如何检测iPhone 5(宽屏设备)?

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 )

2013-11-07 21:39:48

其他回答

+(BOOL)isDeviceiPhone5
{
    BOOL iPhone5 = FALSE;

    CGRect screenBounds = [[UIScreen mainScreen] bounds];
    if (screenBounds.size.height == 568)
    {
        // code for 4-inch screen
        iPhone5 = TRUE;
    }
    else
    {
        iPhone5 = FALSE;
        // code for 3.5-inch screen
    }
    return iPhone5;

}

2014-05-19 14:41:52

if ((int)[[UIScreen mainScreen] bounds].size.height == 568)
{
    // This is iPhone 5 screen
} else {
    // This is iPhone 4 screen
}

2013-09-12 12:03:22

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

     }

2012-12-11 05:39:06

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)

2014-09-24 14:58:18

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 )

2013-11-07 21:39:48

如何检测iPhone 5(宽屏设备)?

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