在Java中解析XML文件后如何调整输出？

如何解决在Java中解析XML文件后如何调整输出？

我正在尝试通过使用递归函数来解析XML文档，但我不确定哪种数据的最佳格式（或者我应该说如何以何种格式保存已解析的文档）以便能够轻松转换它在JSON中。 这是代码：

private static void printNodeList(NodeList nodeList)
{

    for (int count = 0; count < nodeList.getLength(); count++) //getLength of nodeList that is passed inside of function printNodeList
    {
        Node elemNode = nodeList.item(count); //assign name of element we are currently looking at to a Node "elemNode"
        if (elemNode.getNodeType() == Node.ELEMENT_NODE) //if node we are currently looking at has type ELEMENT_NODE do next block of a code
        {
            // get node name and value
            System.out.println("\nNode Name =" + elemNode.getNodeName()+ " [OPEN]");
            System.out.println("Node Content =" + elemNode.getTextContent());

            if (elemNode.hasAttributes()) //if node we are currently looking at has attributes then do this block of a code
            {
                NamedNodeMap nodeMap = elemNode.getAttributes(); //create nodeMap and pass it attributes of a node we are currently looking at
                for (int i = 0; i < nodeMap.getLength(); i++) //getLength of nodeMap or look how many attributes there is for node that we are currently watching
                {
                    Node node = nodeMap.item(i); //every attribute that is stored in nodeMap will be saved in node and his name and value will be printed out
                    System.out.println("attr name : " + node.getNodeName());
                    System.out.println("attr value : " + node.getNodeValue());
                }
            }
            if (elemNode.hasChildNodes()) //if our element (element that we are currently watching at) still has child nodes left -> recursively call "printNodeList" function
            {
                //recursive call if the node has child nodes
                printNodeList(elemNode.getChildNodes());
            }
            //output in console the name of node that we are done with
            System.out.println("Node Name =" + elemNode.getNodeName()+ " [CLOSE]");
        }
    }
}

XML File用于代码，这就是我得到的output。 不知道如何丢失第一个“ Node Content = .....”，因为没有它，它会更干净甚至更易于处理数据？

这个想法是解析XML文件，然后保存它，然后通过使用一组转换规则将保存的文件转换为JSON。还是有一种无需保存已解析的XML文件即可直接转换为JSON的方法？从XML到JSON转换的整个方法也许是错误的？帮助。

解决方法

我有点想出如何修改输出的方法，但是它仍然缺少一些要点以JSON格式显示的要点，因此，如果有人有任何想法，请随时分享。任何帮助都很好，谢谢。 我可以使用以下方法调整输出：

public static void printNode(NodeList nList) {
    //System.out.println(nList.getLength());
    //nList.getLength() gives us cont of <> and </> under that class

    for (int i = 0; i < nList.getLength(); i++) {
        Node node = nList.item(i);
      

        if (node.hasChildNodes()) { //check if node has childNodes
            System.out.println("\"" + node.getNodeName() + "\":");
        }
            if(node.hasAttributes()){ //check if node have any attributes
                System.out.println("->"+node.getAttributes().item(0));
                //getAttributes() is going to get a list of attributes and we want only first one that's why we put "0" as a parameter
            }
            if(node.getChildNodes().getLength()==1){
                //if childNode length is equal to 1 that means this is a leaf node (latest node in the tree)
                System.out.println("---->"+node.getTextContent());
            }
            //recursive call
        printNode(node.getChildNodes());

    }
}

对于this_input，我可以获得this_output。我想尝试创建更像this的东西。