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。

其他回答

如果项目是用Xcode 6创建的,那么使用下面提到的代码来检测设备。

printf("\nDetected Resolution : %d x %d\n\n",(int)[[UIScreen mainScreen] nativeBounds].size.width,(int)[[UIScreen mainScreen] nativeBounds].size.height);

if ([[UIDevice currentDevice] userInterfaceIdiom] == UIUserInterfaceIdiomPhone){
    if ([[UIScreen mainScreen] respondsToSelector: @selector(scale)])
    {
        if([[UIScreen mainScreen] nativeBounds].size.height == 960 || [[UIScreen mainScreen] nativeBounds].size.height == 480){
            printf("Device Type : iPhone 4,4s ");

        }else if([[UIScreen mainScreen] nativeBounds].size.height == 1136){
            printf("Device Type : iPhone 5,5S/iPod 5 ");

        }else if([[UIScreen mainScreen] nativeBounds].size.height == 1334){
            printf("Device Type : iPhone 6 ");

        }else if([[UIScreen mainScreen] nativeBounds].size.height == 2208){
            printf("Device Type : iPhone 6+ ");

        }
    }
}else{
    printf("Device Type : iPad");
}

如果项目是在Xcode 5中创建并在Xcode 6中打开的,那么使用下面提到的代码来检测设备。(如果没有为iPhone 6,6+分配启动图像,此代码有效)

printf("\nDetected Resolution : %d x %d\n\n",(int)[[UIScreen mainScreen] nativeBounds].size.width,(int)[[UIScreen mainScreen] nativeBounds].size.height);
if ([[UIDevice currentDevice] userInterfaceIdiom] == UIUserInterfaceIdiomPhone){
    if ([[UIScreen mainScreen] respondsToSelector: @selector(scale)])
    {
       if([[UIScreen mainScreen] nativeBounds].size.height == 960 || [[UIScreen mainScreen] nativeBounds].size.height == 480){
            printf("Device Type : iPhone 4,4s");
            appType=1;
        }else if([[UIScreen mainScreen] nativeBounds].size.height == 1136 || [[UIScreen mainScreen] nativeBounds].size.height == 1704){
            printf("Device Type : iPhone 5,5S,6,6S/iPod 5 ");
            appType=3;
        }
    }
}else{
    printf("Device Type : iPad");
    appType=2;
}

如果你仍然在使用Xcode 5,那么使用下面的代码来检测设备(iPhone 6和6+将不会被检测到)

printf("\nDetected Resolution : %d x %d\n\n",(int)[[UIScreen mainScreen] bounds].size.width,(int)[[UIScreen mainScreen] bounds].size.height);
if ([[UIDevice currentDevice] userInterfaceIdiom] == UIUserInterfaceIdiomPhone){
    if ([[UIScreen mainScreen] respondsToSelector: @selector(scale)])
    {
        CGSize result = [[UIScreen mainScreen] bounds].size;
        CGFloat scale = [UIScreen mainScreen].scale;
        result = CGSizeMake(result.width * scale, result.height * scale);
        if(result.height == 960 || result.height == 480){
            printf("Device Type : iPhone 4,4S ");

        }else if(result.height == 1136){
            printf("Device Type : iPhone 5s/iPod 5");

        }
    }
}else{
    printf("Device Type : iPad");

}

在Swift, iOS 8+项目中,我喜欢在UIScreen上做一个扩展,比如:

extension UIScreen {

    var isPhone4: Bool {
        return self.nativeBounds.size.height == 960;
    }

    var isPhone5: Bool {
        return self.nativeBounds.size.height == 1136;
    }

    var isPhone6: Bool {
        return self.nativeBounds.size.height == 1334;
    }

    var isPhone6Plus: Bool {
        return self.nativeBounds.size.height == 2208;
    }

}

(注意:nativeBounds是以像素为单位)。

然后代码会是这样的:

if UIScreen.mainScreen().isPhone4 {
    // do smth on the smallest screen
}

因此,代码清楚地表明这是对主屏幕的检查,而不是对设备模型的检查。

我冒昧地将Macmade的宏放入一个C函数中,并正确地命名它,因为它可以检测宽屏可用性,而不一定是iPhone 5。

如果项目不包含Default-568h@2x.png,宏也不会检测到在iPhone 5上运行。如果没有新的默认图像,iPhone 5将显示常规的480x320屏幕大小(以点数计算)。因此,检查不仅仅是宽屏可用性,而是宽屏模式是否启用。

BOOL isWidescreenEnabled()
{
    return (BOOL)(fabs((double)[UIScreen mainScreen].bounds.size.height - 
                                               (double)568) < DBL_EPSILON);
}

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)