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

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) 
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

     }

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

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"