1. 导入依赖

        <dependency>
            <groupId>com.fasterxml.jackson.core</groupId>
            <artifactId>jackson-databind</artifactId>
            <version>2.11.0</version>
        </dependency>

2. 创建实体类

@Data
@AllArgsConstructor
@NoArgsConstructor
public class Myuser {
    private String username;
    private String passowrd;
    private int age;
}

3. 解决JSON中文乱码问题

在springmvc-servlet.xml中的注解驱动支持中添加以下内容

<!-- Spring MVC注解驱动支持(解决JSON乱码问题) -->
    <mvc:annotation-driven>
        <mvc:message-converters register-defaults="true">
            <bean class="org.springframework.http.converter.StringHttpMessageConverter">
                <constructor-arg value="UTF-8"/>
            </bean>
            <bean class="org.springframework.http.converter.json.MappingJackson2HttpMessageConverter">
                <property name="objectMapper">
                    <bean class="org.springframework.http.converter.json.Jackson2ObjectMapperFactoryBean">
                        <property name="failOnEmptyBeans" value="false"/>
                    </bean>
                </property>
            </bean>
        </mvc:message-converters>
    </mvc:annotation-driven>

3. 编写Controller

@ResponseBody 使用此注解当前方法不走视图解析器，而是直接返回一个字符串
@RestController 使用此注解当前类中所有方法不走视图解析器，而是直接返回一个字符串

    @ResponseBody
    @GetMapping("/getalluser")
    public String getAllUser() throws JsonProcessingException {
        //jackson objectMapper
        ObjectMapper objectMapper = new ObjectMapper();
        Myuser myuser = new Myuser("施一帆", "shiyifan1230", 1);
        String jsonstring = objectMapper.writeValueAsString(myuser);
        return jsonstring;
    }

使用jackson的ObjectMapper.writeValueAsString()将对象转换为json字符串返回

4. JSON返回时间

因为objectMapper将时间解析后的默认格式为：Timestamp(时间戳)，因此我们需要将时间格式进行修改得到我们想要的时间格式

1.纯java解决方案

@ResponseBody
    @GetMapping("/gettime")
    public String gettime() throws JsonProcessingException {
        ObjectMapper objectMapper = new ObjectMapper();
        Date date = new Date();

        SimpleDateFormat dateFormat = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");

        String jsonstring = objectMapper.writeValueAsString(dateFormat.format(date));
        return jsonstring;
    }

2.ObjectMapper解决方案

    @ResponseBody
    @GetMapping("/gettime")
    public String gettime() throws JsonProcessingException {
        ObjectMapper objectMapper = new ObjectMapper();
        Date date = new Date();
        //不使用时间戳的方式
        objectMapper.configure(SerializationFeature.WRITE_DATES_AS_TIMESTAMPS, false);
        //自定义日期格式对象
        SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
        //指定日期格式
        objectMapper.setDateFormat(sdf);

        String jsonstring = objectMapper.writeValueAsString(date);
        return jsonstring;
    }
}

5. 将JSON字符串转换为Java对象

  @ResponseBody
    @GetMapping("/adduser")
    public String addUser() throws JsonProcessingException {
        //jackson objectMapper
        ObjectMapper objectMapper = new ObjectMapper();
        //模拟得到前端传入的JSON字符串
        String user = getAllUser();
        Myuser myuser = objectMapper.readValue(user, Myuser.class);
        return myuser.toString();
    }