the main screen which exists in iOS4.x but not iOS3.2, and then check the screen's scale:
我-displayLinkWithTarget:selector:在主屏幕上测试iOS4.x中存在但iOS3.2中不存在的方法，然后检查屏幕的比例：
if ([[UIScreen mainScreen] respondsToSelector:@selector(displayLinkWithTarget:selector:)] &&
([UIScreen mainScreen].scale == 2.0)) {
// Retina display
} else {
// non-Retina display
}
回答by Mick Byrne
@sickp's answer is correct. Just to make things easier, add this line into your Shared.pch file:
@sickp 的回答是正确的。为方便起见，将此行添加到您的 Shared.pch 文件中：
#define IS_RETINA ([[UIScreen mainScreen] respondsToSelector:@selector(displayLinkWithTarget:selector:)] && ([UIScreen mainScreen].scale >= 2.0))
Then in any file you can just do:
然后在任何文件中，您都可以执行以下操作：
if(IS_RETINA)
{
// etc..
}
回答by Mani
+(BOOL)iPhoneRetina{
return ([[UIScreen mainScreen] respondsToSelector:@selector(displayLinkWithTarget:selector:)] && ([UIScreen mainScreen].scale == 2.0))?1:0;
}
回答by primulaveris
Here is a handy swift extension:
这是一个方便的 swift 扩展：
Update for Swift v5:
Swift v5 更新：
extension UIScreen {
public var isRetina: Bool {
guard let scale = screenScale else {
return false
}
return scale >= 2.0
}
public var isRetinaHD: Bool {
guard let scale = screenScale else {
return false
}
return scale >= 3.0
}
private var screenScale: CGFloat? {
guard UIScreen.main.responds(to: #selector(getter: scale)) else {
return nil
}
return UIScreen.main.scale
}
}
Usage:
用法：
if UIScreen.main.isRetina {
// Your code
}
Original:
原来的：
extension UIScreen {
public func isRetina() -> Bool {
return screenScale() >= 2.0
}
public func isRetinaHD() -> Bool {
return screenScale() >= 3.0
}
private func screenScale() -> CGFloat? {
if UIScreen.mainScreen().respondsToSelector(Selector("scale")) {
return UIScreen.mainScreen().scale
}
return nil
}
}
Usage:
用法：
if UIScreen.mainScreen().isRetina() {
// your code
}
回答by Pedro
This snippet...
这个片段...
int d = 0; // standard display
if ([[UIScreen mainScreen] respondsToSelector:@selector(scale)] && [[UIScreen mainScreen] scale] == 2.0) {
d = 1; // is retina display
}
if ([[UIDevice currentDevice] userInterfaceIdiom] == UIUserInterfaceIdiomPad) {
d += 2;
}
Will return... 0 for standard resolution iPhone/iPod touch, 1 for retina iPhone, 2 for standard resolution iPad, 3 for retina iPad.
将返回... 0 代表标准分辨率 iPhone/iPod touch，1 代表视网膜 iPhone，2 代表标准分辨率 iPad，3 代表视网膜 iPad。
回答by Jorge Perez
SSToolkit has a method that does this:
SSToolkit 有一个方法可以做到这一点：
http://sstoolk.it/documentation/Categories/UIScreen(SSToolkitAdditions).html
http://sstoolk.it/documentation/Categories/UIScreen(SSToolkitAdditions).html
It is used in the following way:
它的使用方式如下：
[[UIScreen mainScreen] isRetinaDisplay];
回答by skahlert
It always feels a bit dodgy to compare floating-point values for equality. I prefer going for either
比较浮点值的相等性总是感觉有点狡猾。我更喜欢去
[UIScreen mainScreen].scale > 1.0;
or
或者
[UIScreen mainScreen].scale < 2.0;
回答by Dan Rosenstark
This is a riff on Matt MC's answer above. Just a category on UIScreen.
这是对上述 Matt MC 回答的即兴演奏。只是一个类别UIScreen。
#import "UIScreen+Util.h"
@implementation UIScreen (Util)
+ (BOOL) isRetinaDisplay {
static BOOL retina = NO;
static BOOL alreadyChecked = NO;
if (!alreadyChecked) {
UIScreen *mainScreen = self.mainScreen;
if (mainScreen) {
retina = mainScreen.scale > 1.0;
alreadyChecked = YES;
}
}
return retina;
}
@end
回答by cdf1982
Swift version of the answers above, with >= 2.0 scale so it includes iPhone 6+ and other future devices with higher-than-Retina scale:
上面答案的 Swift 版本，具有 >= 2.0 比例，因此它包括 iPhone 6+ 和其他具有高于 Retina 比例的未来设备：
if UIScreen.mainScreen().respondsToSelector(Selector("scale")) && UIScreen.mainScreen().scale >= 2.0 {
// code executed only on Retina device
}
回答by Roman Solodyashkin
// .h
UIKIT_EXTERN bool isRetinaDisplay();
// .m
bool isRetinaDisplay()
{
static bool flag;
#ifdef __BLOCKS__
static dispatch_once_t onceToken;
dispatch_once(&onceToken, ^{
if([[UIScreen mainScreen] respondsToSelector:@selector(scale)])
{
flag = [[UIScreen mainScreen] scale] > 1.0;
}
else
{
flag = false;
}
});
#else
static bool onceToken;
if(onceToken == false)
{
onceToken = true;
if([[UIScreen mainScreen] respondsToSelector:@selector(scale)])
{
flag = [[UIScreen mainScreen] scale] > 1.0;
}
else
{
flag = false;
}
}
#endif
return flag;
}
