将XML解析为JSON [英] Parsing XML into JSON

问题描述

我有一个XML文件，例如

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

如何我可以将其解析为JSON结构文件吗？

How can I parse this into JSON structure file?

推荐答案

对于一个简单的解决方案，我建议 Jackson ，一个用于生成和读取带有XML扩展的JSON的Java库，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也可以使用）， Jackson XML Databind 0.5.3 （尚未与Jackson 1.8兼容），以及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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