如何将标签从第一个文件导入到第二个文件中的所有标记

Question

如何将标签"mohamed“从第一个文件导入到第二个文件中的所有标记ahmed

<?xml version="1.0" encoding="UTF-8"?>
<map
      xmlns="http://www.sitemaps.org/schemas/sitemap/0.9"
      xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.sitemaps.org/schemas/sitemap/0.9        http://www.sitemaps.org/schemas/sitemap/0.9/sitemap.xsd">
<!-- created with Free Online Sitemap Generator www.xml-sitemaps.com -->


<mohamed>stackover flow</mohamed>
</map>

我的Xml-2文件是

<?xml version="1.0" encoding="UTF-8"?>
<map
      xmlns="http://www.sitemaps.org/schemas/sitemap/0.9"
      xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
      xsi:schemaLocation="http://www.sitemaps.org/schemas/sitemap/0.9
            http://www.sitemaps.org/schemas/sitemap/0.9/sitemap.xsd">
<!-- created with Free Online Sitemap Generator www.xml-sitemaps.com -->

<ahmed></ahmed>
<ahmed></ahmed>
<ahmed></ahmed>

</map>

在这里输入代码

使用这些片段，我可以首先将TagName("mohamed")导入到TagName("ahmed")，但我只想将它导入到第二个文件中的每个TagName(“TagName”)中。

public static void t (int f ,  int g,String z) {
    DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
    DocumentBuilder db = null;
    Document doc = null;
    Document doc2 = null;
String a = "C:\\Users\\chirap\\Desktop\\Startimes\\C.txt" ;
String  c ;
    try {
            db = dbf.newDocumentBuilder();
            doc = db.parse(new File(a));

            doc2 = db.parse(new File("C:\\Users\\chirap\\Desktop\\Startimes\\A (1).txt"));
            NodeList ndListFirstFile = doc.getElementsByTagName("ahmed");

            Node nodeArea = doc.importNode(doc2.getElementsByTagName("mohamed").item(0), true);


        NodeList nList2 = doc2.getElementsByTagName("mohamed");

            for (int i = f; i <g; i++) {
                c = i+"" ;
               doc2 = db.parse(new File("C:\\Users\\chirap\\Desktop\\Startimes\\A ("+c+").txt"));

            for (int temp = 0; temp < nList2.getLength(); temp++) {

                nodeArea = doc.importNode(doc2.getElementsByTagName("mohamed").item(temp), true);
                   ndListFirstFile.item(0).appendChild(nodeArea);

现在的结果是：

<?xml version="1.0" encoding="UTF-8"?>
<map
      xmlns="http://www.sitemaps.org/schemas/sitemap/0.9"
      xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
      xsi:schemaLocation="http://www.sitemaps.org/schemas/sitemap/0.9
            http://www.sitemaps.org/schemas/sitemap/0.9/sitemap.xsd">
<!-- created with Free Online Sitemap Generator www.xml-sitemaps.com -->

<ahmed><mohamed>stackover flow</mohamed></ahmed>
<ahmed></ahmed>
<ahmed></ahmed>

</map>

我想要这个结果：

<?xml version="1.0" encoding="UTF-8"?>
<map
      xmlns="http://www.sitemaps.org/schemas/sitemap/0.9"
      xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
      xsi:schemaLocation="http://www.sitemaps.org/schemas/sitemap/0.9
            http://www.sitemaps.org/schemas/sitemap/0.9/sitemap.xsd">
<!-- created with Free Online Sitemap Generator www.xml-sitemaps.com -->

    <ahmed><mohamed>stackover flow</mohamed></ahmed>
    <ahmed><mohamed>stackover flow</mohamed></ahmed>
    <ahmed><mohamed>stackover flow</mohamed></ahmed>

    </map>

Answer 1

试试这个

try {
     File inputOne = new File("first.xml");
     File inputTwo = new File("second.xml");

     DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
     DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
     Document docOne = dBuilder.parse(inputOne);
     Document docTwo = dBuilder.parse(inputTwo);
     NodeList nodeListAhmed = docTwo.getElementsByTagName("ahmed");

     for (int i = 0; i < nodeListAhmed.getLength(); i++) {
         Node nodeMohamed = docTwo.importNode(docOne.getElementsByTagName("mohamed").item(0), true);
         nodeListAhmed.item(i).appendChild(nodeMohamed);
     }

     DOMSource source = new DOMSource(docTwo);
     TransformerFactory transformerFactory = TransformerFactory.newInstance();
     Transformer transformer = transformerFactory.newTransformer();
     StreamResult result = new StreamResult("output.xml");
     transformer.transform(source, result);

} catch (Exception e) {
     e.printStackTrace();
}

Answer 2

我想要这个结果

可以使用.querySelector()获取具有tagName "mohamed"的元素，将元素的.parentElement存储在变量中，将匹配元素的.textContent存储在变量中，使用.removeChild()从父元素中删除匹配的元素，将父元素.textContent设置为存储的变量。

​

const xml = `<?xml version="1.0" encoding="UTF-8"?>
<ahmed
      xmlns="http://www.sitemaps.org/schemas/sitemap/0.9"
      xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.sitemaps.org/schemas/sitemap/0.9        http://www.sitemaps.org/schemas/sitemap/0.9/sitemap.xsd">
<!-- created with Free Online Sitemap Generator www.xml-sitemaps.com -->


<mohamed>stackover flow</mohamed>
</ahmed>`;
const parser = new DOMParser();
const doc = parser.parseFromString(xml, "text/xml");
let el = doc.querySelector("mohamed");
let parentElement = el.parentElement;
let text = el.textContent;
parentElement.removeChild(el);
parentElement.textContent = text;
console.log(`<?xml version="1.0" encoding="UTF-8"?>${doc.documentElement.outerHTML}`);

​