通过scrollOffset查找最接近的锚点href

Question

我有一个UIWebViewHTML页面完全加载.它UIWebView有一个320 x 480的框架并水平滚动.我可以获得用户当前所在的当前偏移量.我想使用XY偏移找到最近的锚点,这样我就可以"跳转到"锚定位置.这是可能吗？有人能指点我使用Javascript中的资源吗？

<a id="p-1">Text Text Text Text Text Text Text Text Text<a id="p-2">Text Text Text Text Text Text Text Text Text ... 

更新

我超级难过的JS代码:

function posForElement(e)
{
    var totalOffsetY = 0;

    do
    {
        totalOffsetY += e.offsetTop;
    } while(e = e.offsetParent)

    return totalOffsetY;
}

function getClosestAnchor(locationX, locationY)
{
    var a = document.getElementsByTagName('a');

    var currentAnchor;
    for (var idx = 0; idx < a.length; ++idx)
    {
        if(a[idx].getAttribute('id') && a[idx+1])
        {
            if(posForElement(a[idx]) <= locationX && locationX <= posForElement(a[idx+1])) 
            {
                currentAnchor = a[idx];
                break;
            }
            else
            {
                currentAnchor = a[0];
            }
        }
    }

    return currentAnchor.getAttribute('id');
}

Objective-C的

float pageOffset = 320.0f;

NSString *path = [[NSBundle mainBundle] pathForResource:@"GetAnchorPos" ofType:@"js"];
NSString *jsCode = [NSString stringWithContentsOfFile:path encoding:NSUTF8StringEncoding error:nil];
[webView stringByEvaluatingJavaScriptFromString:jsCode];

NSString *execute = [NSString stringWithFormat:@"getClosestAnchor('%f', '0')", pageOffset];
NSString *anchorID = [webView stringByEvaluatingJavaScriptFromString:execute];

Answer 1

[更新]我重写了代码以匹配所有具有id的锚点,并简化了sortByDistance函数中向量的范数的比较.

检查我对jsFiddle的尝试(前一个在这里).

javascript部分:

// findPos : courtesy of @ppk - see http://www.quirksmode.org/js/findpos.html
var findPos = function(obj) {
    var curleft = 0,
        curtop = 0;
    if (obj.offsetParent) {
        curleft = obj.offsetLeft;
        curtop = obj.offsetTop;
        while ((obj = obj.offsetParent)) {
            curleft += obj.offsetLeft;
            curtop += obj.offsetTop;
        }
    }
    return [curleft, curtop];
};

var findClosestAnchor = function (anchors) {

    var sortByDistance = function(element1, element2) {

        var pos1 = findPos( element1 ),
            pos2 = findPos( element2 );

        // vect1 & vect2 represent 2d vectors going from the top left extremity of each element to the point positionned at the scrolled offset of the window
        var vect1 = [
                window.scrollX - pos1[0],
                window.scrollY - pos1[1]
            ],
            vect2 = [
                window.scrollX - pos2[0],
                window.scrollY - pos2[1]
            ];

        // we compare the length of the vectors using only the sum of their components squared
        // no need to find the magnitude of each (this was inspired by Mageek’s answer)
        var sqDist1 = vect1[0] * vect1[0] + vect1[1] * vect1[1],
            sqDist2 = vect2[0] * vect2[0] + vect2[1] * vect2[1];

        if ( sqDist1 <  sqDist2 ) return -1;
        else if ( sqDist1 >  sqDist2 ) return 1;
        else return 0;
    };

    // Convert the nodelist to an array, then returns the first item of the elements sorted by distance
    return Array.prototype.slice.call( anchors ).sort( sortByDistance )[0];
};

当dom准备就绪时,您可以像这样检索和缓存锚点: var anchors = document.body.querySelectorAll('a[id]');

我还没有在智能手机上测试它,但我没有看到任何原因导致它不起作用. 这就是我使用var foo = function() {};表单(更多javascript模式)的原因.

这return Array.prototype.slice.call( anchors ).sort( sortByDistance )[0];条线实际上有点棘手.

document.body.querySelectorAll('a['id']')返回一个NodeList,其中包含当前页面正文中具有属性"id"的所有锚点.遗憾的是,NodeList对象没有"排序"方法,并且不可能使用sortArray原型的方法,因为它与其他一些方法一起使用,例如filter或map(NodeList.prototype.sort = Array.prototype.sort本来真的很棒).

本文更好地解释了为什么我Array.prototype.slice.call以前将NodeList转换为数组的原因.

最后,我使用该Array.prototype.sort方法(以及自定义sortByDistance函数)来比较NodeList的每个元素,并且我只返回第一个项目,这是最接近的项目.

要查找使用固定位置的元素的位置,可以使用以下更新版本findPos:http://www.greywyvern.com/？post = 331.

我的答案可能不是更有效(drdigit必须比我的更多)但我更喜欢简单而不是效率,我认为这是最容易维护的.

[再次更新]

这是一个经过大量修改的findPos版本,适用于webkit css列(没有间隙):

// Also adapted from PPK - this guy is everywhere ! - check http://www.quirksmode.org/dom/getstyles.html
var getStyle = function(el,styleProp)
{
    if (el.currentStyle)
        var y = el.currentStyle[styleProp];
    else if (window.getComputedStyle)
        var y = document.defaultView.getComputedStyle(el,null).getPropertyValue(styleProp);
    return y;
}

// findPos : original by @ppk - see http://www.quirksmode.org/js/findpos.html
// made recursive and transformed to returns the corect position when css columns are used

var findPos = function( obj, childCoords ) {
   if ( typeof childCoords == 'undefined'  ) {
       childCoords = [0, 0];
   }

   var parentColumnWidth,
       parentHeight;

   var curleft, curtop;

   if( obj.offsetParent && ( parentColumnWidth = parseInt( getStyle( obj.offsetParent, '-webkit-column-width' ) ) ) ) {
       parentHeight = parseInt( getStyle( obj.offsetParent, 'height' ) );
       curtop = obj.offsetTop;
       column = Math.ceil( curtop / parentHeight );
       curleft = ( ( column - 1 ) * parentColumnWidth ) + ( obj.offsetLeft % parentColumnWidth );
       curtop %= parentHeight;
   }
   else {
       curleft = obj.offsetLeft;
       curtop = obj.offsetTop;
   }

   curleft += childCoords[0];
   curtop += childCoords[1];

   if( obj.offsetParent ) {
       var coords = findPos( obj.offsetParent, [curleft, curtop] );
       curleft = coords[0];
       curtop = coords[1];
   }
   return [curleft, curtop];
}