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

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");

}

其他回答

针对SDK和OS的任何组合进行测试和设计:

斯威夫特

增加了iPad类型。iPad 2和iPad mini是非视网膜iPad。而iPad Mini 2及以上,iPad 3,4, iPad Air, Air 2, Air 3和iPad Pro 9.7具有相同的1024逻辑分辨率。iPad Pro的maxLength为1366。参考

import UIKit

public enum DisplayType {
    case unknown
    case iphone4
    case iphone5
    case iphone6
    case iphone6plus
    case iPadNonRetina
    case iPad
    case iPadProBig
    static let iphone7 = iphone6
    static let iphone7plus = iphone6plus
}

public final class Display {
    class var width:CGFloat { return UIScreen.main.bounds.size.width }
    class var height:CGFloat { return UIScreen.main.bounds.size.height }
    class var maxLength:CGFloat { return max(width, height) }
    class var minLength:CGFloat { return min(width, height) }
    class var zoomed:Bool { return UIScreen.main.nativeScale >= UIScreen.main.scale }
    class var retina:Bool { return UIScreen.main.scale >= 2.0 }
    class var phone:Bool { return UIDevice.current.userInterfaceIdiom == .phone }
    class var pad:Bool { return UIDevice.current.userInterfaceIdiom == .pad }
    class var carplay:Bool { return UIDevice.current.userInterfaceIdiom == .carPlay }
    class var tv:Bool { return UIDevice.current.userInterfaceIdiom == .tv }
    class var typeIsLike:DisplayType {
        if phone && maxLength < 568 {
            return .iphone4
        }
        else if phone && maxLength == 568 {
                return .iphone5
        }
        else if phone && maxLength == 667 {
            return .iphone6
        }
        else if phone && maxLength == 736 {
            return .iphone6plus
        }
        else if pad && !retina {
            return .iPadNonRetina
        }
        else if pad && retina && maxLength == 1024 {
            return .iPad
        }
        else if pad && maxLength == 1366 {
            return .iPadProBig
        }
        return .unknown
    }
}

看看它的实际应用 https://gist.github.com/hfossli/bc93d924649de881ee2882457f14e346

注意:例如,如果iPhone 6处于放大模式,UI就是放大版的iPhone 5。这些功能不是决定设备类型,而是决定显示模式,因此iPhone 5是本例中的理想结果。

objective - c

#define IS_IPAD (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPad)
#define IS_IPHONE (UI_USER_INTERFACE_IDIOM() == UIUserInterfaceIdiomPhone)
#define IS_RETINA ([[UIScreen mainScreen] scale] >= 2.0)

#define SCREEN_WIDTH ([[UIScreen mainScreen] bounds].size.width)
#define SCREEN_HEIGHT ([[UIScreen mainScreen] bounds].size.height)
#define SCREEN_MAX_LENGTH (MAX(SCREEN_WIDTH, SCREEN_HEIGHT))
#define SCREEN_MIN_LENGTH (MIN(SCREEN_WIDTH, SCREEN_HEIGHT))
#define IS_ZOOMED (IS_IPHONE && SCREEN_MAX_LENGTH == 736.0)

#define IS_IPHONE_4_OR_LESS (IS_IPHONE && SCREEN_MAX_LENGTH < 568.0)
#define IS_IPHONE_5 (IS_IPHONE && SCREEN_MAX_LENGTH == 568.0)
#define IS_IPHONE_6 (IS_IPHONE && SCREEN_MAX_LENGTH == 667.0)
#define IS_IPHONE_6P (IS_IPHONE && SCREEN_MAX_LENGTH == 736.0)

用法:http://pastie.org/9687735

注意:例如,如果iPhone 6处于放大模式,UI就是放大版的iPhone 5。这些功能不是决定设备类型,而是决定显示模式,因此iPhone 5是本例中的理想结果。

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

我认为它应该很好,如果这个宏将在设备和模拟器中工作,下面是解决方案。

#define IS_WIDESCREEN (fabs((double)[[UIScreen mainScreen]bounds].size.height - (double)568) < DBL_EPSILON)
#define IS_IPHONE (([[[UIDevice currentDevice] model] isEqualToString:@"iPhone"]) || ([[[UIDevice currentDevice] model] isEqualToString: @"iPhone Simulator"]))
#define IS_IPOD   ([[[UIDevice currentDevice]model] isEqualToString:@"iPod touch"])
#define IS_IPHONE_5 ((IS_IPHONE || IS_IPOD) && IS_WIDESCREEN)

依赖规模在很多层面上都是错误的。如果我们问系统呢?

- (NSString *) getDeviceModel
{
    struct utsname systemInfo;
    uname(&systemInfo);
    return [NSString stringWithCString:systemInfo.machine encoding:NSUTF8StringEncoding];
}

摘自《检测硬件类型的最佳方法,iPhone4还是iPhone5?》edzio27回答。

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)
}