fastJson中常用方法以及遇到的“坑”

1.使用fastJson，首先引入fastJson依赖

<!-- https://mvnrepository.com/artifact/com.alibaba/fastjson -->

<dependency>

    <groupId>com.alibaba</groupId>

    <artifactId>fastjson</artifactId>

    <version>1.2.54</version>

</dependency>

2.JSON String to Java Bean

/**

 * JSON->Java Bean

 */

@Test

public void test1(){

    Person person = new Person().setId("1").setName("fastJson").setAge(1);

    String jsonString = JSON.toJSONString(person);

    System.out.println(jsonString);
    //传入字节码，调用parseObject

    Person person1 = JSON.parseObject(jsonString, Person.class);

    System.out.println(person1);

}

3.JSON String to List Java Bean

/**

 * JSON->List Java Bean

 */

@Test

public void test2(){

    List<Person> list = Arrays.asList(new Person().setId("1").setName("fastJson1").setAge(1),

            new Person().setId("2").setName("fastJson2").setAge(2),

            new Person().setId("3").setName("fastJson3").setAge(3));

    String jsonString = JSON.toJSONString(list);

    System.out.println(jsonString);
    //传入字节码，调用parseArray

    List<Person> person1 = JSON.parseArray(jsonString, Person.class);

    System.out.println(person1);

}

4.JSON String to List String

/**

 * JSON->List String

 */

@Test

public void test3(){

    List<String> list = Arrays.asList("hello","world","hello world");

    String jsonString = JSON.toJSONString(list);

    System.out.println(jsonString);
　　 // new TypeReference<List<String>>() {}

    List<String> list1 = JSON.parseObject(jsonString, new TypeReference<List<String>>() {});

    System.out.println(list1);

}

5.JSON String to List<Map<String,Object>>

/**

 * JSON->List<Map<String,Object>>

 */

@Test

public void test4(){

    List<Map<String,Object>> list = new ArrayList<>();

    Map<String,Object> map = new HashMap<>();

    map.put("key1","value1");

    map.put("key2","value2");

    list.add(map);

    Map<String,Object> map1 = new HashMap<>();

    map1.put("key11","value11");

    map1.put("key22","value22");

    list.add(map1);

    Map<String,Object> map2 = new HashMap<>();

    map2.put("key111","value111");

    map2.put("key222","value222");

    list.add(map2);

    String jsonString = JSON.toJSONString(list);

    System.out.println(jsonString);
　　// new TypeReference<T>

    List<Map<String, Object>> maps = JSON.parseObject(jsonString, new TypeReference<List<Map<String, Object>>>() {});

    System.out.println(maps);

}

6.JSON.toJSONString中序列化map中的空字符串会出现空对象问题

@Test

public void testJSON() throws Exception {

	Map<String,String> map = new HashMap<String, String>();

	map.put("aaa",null);

	map.put("bbb",null);

	map.put("ccc",null);

	System.out.println(map);

	String s = JSON.toJSONString(map);

	System.out.println(s);

	Map<String, String> stringMap = JSON.parseObject(s, new TypeReference<Map<String, String>>() {

	});

	System.out.println(stringMap);

}

上面的代码需要经过如下修改，才不会出现空对象问题

String s = JSON.toJSONString(map, SerializerFeature.WriteMapNullValue);

7.总结：如果大家在使用fastJSON序列化时出现问题，可以考虑朝着序列化这个方向考虑问题，多了解了解SerializerFeature。我们在学习并使用某一项api时，不仅仅要会这个api，同时

要更加注意这个框架，这个工具类所存在的问题，会有哪些坑！

巴特西

fastJson中常用方法以及遇到的“坑”

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