【发布时间】:2016-09-12 18:24:41
【问题描述】:
当我尝试将对象保存到数据库时出现错误:
java.sql.SQLIntegrityConstraintViolationException: Cannot add or update a child row: a foreign key constraint fails (`smartphones`.`smartphone`, CONSTRAINT `fk_smartphone_resolution1` FOREIGN KEY (`resolution_id`) REFERENCES `resolution` (`id`) ON DELETE NO ACTION ON UPDATE NO ACTION)
第一件事是我在Smartphone 类中的引用列名称错误,但我检查了一下,它看起来不错。也许有人弄清楚这个问题的原因是什么?
数据库截图
用于创建智能手机表的 SQL 脚本
CREATE TABLE IF NOT EXISTS `smartphones`.`smartphone` (
`id` INT(11) NOT NULL AUTO_INCREMENT,
`name` VARCHAR(45) NULL DEFAULT NULL,
`resolution_id` INT(11) NOT NULL,
...other
PRIMARY KEY (`id`, `resolution_id`),
UNIQUE INDEX `id_UNIQUE` (`id` ASC),
INDEX `fk_smartphone_resolution1_idx` (`resolution_id` ASC),
CONSTRAINT `fk_smartphone_resolution1`
FOREIGN KEY (`resolution_id`)
REFERENCES `smartphones`.`resolution` (`id`)
ON DELETE NO ACTION
ON UPDATE NO ACTION)
Smartphone 类,但只有一个关系对象。
package com.project.model;
import javax.persistence.*;
@Entity
public class Smartphone {
private int id;
private String name;
private Resolution resolutionId;
@Id
@Column(name = "id")
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
@Basic
@Column(name = "name")
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
@ManyToOne(fetch=FetchType.EAGER, cascade=CascadeType.PERSIST)
@JoinColumn(name = "resolution_id", referencedColumnName = "id", nullable = false)
public Resolution getResolutionId() {
return resolutionId;
}
public void setResolutionId(Resolution resolutionId) {
this.resolutionId = resolutionId;
}
}
[编辑:解析智能手机型号并保存到数据库]
@RequestMapping(value = { "apple" }, method = RequestMethod.GET)
public String parseApple(ModelMap model) {
try {
String appleData = Utilities.getResourceAsString(this, "json/apple.json");
JSONArray array = new JSONArray(appleData);
Session session = sessionFactory.openSession();
Transaction transaction = session.beginTransaction();
for (int i = 0; i < array.length(); i++) {
Smartphone smartphone = new Smartphone();
String resolutionValue = array.getJSONObject(i).getString("resolution");
String resolution_w = resolutionValue.split(" ")[0];
String resolution_h = resolutionValue.split(" ")[2];
Resolution resolution = new Resolution();
resolution.setHeight(Integer.valueOf(resolution_h));
resolution.setWidth(Integer.valueOf(resolution_w));
resolution.setTypeId(typeService.findByCode(session, Resolution.serialId));
session.save(resolution);
smartphone.setResolutionId(resolution);
//other
session.save(smartphone);
break;
}
transaction.commit();
sessionFactory.close();
} catch (IOException e) {
e.printStackTrace();
}
return "index";
}
[编辑:添加]分辨率类:
@Entity
public class Resolution {
public static final int serialId = 106;
private int id;
private Integer height;
private Integer width;
private Type typeId;
private Collection<Smartphone> resolutionId;
@Id
@Column(name = "id")
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
@Basic
@Column(name = "height")
public Integer getHeight() {
return height;
}
public void setHeight(Integer height) {
this.height = height;
}
@Basic
@Column(name = "width")
public Integer getWidth() {
return width;
}
public void setWidth(Integer width) {
this.width = width;
}
@ManyToOne(fetch=FetchType.EAGER)
@JoinColumn(name = "type_id", referencedColumnName = "id", nullable = false)
public Type getTypeId() {
return typeId;
}
public void setTypeId(Type typeId) {
this.typeId = typeId;
}
@LazyCollection(LazyCollectionOption.FALSE)
@OneToMany(mappedBy = "resolutionId")
public Collection<Smartphone> getResolutionId() {
return resolutionId;
}
public void setResolutionId(Collection<Smartphone> resolutionId) {
this.resolutionId = resolutionId;
}
}
【问题讨论】:
-
当你保存“智能手机”时,它引用的“解决方案”应该是数据库中的持久实体。
-
是的,我正在使用 session.save()。我应该使用新事务来保存新对象吗?请检查。我添加了源代码。
-
对我来说也是,但它不起作用。
-
我添加了
Resolution类源代码。
标签: java mysql spring hibernate