java中json字符串与对象转换
常见的转换工具有:
Jackson:SpringMVC内置的转换工具
jsonlib:Java提供的转换工具(一般不用)
gson:google提供的转换工具(轻量级的框架)
fastjson:Alibaba提供的转换工具(效率高速度快)
Jackson:
相关jar包:
jackson-annotations-2.2.3.jar
jackson-core.2.2.3.jar
jackson-databind-2.2.3.jar
主要方法:
//对象----->字符串
String jsonStr = new ObjectMapper().**writeValueAsString**(resultBean);
//字符串--->map
Map map = JSON.**parseObject**(jsonStr, Map.class);
//字符串---> list
TypeReference<List<User>> ref = new TypeReference<List<User>>(){};
List<User> list = objectMapper.readValue(jsonStr, ref);
java对象转json:
//Map对象或者JavaBean对象转换成json的时候会得到一个json字符串
//List<Map>或者List<JavaBean>转换成json的时候会得到一个json数组的字符串
@Test
public void test04() throws JsonProcessingException {
List<User> userList = new ArrayList<>();
userList.add(new User(1, "娜扎", "123456", "18999999999@163.com", "18999999999"));
userList.add(new User(2, "热巴", "123456", "18999999999@163.com", "19898989898"));
userList.add(new User(3, "哈尼", "123456", "18999999999@163.com", "19866666666"));
ResultBean resultBean = new ResultBean(true, userList);
String jsonStr = new ObjectMapper().writeValueAsString(resultBean);
System.out.println(jsonStr);
}
json字符串转对象:
使用jackson将json字符串转换JavaBean对象或者Map
@Test
//把json转成JavaBean(user对象)
public void test06() throws IOException {
String jsonStr = "{\"id\":1,\"username\":\"zs\",\"password\":\"123456\",\"email\":\"zs@163.com\",\"phone\":\"1386789898\"}";
//1.调用JSON.parseObject(String json,Class clazz);
//转换成user
User user = JSON.parseObject(jsonStr, User.class);
System.out.println(user);
//转换成map
Map map = JSON.parseObject(jsonStr, Map.class);
System.out.println(map);
}
使用jackson将json数组字符串转换成List
@Test
//把json转成List<JavaBean>对象
public void test07() throws Exception {
String jsonStr = "[{\"id\":1,\"username\":\"zs\",\"password\":\"123456\",\"email\":\"zs@163.com\",\"phone\":\"1386789898\"},{\"id\":2,\"username\":\"ls\",\"password\":\"123456\",\"email\":\"ls@163.com\",\"phone\":\"1386781898\"},{\"id\":3,\"username\":\"ww\",\"password\":\"123456\",\"email\":\"ww@163.com\",\"phone\":\"1386782898\"}]";
//1.创建ObjectMapper对象
ObjectMapper objectMapper = new ObjectMapper();
//2.调用readValue()
TypeReference<List<User>> ref = new TypeReference<List<User>>(){};
List<User> list = objectMapper.readValue(jsonStr, ref);
System.out.println(list);
}
fastjson
jar包:fastjson-1.2.39.jar
主要方法:
String jsonStr = JSON.toJSONString(user) //user(对象) -----> jsonStr(字符串)
Map map = JSON.parseObject(jsonStr, Map.class); //jsonStr(字符串) --> map
List<User> userList = JSON.parseArray(jsonArr, User.class); //jsonStr(字符串) --> List
使用fastjson将java对象转成json字符串
@Test
public void test06(){
//使用fastjson将user对象转换成json字符串
User user = new User(1,"张三","123456","123456@qq.com","18999999999");
String jsonStr = JSON.toJSONString(user);
System.out.println(jsonStr);
}
使用fastjson将json字符串转换成JavaBean对象或者Map
@Test
//把json转成JavaBean(user对象)
public void test08() throws IOException {
String jsonStr = "{\"id\":1,\"username\":\"zs\",\"password\":\"123456\",\"email\":\"zs@163.com\",\"phone\":\"1386789898\"}";
//1.调用JSON.parseObject(String json,Class clazz);
User user = JSON.parseObject(jsonStr, User.class);
System.out.println(user);
}
@Test
//把json转成Map
public void test09() throws IOException {
String jsonStr = "{\"id\":1,\"username\":\"zs\",\"password\":\"123456\",\"email\":\"zs@163.com\",\"phone\":\"1386789898\"}";
//1.调用JSON.parseObject(String json,Class clazz);
Map map = JSON.parseObject(jsonStr, Map.class);
System.out.println(map);
}
@Test
//使用fastjson将json字符串转换成List<JavaBean>
public void test11() {
//使用fastjson将json数组的字符串,转换成List<User>
String jsonArr = "[{\"id\":1,\"username\":\"张三\",\"password\":\"123456\",\"email\":\"123456@qq.com\",\"phone\":\"18999999999\"},{\"id\":2,\"username\":\"李四\",\"password\":\"654321\",\"email\":\"654321@qq.com\",\"phone\":\"18666666666\"},{\"id\":3,\"username\":\"王五\",\"password\":\"777777\",\"email\":\"777777@qq.com\",\"phone\":\"18777777777\"}]";
List<User> userList = JSON.parseArray(jsonArr, User.class);
for (User user : userList) {
System.out.println(user.getUsername());
}
}
fastjson解析复杂json数据:
1、如何从字符串String获得JSONObject对象和JSONArray对象
JSONObject jsonObject = new JSONObject ( String str);
JSONArray jsonArray = new JSONArray(String str ) ;
2、如何从JSONArray中获得JSONObject对象
可以把JSONArray当成一般的数组来对待,只是获取的数据内数据的方法不一样
JSONObject jsonObject = jsonArray.getJSONObject(i) ;
3、获取JSON内的数据
int mid= jsonObject.getInt ( "id" ) ;
String mcourse=jsonObject.getString( " courseID") ;