使用XPath創建XML屬性

我想創建一個 xml 屬性，但為了找到要添加查詢的元素，我需要使用 xpath。我怎樣才能做到這一點？

示例 =

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);

我想為這個 r 創建一個屬性；

uj5u.com熱心網友回復：

首先，xmlText字串應該是一個模板文字，現在它有新的行。

evaluate()接縫很好，但結果是需要使用 XPathResult.iterateNext() 迭代的XPathResult。

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

根據迭代器，您需要收集您感興趣的節點，然后再更改它們。在下文中，我創建了可以基于 XPath 運算式和表示新資料的物件創建子元素和屬性的函式。

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);

uj5u.com熱心網友回復：

使用XPathResult.snapshotItem()找到了解決方案。

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);

標籤：javascript xml 路径

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