将 XML 解析为 JSON [英] Parsing XML into JSON

问题描述

我有一个 XML 文件，比如

I have an XML file, like

<stock><name>AXL</name><time>19-07</time><price>11.34</price></stock>
<stock><name>AIK</name><time>19-07</time><price>13.54</price></stock>
<stock><name>ALO</name><time>19-07</time><price>16.32</price></stock>
<stock><name>APO</name><time>19-07</time><price>13.56</price></stock>
...............more

如何将其解析为 JSON 结构文件?

How can I parse this into JSON structure file?

推荐答案

对于一个简单的解决方案，我推荐 Jackson，一个 Java 库生成和读取带有 XML 扩展的 JSON，因为它可以通过几行简单的代码将任意复杂的 XML 转换为 JSON.

For a simple solution, I recommend Jackson, a Java library for generating and reading JSON with an extension for XML, as it can transform arbitrarily complex XML into JSON with just a few simple lines of code.

input.xml

<entries>
  <stock><name>AXL</name><time>19-07</time><price>11.34</price></stock>
  <stock><name>AIK</name><time>19-07</time><price>13.54</price></stock>
  <stock><name>ALO</name><time>19-07</time><price>16.32</price></stock>
  <stock><name>APO</name><time>19-07</time><price>13.56</price></stock>
</entries>

Java 代码:

import java.io.File;
import java.util.List;

import org.codehaus.jackson.map.ObjectMapper;

import com.fasterxml.jackson.xml.XmlMapper;

public class Foo
{
  public static void main(String[] args) throws Exception
  {
    XmlMapper xmlMapper = new XmlMapper();
    List entries = xmlMapper.readValue(new File("input.xml"), List.class);

    ObjectMapper jsonMapper = new ObjectMapper();
    String json = jsonMapper.writeValueAsString(entries);
    System.out.println(json);
    // [{"name":"AXL","time":"19-07","price":"11.34"},{"name":"AIK","time":"19-07","price":"13.54"},{"name":"ALO","time":"19-07","price":"16.32"},{"name":"APO","time":"19-07","price":"13.56"}]
  }
}

此演示使用 Jackson 1.7.7(较新的 1.7.8 也应该可以使用)，Jackson XML Databind 0.5.3(还不兼容 Jackson 1.8)和 Stax2 3.1.1.

This demo uses Jackson 1.7.7 (the newer 1.7.8 should also work), Jackson XML Databind 0.5.3 (not yet compatible with Jackson 1.8), and Stax2 3.1.1.

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