Spring Boot启动后自动发起POST请求报JsonParseException问题

你碰到的400 Bad Request错误本质是请求数据格式不匹配——你的后端API通过@RequestBody声明要接收JSON格式的数据，但你用UrlEncodedFormEntity发送的是表单编码（application/x-www-form-urlencoded）的参数，这直接导致Jackson解析JSON时失败，抛出了"Unrecognized token 'name_of_doctor'"的错误。

下面给你两种正确的实现方式，以及一个优化建议：

方式一：手动构造JSON请求体 #

直接拼接符合要求的JSON字符串，用StringEntity封装并设置正确的Content-Type头：

@SpringBootApplication
@EnableTransactionManagement
public class AppointmentApplication {
    public static void main(String[] args) {
        System.out.println("\n\nAppointment Manager\n\n");
        SpringApplication.run(AppointmentApplication.class, args);
        
        HttpClient client = HttpClientBuilder.create().build();
        HttpPost post = new HttpPost("http://localhost:8080/api/v1/appointments/createAppointment");
        
        // 构造JSON格式的请求体，注意转义引号或者用双引号包裹字符串
        String jsonRequestBody = "{\"name_of_doctor\": \"Monika\", \"price\": \"12.5\"}";
        
        try {
            // 使用StringEntity封装JSON，指定Content-Type为application/json
            StringEntity requestEntity = new StringEntity(
                jsonRequestBody, 
                ContentType.APPLICATION_JSON
            );
            post.setEntity(requestEntity);
            
            HttpResponse response = client.execute(post);
            System.out.println("请求响应结果：" + EntityUtils.toString(response.getEntity()));
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}

方式二：用Jackson序列化对象（更优雅） #

如果你的Appointment实体类已经定义好，推荐用Jackson把对象直接序列化为JSON，避免手动拼接字符串出错：

@SpringBootApplication
@EnableTransactionManagement
public class AppointmentApplication {
    public static void main(String[] args) {
        System.out.println("\n\nAppointment Manager\n\n");
        SpringApplication.run(AppointmentApplication.class, args);
        
        HttpClient client = HttpClientBuilder.create().build();
        HttpPost post = new HttpPost("http://localhost:8080/api/v1/appointments/createAppointment");
        
        // 实例化Appointment对象并设置属性
        Appointment appointment = new Appointment();
        appointment.setName_of_doctor("Monika");
        appointment.setPrice("12.5");
        
        // 用Jackson的ObjectMapper序列化对象为JSON
        ObjectMapper objectMapper = new ObjectMapper();
        try {
            String jsonRequestBody = objectMapper.writeValueAsString(appointment);
            StringEntity requestEntity = new StringEntity(
                jsonRequestBody, 
                ContentType.APPLICATION_JSON
            );
            post.setEntity(requestEntity);
            
            HttpResponse response = client.execute(post);
            System.out.println("请求响应结果：" + EntityUtils.toString(response.getEntity()));
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}

额外优化：用ApplicationRunner确保应用就绪后再请求 #

你当前在main方法里直接发送请求，可能会碰到Spring Boot刚启动、容器还没完全就绪的情况，导致连接失败。推荐用Spring提供的ApplicationRunner来执行初始化逻辑，它会在应用上下文完全加载后自动执行：

@SpringBootApplication
@EnableTransactionManagement
public class AppointmentApplication {

    public static void main(String[] args) {
        System.out.println("\n\nAppointment Manager\n\n");
        SpringApplication.run(AppointmentApplication.class, args);
    }

    @Bean
    public ApplicationRunner initAppointmentData() {
        return args -> {
            HttpClient client = HttpClientBuilder.create().build();
            HttpPost post = new HttpPost("http://localhost:8080/api/v1/appointments/createAppointment");
            
            String jsonRequestBody = "{\"name_of_doctor\": \"Monika\", \"price\": \"12.5\"}";
            StringEntity requestEntity = new StringEntity(jsonRequestBody, ContentType.APPLICATION_JSON);
            post.setEntity(requestEntity);
            
            HttpResponse response = client.execute(post);
            System.out.println("初始化预约数据响应：" + EntityUtils.toString(response.getEntity()));
        };
    }
}

最后提个小细节：你的后端代码里有个拼写错误Craeted_at，应该是Created_at，虽然这不是当前错误的原因，但后续可能会导致数据映射问题，建议修正~

内容的提问来源于stack exchange，提问作者Arefe