Question

专为480x960像素分辨率的设备设计的应用程序在480x1136设备上执行。在480x1136设备上运行时，我需要能够检测到真正的分辨率。遗憾的是，我没有获得设备分辨率（'480x1136'），而是获得了应用分辨率（var len = pathOne.getTotalLength(); paper.customAttributes.along = function (v) { var point = pathOne.getPointAtLength(v * len); return { transform: "t" + [point.x, point.y] + "r" + point.alpha }; }; circle.attr({ along: 0 }); function animateThere( val ) { val = +!val; // toggle pathOne.click( function() { animateThere( val ) } ); circle.animate({ along: val }, 2000 ); }; pathOne.click( function() { animateThere(0) } );），因此== '480x960'返回[[UIScreen mainScreen] bounds].size * [[UIScreen mainScreen] scale]而不是预期的'480x960'

如何解析真实的设备物理分辨率？

Answer 1

//Get device screen resolution for this specific case     

CGRect boundsOfScreen=[[UIScreen mainScreen] bounds];

        if (boundsOfScreen.size.width == 480 && boundsOfScreen.size.height == 960 ) {
            //iPod Touch 3.5-inch
        }
        else if (boundsOfScreen.size.width == 480 && boundsOfScreen.size.height == 1136) {
             //iPod Touch 4-inch
        }

Answer 2

这很难看，但是，这是我找到解决真实物理主显示分辨率的唯一解决方案（独立于正在执行的应用程序）：

#import <sys/utsname.h>

bool getDeviceNativeResolution(CGSize& res) {
    utsname si;
    uname(&si);

    if((0 == strncmp(si.machine, "iPod7", 5)) || (0 == strncmp(si.machine, "iPhone7", 7)))
        res = CGSizeMake(1080, 1920);
    else if((0 == strncmp(si.machine, "iPod6", 5)) || (0 == strncmp(si.machine, "iPhone6", 7)))
        res = CGSizeMake(750, 1334);
    else if((0 == strncmp(si.machine, "iPod5", 5)) || (0 == strncmp(si.machine, "iPhone5", 7)))
        res = CGSizeMake(640, 1136);
    else if((0 == strncmp(si.machine, "iPod4", 5)) || (0 == strncmp(si.machine, "iPhone4", 7)))
        res = CGSizeMake(640, 960);
    else
        return false;

    return true;
}

获得iOS显示分辨率

2 个答案: