Question

我正在尝试获取范围对象中的所有DOM节点，这是最好的方法吗？

var selection = window.getSelection(); //what the user has selected
var range = selection.getRangeAt(0); //the first range of the selection
var startNode = range.startContainer;
var endNode = range.endContainer;
var allNodes = /*insert magic*/;

我一直在考虑过去几个小时的方法，并提出了这个：

var getNextNode = function(node, skipChildren){
    //if there are child nodes and we didn't come from a child node
    if (node.firstChild && !skipChildren) {
        return node.firstChild;
    }
    if (!node.parentNode){
        return null;
    }
    return node.nextSibling 
        || getNextNode(node.parentNode, true);
};

var getNodesInRange = function(range){
    var startNode = range.startContainer.childNodes[range.startOffset]
            || range.startContainer;//it's a text node
    var endNode = range.endContainer.childNodes[range.endOffset]
            || range.endContainer;

    if (startNode == endNode && startNode.childNodes.length === 0) {
        return [startNode];
    };

    var nodes = [];
    do {
        nodes.push(startNode);
    }
    while ((startNode = getNextNode(startNode)) 
            && (startNode != endNode));
    return nodes;
};

但是，当结束节点是起始节点的父节点时，它将返回页面上的所有内容。我敢肯定我忽略了一些明显的东西？或者可能以完全错误的方式进行。

MDC/DOM/range

Answer 1

如果getNextNode是父节点，则会以递归方式跳过所需的endNode。

执行getNextNode内部的条件中断检查：

var getNextNode = function(node, skipChildren, endNode){
  //if there are child nodes and we didn't come from a child node
  if (endNode == node) {
    return null;
  }
  if (node.firstChild && !skipChildren) {
    return node.firstChild;
  }
  if (!node.parentNode){
    return null;
  }
  return node.nextSibling 
         || getNextNode(node.parentNode, true, endNode); 
};

并在while声明中：

while (startNode = getNextNode(startNode, false , endNode));

Answer 2

这是我想出来解决这个问题的一个实现：

function getNextNode(node)
{
    if (node.firstChild)
        return node.firstChild;
    while (node)
    {
        if (node.nextSibling)
            return node.nextSibling;
        node = node.parentNode;
    }
}

function getNodesInRange(range)
{
    var start = range.startContainer;
    var end = range.endContainer;
    var commonAncestor = range.commonAncestorContainer;
    var nodes = [];
    var node;

    // walk parent nodes from start to common ancestor
    for (node = start.parentNode; node; node = node.parentNode)
    {
        nodes.push(node);
        if (node == commonAncestor)
            break;
    }
    nodes.reverse();

    // walk children and siblings from start until end is found
    for (node = start; node; node = getNextNode(node))
    {
        nodes.push(node);
        if (node == end)
            break;
    }

    return nodes;
}

Answer 3

Rangy library有一个Range.getNodes([Array nodeTypes[, Function filter]]) function。

Answer 4

以下代码解决您的问题

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>payam jabbari</title>
<script src="http://code.jquery.com/jquery-2.0.2.min.js" type="text/javascript"></script>
<script type="text/javascript">

$(document).ready(function(){
    var startNode = $('p.first').contents().get(0);
var endNode = $('span.second').contents().get(0);
var range = document.createRange();
range.setStart(startNode, 0);
range.setEnd(endNode, 5);
var selection = document.getSelection();
selection.addRange(range);
// below code return all nodes in selection range. this code work in all browser
var nodes = range.cloneContents().querySelectorAll("*");
for(var i=0;i<nodes.length;i++)
{
   alert(nodes[i].innerHTML);
}
});
</script>
</head>

<body>
<div>

<p class="first">Even a week ago, the idea of a Russian military intervention in Ukraine seemed far-fetched if not totally alarmist. But the arrival of Russian troops in Crimea over the weekend has shown that he is not averse to reckless adventures, even ones that offer little gain. In the coming days and weeks</p>

<ol>
    <li>China says military will respond to provocations.</li>
    <li >This Man Has Served 20 <span class="second"> Years—and May Die—in </span> Prison for Marijuana.</li>
    <li>At White House, Israel's Netanyahu pushes back against Obama diplomacy.</li>
</ol>
</div>
</body>
</html>

Answer 5

我根据MikeB的答案做了2个额外的修复，以提高所选节点的准确性。

我特别在选择所有操作时对此进行测试，而不是通过跨越多个元素的文本拖动光标进行范围选择。

在Firefox中，点击全选（CMD + A）会返回一个范围，其中包含startContainer＆amp; endContainer是contenteditable div，区别在于startOffset＆amp; endOffset，它分别是第一个和最后一个子节点的索引。

在Chrome中，点击全选（CMD + A）会返回一个范围，其中startContainer是contenteditable div的第一个子节点，endContainer是contenteditable div的最后一个子节点。

我添加的修改工作围绕两者之间的差异。您可以在代码中查看注释以获得其他说明。

function getNextNode(node) {
    if (node.firstChild)
        return node.firstChild;

    while (node) {
        if (node.nextSibling) return node.nextSibling;
        node = node.parentNode;
    }
}

function getNodesInRange(range) {

    // MOD #1
    // When the startContainer/endContainer is an element, its
    // startOffset/endOffset basically points to the nth child node
    // where the range starts/ends.
    var start = range.startContainer.childNodes[range.startOffset] || range.startContainer;
    var end = range.endContainer.childNodes[range.endOffset] || range.endContainer;
    var commonAncestor = range.commonAncestorContainer;
    var nodes = [];
    var node;

    // walk parent nodes from start to common ancestor
    for (node = start.parentNode; node; node = node.parentNode)
    {
        nodes.push(node);
        if (node == commonAncestor)
            break;
    }
    nodes.reverse();

    // walk children and siblings from start until end is found
    for (node = start; node; node = getNextNode(node))
    {
        // MOD #2
        // getNextNode might go outside of the range
        // For a quick fix, I'm using jQuery's closest to determine
        // when it goes out of range and exit the loop.
        if (!$(node.parentNode).closest(commonAncestor)[0]) break;

        nodes.push(node);
        if (node == end)
            break;
    }

    return nodes;
};

Answer 6

Annon，很棒的工作。我修改了原始版本，包括以下Stefan的修改。

此外，我删除了对Range的依赖，它将函数转换为通用算法以在两个节点之间移动。另外，我已将所有内容整合到一个功能中。

关于其他解决方案的想法：

对依赖jquery不感兴趣
使用cloneNode将结果提升为片段，这可以防止在过滤期间可能要进行的许多操作。
在克隆片段上使用querySelectAll是不可能的，因为开始或结束节点可能在包装节点内，因此解析器可能没有结束标记？

示例：

<div>
    <p>A</p>
    <div>
        <p>B</p>
        <div>
            <p>C</p>
        </div>
    </div>
</div>

假设起始节点是＆＃34; A＆＃34;段落，结束节点是＆＃34; C＆＃34;段。得到的克隆片段为：

<p>A</p>
    <div>
        <p>B</p>
        <div>
            <p>C</p>

我们错过了结束标记？导致时髦的DOM结构？

无论如何，这是函数，其中包含一个过滤器选项，该选项应返回TRUE或FALSE以包含/排除结果。

var getNodesBetween = function(startNode, endNode, includeStartAndEnd, filter){
    if (startNode == endNode && startNode.childNodes.length === 0) {
        return [startNode];
    };

    var getNextNode = function(node, finalNode, skipChildren){
        //if there are child nodes and we didn't come from a child node
        if (finalNode == node) {
            return null;
        }
        if (node.firstChild && !skipChildren) {
            return node.firstChild;
        }
        if (!node.parentNode){
            return null;
        }
        return node.nextSibling || getNextNode(node.parentNode, endNode, true);
    };

    var nodes = [];

    if(includeStartAndEnd){
        nodes.push(startNode);
    }

    while ((startNode = getNextNode(startNode, endNode)) && (startNode != endNode)){
        if(filter){
            if(filter(startNode)){
                nodes.push(startNode);
            }
        } else {
            nodes.push(startNode);
        }
    }

    if(includeStartAndEnd){
        nodes.push(endNode);
    }

    return nodes;
};

Answer 7

鲍勃。该函数仅返回startNode和endNode。中间的节点不会被推送到数组。

似乎while循环在getNextNode（）上返回null，因此该块永远不会被执行。

Answer 8

这里是函数返回子范围数组

function getSafeRanges(range) {

var doc = document;

var commonAncestorContainer = range.commonAncestorContainer;
var startContainer = range.startContainer;
var endContainer = range.endContainer;
var startArray = new Array(0),
    startRange = new Array(0);
var endArray = new Array(0),
    endRange = new Array(0);
// @@@@@ If start container and end container is same
if (startContainer == endContainer) {
    return [range];
} else {
    for (var i = startContainer; i != commonAncestorContainer; i = i.parentNode) {
        startArray.push(i);
    }
    for (var i = endContainer; i != commonAncestorContainer; i = i.parentNode) {
        endArray.push(i);
    }
}
if (0 < startArray.length) {
    for (var i = 0; i < startArray.length; i++) {
        if (i) {
            var node = startArray[i - 1];
            while ((node = node.nextSibling) != null) {
                startRange = startRange.concat(getRangeOfChildNodes(node));
            }
        } else {
            var xs = doc.createRange();
            var s = startArray[i];
            var offset = range.startOffset;
            var ea = (startArray[i].nodeType == Node.TEXT_NODE) ? startArray[i] : startArray[i].lastChild;
            xs.setStart(s, offset);
            xs.setEndAfter(ea);
            startRange.push(xs);
        }
    }
}
if (0 < endArray.length) {
    for (var i = 0; i < endArray.length; i++) {
        if (i) {
            var node = endArray[i - 1];
            while ((node = node.previousSibling) != null) {
                endRange = endRange.concat(getRangeOfChildNodes(node));
            }
        } else {
            var xe = doc.createRange();
            var sb = (endArray[i].nodeType == Node.TEXT_NODE) ? endArray[i] : endArray[i].firstChild;
            var end = endArray[i];
            var offset = range.endOffset;
            xe.setStartBefore(sb);
            xe.setEnd(end, offset);
            endRange.unshift(xe);
        }
    }
}
var topStartNode = startArray[startArray.length - 1];
var topEndNode = endArray[endArray.length - 1];
var middleRange = getRangeOfMiddleElements(topStartNode, topEndNode);
startRange = startRange.concat(middleRange);
response = startRange.concat(endRange);
return response;

}

Answer 9

我为此编写了完美的代码，它对每个节点都有效100％：

function getNodesInSelection() {
    
var range = window.getSelection().getRangeAt(0);
var node = range.startContainer;

var ranges = []
var nodes = []
        
while (node != null) {        
    
    var r = document.createRange();
    r.selectNode(node)
    
    if(node == range.startContainer){
        r.setStart(node, range.startOffset)
    }
    
    if(node == range.endContainer){
        r.setEnd(node, range.endOffset)
    }
    
    
    ranges.push(r)
    nodes.push(node)
    
    node = getNextElementInRange(node, range)
}
     
// do what you want with ranges and nodes
}

这是一些帮助功能

function getClosestUncle(node) {

var parent = node.parentElement;

while (parent != null) {
    var uncle = parent.nextSibling;
    if (uncle != null) {
        return uncle;
    }
    
    uncle = parent.nextElementSibling;
    if (uncle != null) {
        return uncle;
    }
    
    parent = parent.parentElement
}

return null
}
 

                    
function getFirstChild(_node) {

var deep = _node

while (deep.firstChild != null) {
    
    deep = deep.firstChild
}

return deep
}
  

                    
function getNextElementInRange(currentNode, range) {

var sib = currentNode.nextSibling;

if (sib != null && range.intersectsNode(sib)) {
    return getFirstChild(sib)
}
        
var sibEl = currentNode.nextSiblingElemnent;

if (sibEl != null && range.intersectsNode(sibEl)) {
    return getFirstChild(sibEl)
}

var uncle = getClosestUncle(currentNode);
var nephew = getFirstChild(uncle)

if (nephew != null && range.intersectsNode(nephew)) {
    return nephew
}

return null
}

如何使用javascript获取位于范围内的节点？

9 个答案: