我面对 "java.lang.IllegalArgumentException: Can not set int field com.example.demo.model.Customer.customerId to java.util.LinkedHashMap"

Question

我得到：

java.lang.IllegalArgumentException: Can not set int field com.example.demo.model.Customer.customerId to java.util.LinkedHashMap

在我的 Spring 引导项目中，我在其中尝试使用 Hibernate One-To-Many 关系 来保存两个 Pojo classes 的数据。我正在尝试保存已定义 Collection 元素集的持久 class 的值。

Parent Pojo Class:-

package com.example.demo.model;

import java.util.Set;

import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.OneToMany;
import javax.persistence.Table;

@Entity
@Table(name = "vendor")
public class Vendor {

    @Id
    int vendorId;

    @Column
    String vendorName;

    @OneToMany(fetch = FetchType.LAZY, targetEntity = Customer.class, cascade = CascadeType.ALL)
    @JoinColumn(name = "vendorId")

    Set children;

    public int getVendorId() {
        return vendorId;
    }

    public void setVendorId(int vendorId) {
        this.vendorId = vendorId;
    }

    public String getVendorName() {
        return vendorName;
    }

    public void setVendorName(String vendorName) {
        this.vendorName = vendorName;
    }

    public Set getChildren() {
        return children;
    }

    public void setChildren(Set children) {
        this.children = children;
    }
}

Child 波乔 Class:-

package com.example.demo.model;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.Table;

import org.hibernate.annotations.GeneratorType;

import com.fasterxml.jackson.databind.annotation.JsonDeserialize;

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

    @Id
    int customerId;

    @Column
    String customerName;

    public int getCustomerId() {
        return customerId;
    }

    public void setCustomerId(int customerId) {
        this.customerId = customerId;
    }

    public String getCustomerName() {
        return customerName;
    }

    public void setCustomerName(String customerName) {
        this.customerName = customerName;
    }

}

控制器Class:-

package com.example.demo.controller;

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.web.bind.annotation.PostMapping;
import org.springframework.web.bind.annotation.RequestBody;
import org.springframework.web.bind.annotation.RestController;

import com.example.demo.model.Vendor;
import com.example.demo.service.VendorDataSaveService;

@RestController
public class VendorSaveController {

    @Autowired
    private VendorDataSaveService dataSaveService;

    @PostMapping("/save")
    public void saveVendor(@RequestBody Vendor vendor) {
        dataSaveService.saveVendorRecord(vendor);
    }
}

服务Class:-

package com.example.demo.service;

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Service;

import com.example.demo.model.Vendor;
import com.example.demo.repository.VendorDataSaveRepository;

@Service
public class VendorDataSaveService {

    @Autowired
    private VendorDataSaveRepository repository;

    public void saveVendorRecord(Vendor vendor) {
        repository.save(vendor);
    }
}

存储库Class:-

package com.example.demo.repository;

import org.springframework.data.jpa.repository.JpaRepository;

import com.example.demo.model.Vendor;

public interface VendorDataSaveRepository extends JpaRepository<Vendor, Integer> {

}

我从 Postman 发送的 JSON 格式：-

{
    "vendorId" : 101,
    "vendorName" : "JAIN BOOKS",
    "children" : [{
                    "customerId" : 1,
                    "customerName" : "AMIT"
                 }]
}

来自控制台的错误消息：-

java.lang.IllegalArgumentException: Can not set int field com.example.demo.model.Customer.customerId to java.util.LinkedHashMap

邮递员的错误信息：-

Error accessing field [int com.example.demo.model.Customer.customerId] by reflection for persistent property [com.example.demo.model.Customer#customerId] : {customerId=1, customerName=AMIT}; nested exception is org.hibernate.property.access.spi.PropertyAccessException: Error accessing field [int com.example.demo.model.Customer.customerId] by reflection for persistent property [com.example.demo.model.Customer#customerId] : {customerId=1, customerName=AMIT}"

Answer 1

需要在设置子集合类型中添加通用类型 Customer 参数。

Set<Customer> children;

我面对 "java.lang.IllegalArgumentException: Can not set int field com.example.demo.model.Customer.customerId to java.util.LinkedHashMap"

I Am Facing "java.lang.IllegalArgumentException: Can not set int field com.example.demo.model.Customer.customerId to java.util.LinkedHashMap"

java

hibernate-mapping

hibernate-onetomany

spring-data-jpa

spring-boot