跳过URI变量'id'，因为请求包含具有相同名称的绑定值

Question

[存在PUT方法，其任务是更换Customer的卡。 Customer类本身与User类之间的@ OneToOne关系（字段name，最后一个name等）一起工作。一切正常，该方法执行其任务，并且没有问题。但是出于某种原因，它会发出警告：

2020-04-26 14:15:14.160  WARN 5340 --- [nio-8080-exec-2] o.springframework.validation.DataBinder  : Skipping URI variable 'id' because request contains bind value with same name.

如何解决？

我使用Spring-MVC +Hibernate+Jpa+ PostgreSQL；

班级用户：

package com.tinychiefdelights.model;

import io.swagger.annotations.ApiModel;
import lombok.Data;

import javax.persistence.*;

@ApiModel
@Data
@Entity
@Table(name = "pg_user", schema = "public")
public class User {

    public User() { // Пустой конструктор для Hibernate

    }

    public User(String name, String lastName, String role,
                String login, String password) { // Базовый конструктор

        this.name = name;
        this.lastName = lastName;
        this.role = role;
        this.login = login;
        this.password = password;
    }


    // Поля
    private @Id
    @GeneratedValue
    Long id;

    @Column(name = "login")
    private String login;

    @Column(name = "password")
    private String password;

    @Column(name = "role")
    private String role;

    @Column(name = "name")
    private String name;

    @Column(name = "last_name")
    private String lastName;
}

客户：

package com.tinychiefdelights.model;

import com.fasterxml.jackson.annotation.*;
import lombok.Data;
import org.hibernate.annotations.NotFound;
import org.hibernate.annotations.NotFoundAction;

import javax.persistence.*;
import java.util.List;

@Data
@Entity
@Table(name = "customer", schema = "public")
public class Customer {

    public Customer() { // Пустой конструктор для Hibernate

    }


    // Поля

    // name, lastName, login, password берем от класса User через связи;

    private @Id
    @GeneratedValue
    Long id;

    @Column(name = "wallet")
    private double wallet;


    //Relationships
    //
    @OneToOne(cascade = CascadeType.ALL)
    @JoinColumn(name = "user_id", referencedColumnName = "id") // Join without Customer in User class
    @NotFound(action = NotFoundAction.IGNORE)
    private User user;

    //Лист заказов
    @OneToMany(mappedBy = "customer", cascade = CascadeType.ALL)
    @JsonIgnore // Таким образом я предотвратил рекурсию
    private List<Order> orderList;
}

客户服务：

    public Customer editCustomer(Long id, User user, double wallet){
        Customer customer = customerRepository.getByIdAndUserRole(id, "customer");
        customer.setUser(user);
        customer.setWallet(wallet);
        return customerRepository.save(customer);
    }

客户总监：

 @PutMapping("/customer/{id}")
    Customer editCustomer(@PathVariable Long id, User user, @RequestParam double wallet) {
        return customerService.editCustomer(id, user, wallet);
    }

Answer 1

由于您将具有ID的用户传递给@PutMapping，因此也有一个路径变量ID。从Spring MVC模型属性中检查以下代码：

private void bind(ServletRequest request, ServletRequestDataBinder dataBinder) {
    MutablePropertyValues mpvs = new ServletRequestParameterPropertyValues(request);
    MultipartRequest multipartRequest = WebUtils.getNativeRequest(request, MultipartRequest.class);
    if (multipartRequest != null) {
        bindMultipart(multipartRequest.getMultiFileMap(), mpvs);
    }

    String attr = HandlerMapping.URI_TEMPLATE_VARIABLES_ATTRIBUTE;
    Map<String, String> uriVars = (Map<String, String>) request.getAttribute(attr);
    if (uriVars != null) {
        for (Map.Entry<String, String> entry : uriVars.entrySet()) {
            if (mpvs.contains(entry.getKey())) {
                logger.warn("Skipping URI variable '" + entry.getKey()
                        + "' since the request contains a bind value with the same name.");
            } else {
                mpvs.addPropertyValue(entry.getKey(), entry.getValue());
            }
        }
    }

    this.extendDataBinder.doExtendBind(mpvs, dataBinder);

    dataBinder.bind(mpvs);
}