Java 中如何把xml转化为json 呢?
常规思路是:
(1)通过第三方库 把xml 转换为java bean;
(2)把java bean 序列化为json 字符串
但是上述方式有一个缺点,那就是需要java bean来中转.
以下提供两种方式 不需要java bean
方式一:使用json-lib
Java代码
- XMLSerializer xmlSerializer = new XMLSerializer();
- JSON jsonObj = xmlSerializer.read(responseTextObj);
- String jsonStr = jsonObj.toString();
- jsonStr = jsonStr.replace("[]", "\"\"");
依赖:
Java代码
- <dependency>
- <groupId>net.sf.json-lib</groupId>
- <artifactId>json-lib</artifactId>
- <version>2.4</version>
- <classifier>jdk15</classifier>
- </dependency>
- <dependency>
- <groupId>xom</groupId>
- <artifactId>xom</artifactId>
- <version>1.2.5</version>
- </dependency>
- <dependency>
- <groupId>xom</groupId>
- <artifactId>xom</artifactId>
- <version>1.2.5</version>
- <classifier>sources</classifier>
- </dependency>
注意:通过json-lib 把xml转化为json时,空节点都会转化为空数组,即[],这是非常不好的,所以需要把[]转化为空字符串:jsonStr.replace("[]", "\"\"")
参考:http://hw1287789687.iteye.com/blog/2224407
方式二:使用github 上开源的库
Java代码
- package com.JSON_java;
- public class Main {
- public static int PRETTY_PRINT_INDENT_FACTOR = 4;
- public static String TEST_XML_STRING =
- "<breakfast_menu>\n" +
- "<food>\n" +
- "<name>Belgian Waffles</name>\n" +
- "<price>$5.95</price>\n" +
- "<description>\n" +
- "Two of our famous Belgian Waffles with plenty of real maple syrup\n" +
- "</description>\n" +
- "<calories>650</calories>\n" +
- "</food>\n" +
- "<food>\n" +
- "<name>Strawberry Belgian Waffles</name>\n" +
- "<price>$7.95</price>\n" +
- "<description>\n" +
- "Light Belgian waffles covered with strawberries and whipped cream\n" +
- "</description>\n" +
- "<calories>900</calories>\n" +
- "</food>\n" +
- "</breakfast_menu>";
- public static void main(String[] args) {
- try {
- JSONObject xmlJSONObj = XML.toJSONObject(TEST_XML_STRING);
- String jsonPrettyPrintString = xmlJSONObj.toString(PRETTY_PRINT_INDENT_FACTOR);
- System.out.println(jsonPrettyPrintString);
- } catch (JSONException je) {
- System.out.println(je.toString());
- }
- }
- }
实际应用:
