Question

我有一个格式如下的文件

<reports>
  <report>
   <text>....</text>
   <author>...</author>
  </report>
  <report>
   <text>....</text>
   <author>...</author>
  </report>
</reports>

报告根节点包含多个report个节点。我想将所有单个报告节点转换为json对象（我将保存为单个文件）。

我是XML新手并尝试使用以下代码

    List<String> readLines = IOUtils.readLines(new FileInputStream(new File("file.xml")));
    String out = String.join("\n", readLines);

    //convert to XML
    DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
    DocumentBuilder builder;
    try {
        builder = factory.newDocumentBuilder();
        Document doc = builder.parse(new InputSource(new StringReader(out)));
        if (doc != null) {
            XPathExpression expression = XPathFactory.newInstance().newXPath().compile("/reports");
            Node reportsRoot = (Node) expression.evaluate(doc, XPathConstants.NODE);
            NodeList reports = reportsRoot.getChildNodes();
            for (int i = 0; i < reports.getLength(); i++) {
               //I am stuck here. How can I convert the report node into a json object?
            }
        }
    } catch (Exception e) {
        e.printStackTrace();
    }

Answer 1

我无法弄清楚如何提取XML子节点并将它们转换为JSON。所以我将整个XML文件转换为JSON，然后提取单个报告。那更容易。这是我使用的代码：

package test;

import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.util.List;
import java.util.logging.Level;
import java.util.logging.Logger;
import org.apache.commons.io.IOUtils;
import org.json.JSONArray;
import org.json.JSONObject;
import org.json.XML;

public class Main {

    public static void main(String[] args) {
        try {

            List<String> readLines = IOUtils.readLines(new FileInputStream(new File("test.xml")));
            String out = String.join("\n", readLines);
            JSONObject jsonObj = XML.toJSONObject(out);

            JSONObject reportsObj = (JSONObject) jsonObj.get("reports");
            JSONArray reports = (JSONArray) reportsObj.get("report");
            int reportCount = reports.length();

            for (int i = 0; i < reportCount; i++) {
                JSONObject report = (JSONObject) reports.get(i);
                System.out.println(report);
            }
        } catch (IOException ex) {
            Logger.getLogger(Main.class.getName()).log(Level.SEVERE, null, ex);
        }
    }
}

如何将根XML节点的子节点转换为json？

1 个答案: