SpringBoot 不會在實體中保存 Set

Question

我的 TimeTrail 類有一組 TimeTrailStage。 保存后，如果查詢對象，舞台上的 timetrail_id 為空。 所以階段是空的。 我不知道我做錯了什么。 我查看了 OneToMany/ManyToOne 關系，但據我所知它是正確的。

時間軌跡類

package com.verhees.cm.model.competition;

import com.verhees.cm.model.stage.TimeTrialStage;
import lombok.AllArgsConstructor;
import lombok.Getter;
import lombok.NoArgsConstructor;
import lombok.Setter;
import lombok.experimental.SuperBuilder;

import javax.persistence.Entity;
import javax.persistence.OneToMany;
import java.util.Set;

@Entity
@NoArgsConstructor
@AllArgsConstructor
@SuperBuilder
@Getter
@Setter
public class TimeTrail extends Competition {
    @OneToMany(mappedBy = "competitie")
    private Set<TimeTrialStage> stages;
}

競賽類

package com.verhees.cm.model.competition;

import lombok.*;
import lombok.experimental.SuperBuilder;

import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.Table;
import javax.validation.constraints.Size;

@Entity
@SuperBuilder
@NoArgsConstructor
@AllArgsConstructor
@Getter
@Setter
public abstract class Competition {
    @Id
    @Size(min = 3, max = 15)
    private String name;
}

TimeTrailStage.class

public class TimeTrialStage extends Stage {

    @ManyToOne
    @JoinColumn(name="timetrail_id", nullable = false)
    private TimeTrail competitie;

    @ManyToMany
    @JoinTable(
            name = "team_stage",
            joinColumns = @JoinColumn(name = "stage_id"),
            inverseJoinColumns = @JoinColumn(name = "team_id"))
    private List<Team> teams = new ArrayList<>();

    @OneToMany(cascade = CascadeType.ALL)
    private List<Score> scores;

    @OneToMany(cascade = CascadeType.ALL)
    private List<TimeTrialStagePrediction> predictions;

}

舞台.class

package com.verhees.cm.model.stage;

import com.fasterxml.jackson.annotation.JsonAutoDetect;
import lombok.AllArgsConstructor;
import lombok.Getter;
import lombok.NoArgsConstructor;
import lombok.Setter;
import lombok.experimental.SuperBuilder;

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

import java.util.Date;

import static javax.persistence.GenerationType.IDENTITY;

@Entity
@AllArgsConstructor
@NoArgsConstructor
@SuperBuilder
@Getter
@Setter
public abstract class Stage {
    @Id
    @GeneratedValue(strategy = IDENTITY)
    private Long id;

    private Date date;
}

數據庫運行在 Mysql、MySQL5InnoDBDialect 上。

Answer 1

您應該嘗試在競賽類上使用@MappedSuperclass 。 現在休眠不知道如何將 id（名稱）映射到數據庫模型。

在這里您可以找到示例/文檔： https : //docs.oracle.com/javaee/5/api/javax/persistence/MappedSuperclass.html