使用 XPath 创建 XML 属性

Question

我想创建一个 xml 属性，但为了找到我想要添加查询的元素，我需要使用 xpath。我该怎么做？

例子=

const xmlText = `<?xml version="1.0" encoding="ISO-8859-1"?>
<bookstore>
<book>
  <title lang="eng">Harry Potter</title>
  <price>29.99</price>
</book>
<book>
  <title id="somethingeng">Learning XML</title>
  <price>39.95</price>
</book>
</bookstore>`;

var doc = new DOMParser().parseFromString(xmlText,'text/xml');

var r = doc.evaluate("//*[@lang[contains(.,'eng')]]", doc, null, XPathResult.ANY_TYPE, null);

我想为这个 r 创建一个属性；

Answer 1

首先，xmlText 字符串应该是一个 template literals，现在它有新的行。

evaluate() 接缝很好，但结果是 XPathResult that needs to be iterated with XPathResult.iterateNext().

const xmlText = `<?xml version="1.0" encoding="ISO-8859-1"?>
<bookstore>
<book>
  <title lang="eng">Harry Potter</title>
  <price>29.99</price>
</book>
<book>
  <title id="somethingeng">Learning XML</title>
  <price>39.95</price>
</book>
</bookstore>`;

var doc = new DOMParser().parseFromString(xmlText,'text/xml');

var r = doc.evaluate("//*[@lang[contains(.,'eng')]]", doc, null, XPathResult.ANY_TYPE, null);

var next = r.iterateNext();
while (next) {
  console.log(next.textContent);
  next = r.iterateNext();
}

更新

基于迭代器，您需要收集您感兴趣的节点，然后再对其进行更改。在下文中，我创建了可以基于 XPath 表达式和表示新数据的对象创建子元素和属性的函数。

const xmlText = `<?xml version="1.0" encoding="ISO-8859-1"?>
<bookstore>
<book>
  <title lang="eng">Harry Potter</title>
  <price>29.99</price>
</book>
<book>
  <title id="somethingeng">Learning XML</title>
  <price>39.95</price>
</book>
</bookstore>`;

var doc = new DOMParser().parseFromString(xmlText, 'text/xml');

function addAttribute(doc, xpath, obj) {
  let r = doc.evaluate(xpath, doc, null, XPathResult.ANY_TYPE, null);

  let nodes = [];
  let next = r.iterateNext();
  while (next) {
    nodes.push(next);
    next = r.iterateNext();
  }

  nodes.forEach(node => {
    Object.keys(obj).forEach(key => {
      let newattr = doc.createAttribute(key);
      newattr.value = obj[key];
      node.setAttributeNode(newattr);
    });
  });
}

function addChildNode(doc, xpath, obj) {
  let r = doc.evaluate(xpath, doc, null, XPathResult.ANY_TYPE, null);

  let nodes = [];
  let next = r.iterateNext();
  while (next) {
    console.log(next.textContent);
    nodes.push(next);
    next = r.iterateNext();
  }

  nodes.forEach(node => {
    Object.keys(obj).forEach(key => {
      let newnode = doc.createElement(key);
      newnode.textContent = obj[key];
      node.appendChild(newnode);
    });
  });
}



addAttribute(doc, "//title[@lang[contains(.,'eng')]]", {"data-lang":"eng", index: 2});
addChildNode(doc, "//book[number(price) < 30]", {sale: true});

console.log(doc.documentElement.outerHTML);

Answer 2

使用 XPathResult.snapshotItem() 找到解决方案。

var xmlText = `<?xml version="1.0" encoding="ISO-8859-1"?>
  <bookstore>
    <book>
      <title lang="eng">Harry Potter</title>
      <price>29.99</price>
    </book>
    
    <book>
      <title id=\"somethingeng\">Learning XML</title>
      <price>39.95</price>
    </book>
    
  </bookstore>`;

var doc = new DOMParser().parseFromString(xmlText, 'text/xml');
var r = doc.evaluate("//*[@id[contains(.,'eng')]]", doc, null, XPathResult.ORDERED_NODE_SNAPSHOT_TYPE, null);

var index = 0;
while (index < r.snapshotLength) {
  var next = r.snapshotItem(index);
  next.setAttribute("value", "val");
  index++;
}

var xmlSerializer = new XMLSerializer();
const updatedDoc = xmlSerializer.serializeToString(doc);

console.log(updatedDoc);

使用 XPath 创建 XML 属性

Create XML attribute using XPath

javascript

xml

xpath

更新