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
我们现在需要考虑iPhone 6和6Plus的屏幕尺寸。以下是最新的答案
if(UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPhone)
{
//its iPhone. Find out which one?
CGSize result = [[UIScreen mainScreen] bounds].size;
if(result.height == 480)
{
// iPhone Classic
}
else if(result.height == 568)
{
// iPhone 5
}
else if(result.height == 667)
{
// iPhone 6
}
else if(result.height == 736)
{
// iPhone 6 Plus
}
}
else
{
//its iPad
}
一些有用的信息
iPhone 6 Plus 736x414 points 2208x1242 pixels 3x scale 1920x1080 physical pixels 401 physical ppi 5.5"
iPhone 6 667x375 points 1334x750 pixels 2x scale 1334x750 physical pixels 326 physical ppi 4.7"
iPhone 5 568x320 points 1136x640 pixels 2x scale 1136x640 physical pixels 326 physical ppi 4.0"
iPhone 4 480x320 points 960x640 pixels 2x scale 960x640 physical pixels 326 physical ppi 3.5"
iPhone 3GS 480x320 points 480x320 pixels 1x scale 480x320 physical pixels 163 physical ppi 3.5"
其他回答
通过这种方式,您可以检测设备系列。
#import <sys/utsname.h>
NSString* deviceName()
{
struct utsname systemInformation;
uname(&systemInformation);
NSString *result = [NSString stringWithCString:systemInformation.machine
encoding:NSUTF8StringEncoding];
return result;
}
#define isIPhone5 [deviceName() rangeOfString:@"iPhone5,"].location != NSNotFound
#define isIPhone5S [deviceName() rangeOfString:@"iPhone6,"].location != NSNotFound
if ((int)[[UIScreen mainScreen] bounds].size.height == 568)
{
// This is iPhone 5 screen
} else {
// This is iPhone 4 screen
}
使用下列守则:
CGFloat screenScale = [[UIScreen mainScreen] scale];
CGRect screenBounds = [[UIScreen mainScreen] bounds];
CGSize screenSize = CGSizeMake(screenBounds.size.width * screenScale, screenBounds.size.height * screenScale);
if (screenSize.height==1136.000000)
{
// Here iPhone 5 View
// Eg: Nextview~iPhone5.Xib
} else {
// Previous Phones
// Eg : Nextview.xib
}
我发现答案并不包括模拟器的特殊情况。
#define IS_WIDESCREEN ( [ [ UIScreen mainScreen ] bounds ].size.height == 568 )
#define IS_IPHONE ([[ [ UIDevice currentDevice ] model ] rangeOfString:@"iPhone"].location != NSNotFound)
#define IS_IPAD ([[ [ UIDevice currentDevice ] model ] rangeOfString:@"iPad"].location != NSNotFound)
#define IS_IPHONE_5 ( IS_IPHONE && IS_WIDESCREEN )
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)