将Json转换为Xml的最简单方法

Question

我在.net中有网络服务.当我从数据库中检索数据时,它返回Android Mobile中的JSON文件.如何将JSON文件转换为XML或文本.

Answer 1

对于一个简单的解决方案,我推荐Jackson,因为它可以通过几行简单的代码将任意复杂的JSON转换为XML.

import org.codehaus.jackson.map.ObjectMapper;

import com.fasterxml.jackson.xml.XmlMapper;

public class Foo
{
  public String name;
  public Bar bar;

  public static void main(String[] args) throws Exception
  {
    // JSON input: {"name":"FOO","bar":{"id":42}}
    String jsonInput = "{\"name\":\"FOO\",\"bar\":{\"id\":42}}";

    ObjectMapper jsonMapper = new ObjectMapper();
    Foo foo = jsonMapper.readValue(jsonInput, Foo.class);

    XmlMapper xmlMapper = new XmlMapper();
    System.out.println(xmlMapper.writeValueAsString(foo));
    // <Foo xmlns=""><name>FOO</name><bar><id>42</id></bar></Foo>
  }
}

class Bar
{
  public int id;
}

这个演示使用Jackson 1.7.7(较新的1.7.8也应该工作),Jackson XML Databind 0.5.3(尚未与Jackson 1.8兼容)和Stax2 3.1.1.

Answer 2

这是如何生成有效XML的示例。我还在Maven项目中使用Jackson库。

Maven设置：

<!-- https://mvnrepository.com/artifact/com.fasterxml/jackson-xml-databind -->
    <dependency>
        <groupId>com.fasterxml</groupId>
        <artifactId>jackson-xml-databind</artifactId>
        <version>0.6.2</version>
    </dependency>
    <!-- https://mvnrepository.com/artifact/com.fasterxml.jackson.core/jackson-databind -->
    <dependency>
        <groupId>com.fasterxml.jackson.core</groupId>
        <artifactId>jackson-databind</artifactId>
        <version>2.8.6</version>
    </dependency>

以下是一些Java代码，该代码首先将JSON字符串转换为对象，然后将具有XMLMapper的对象转换为XML，还删除任何错误的元素名称。替换XML元素名称中错误字符的原因是，您可以在$ oid这样的JSON元素名称中使用XML不允许的字符。Jackson库没有考虑到这一点，因此我最终添加了一些代码，该代码从元素名称和名称空间声明中删除了非法字符。

import com.fasterxml.jackson.core.type.TypeReference;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.xml.XmlMapper;

import java.io.IOException;
import java.util.Map;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

/**
 * Converts JSON to XML and makes sure the resulting XML 
 * does not have invalid element names.
 */
public class JsonToXMLConverter {

    private static final Pattern XML_TAG =
            Pattern.compile("(?m)(?s)(?i)(?<first><(/)?)(?<nonXml>.+?)(?<last>(/)?>)");

    private static final Pattern REMOVE_ILLEGAL_CHARS = 
            Pattern.compile("(i?)([^\\s=\"'a-zA-Z0-9._-])|(xmlns=\"[^\"]*\")");

    private ObjectMapper mapper = new ObjectMapper();

    private XmlMapper xmlMapper = new XmlMapper();

    String convertToXml(Object obj) throws IOException {
        final String s = xmlMapper.writeValueAsString(obj);
        return removeIllegalXmlChars(s);
    }

    private String removeIllegalXmlChars(String s) {
        final Matcher matcher = XML_TAG.matcher(s);
        StringBuffer sb = new StringBuffer();
        while(matcher.find()) {
            String elementName = REMOVE_ILLEGAL_CHARS.matcher(matcher.group("nonXml"))
                    .replaceAll("").trim();
            matcher.appendReplacement(sb, "${first}" + elementName + "${last}");
        }
        matcher.appendTail(sb);
        return sb.toString();
    }

    Map<String, Object> convertJson(String json) throws IOException {
        return mapper.readValue(json, new TypeReference<Map<String, Object>>(){});
    }

    public String convertJsonToXml(String json) throws IOException {
        return convertToXml(convertJson(json));
    }
}

这是convertJsonToXml的JUnit测试：

@Test
void convertJsonToXml() throws IOException, ParserConfigurationException, SAXException {
    try(InputStream in = Thread.currentThread().getContextClassLoader().getResourceAsStream("json/customer_sample.json")) {
        String json = new Scanner(in).useDelimiter("\\Z").next();
        String xml = converter.convertJsonToXml(json);
        DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
        DocumentBuilder db = dbf.newDocumentBuilder();
        Document doc = db.parse(new ByteArrayInputStream(xml.getBytes("UTF-8")));
        Node first = doc.getFirstChild();
        assertNotNull(first);
        assertTrue(first.getChildNodes().getLength() > 0);
    }
}

Answer 3

Android 中没有直接转换 API 可以将 JSON 转换为 XML。您需要首先解析 JSON，然后必须编写将其转换为 xml 的逻辑。