Question

我整天都被困在这一天。我有一个表单对API进行POST，我希望将数据保存到3个表中。

记录保存在一个具有自动生成ID的表格中（ Squad ）。在此表中插入时，我想读取自动生成的记录ID并将其插入到不同的表（ SquadPlayers ）中，另外还插入由第2个表格中的表单提交的额外字段（< strong> SquadPlayers ： GenericPlayerId ）。
还有一点关于我想从前端表单提交的内容。我想要提交最多11名球员的球队加上Ids的所有信息（这些ID是我希望在 SquadPlayers 表中提交的 GenericPlayerId 字段中保存的）。

我是后期编码的新手，特别是数据库和我为了学习目的而选择的这个新堆栈，所以如果你在这里看到任何愚蠢的东西，现在你知道为什么:-)所以如果你认为我完全关闭或错了我的数据库设计让我知道。

到目前为止，我在Squad和SquadPlayers的两个课程中都有这个。

Squad.java

package com.FUT.track.web.FUTtrackapplication.squads;

import javax.persistence.*;

@Entity
@Table(name="Squad")
public class Squad {

    @Id
    @GeneratedValue( strategy = GenerationType.IDENTITY )
    private int squadId;
    private String squadName;
    private String squadDescription;
    private String primaryFormation;

    @OneToOne(cascade = CascadeType.ALL)
    @JoinColumn(name = "playerId")
    private SquadPlayers squadPlayers;


    public Squad() {

    }

    public Squad(String squadName, String squadDescription, String primaryFormation, SquadPlayers squadPlayers) {
        super();
        this.squadName = squadName;
        this.squadDescription = squadDescription;
        this.primaryFormation = primaryFormation;
        this.squadPlayers = squadPlayers;

    }

    public int getSquadId() {
        return squadId;
    }

    public void setSquadId(int squadId) {
        this.squadId = squadId;
    }

    public String getSquadName() {
        return squadName;
    }

    public void setSquadName(String squadName) {
        this.squadName = squadName;
    }

    public String getSquadDescription() {
        return squadDescription;
    }

    public void setSquadDescription(String squadDescription) {
        this.squadDescription = squadDescription;
    }

    public String getPrimaryFormation() {
        return primaryFormation;
    }

    public void setPrimaryFormation(String primaryFormation) {
        this.primaryFormation = primaryFormation;
    }

    public SquadPlayers getSquadPlayers() {
        return squadPlayers;
    }

    public void setSquadPlayers(SquadPlayers squadPlayers) {
        this.squadPlayers = squadPlayers;
    }

}

SquadPlayers.java

 package com.FUT.track.web.FUTtrackapplication.squads;

import javax.persistence.*;

@Entity
@Table(name="SquadPlayers")
public class SquadPlayers {

    @Id
    private Integer playerId;
    private Integer squadId;
    private Integer genericPlayerId;

    @OneToOne(mappedBy = "squadPlayers")
    private Squad squad;

    public Integer getPlayerId() {
        return playerId;
    }

    public void setPlayerId(Integer playerId) {
        this.playerId = playerId;
    }

    public Integer getSquadId() {
        return squadId;
    }

    public void setSquadId(Integer squadId) {
        this.squadId = squadId;
    }


    public Squad getSquad() {
        return squad;
    }

    public void setSquad(Squad squad) {
        this.squad = squad;
    }

    public Integer getGenericPlayerId() {
        return genericPlayerId;
    }

    public void setGenericPlayerId(Integer genericPlayerId) {
        this.genericPlayerId = genericPlayerId;
    }

}

Answer 1

假设您希望如何执行数据流

您有预定义的player列表。 player可以属于多个小队唯一
您将通过为表单提交squad来为表单提交创建player。指定的玩家将拥有唯一的识别码。
每个squadPlayer都有playerStat

注意：

Player与Squad之间的关系为 ManyToMany

SquadPlayer与Player
之间的Squad是加入表
因此，Player和SquadPlayer之间的关系是 OneToMany

Squad和SquadPlayer之间的关系是 OneToMany

<强>播放器

@Entity @Table(name = "player") public class Player { @Id @GeneratedValue(strategy = GenerationType.IDENTITY) private int id; private String name; private String position; private String country; private String club; @OneToMany(mappedBy = "player") private Set<SquadPlayer> squadSet = new HashSet<>(); .... }

此处，您的Player实体与squadSet字段的 OneToMany 关联描述了玩家可以包含在多个小队中的事实。

<强>队

@Entity @Table(name="squad") public class Squad { @Id @GeneratedValue( strategy = GenerationType.IDENTITY ) private int id; private String name; private String description; private String primaryFormation; @OneToMany(mappedBy = "squad", cascade = {CascadeType.MERGE, CascadeType.PERSIST}) private Set<SquadPlayer> playerSet = new HashSet<>(); ... }

此处Squad实体与playerSet字段的 OneToMany 关联描述了一个Squad可能有多个玩家。请注意，与Player实体不同， OneToMany 注释定义了Merge和Persist的级联类型。当持久化Squad时，这会告诉hibernate 持久存在这种关系。

<强> SquadPlayer

@Entity @Table(name = "squad_player") public class SquadPlayer { @Id @GeneratedValue(strategy = GenerationType.IDENTITY) private int id; @Column(name = "generic_player_id") private int genericPlayerId; @ManyToOne @JoinColumn(name = "squad_id") private Squad squad; @ManyToOne @JoinColumn(name = "player_id") private Player player; @OneToOne(mappedBy = "squadPlayer", orphanRemoval = true, cascade = CascadeType.ALL) private PlayerStat playerStat; ... }

这里只有Player和Squad与相应加入列的映射的另一端

现在，您的PlayerStat实体与 SquadPlayer 具有 OneToOne 关系。当orphanRemoval=true被移除时，PlayerStat将SquadPlayer删除该条目（尽管可选）。我们还为此关系定义了级联规则。

<强> PlayerStat

@Entity @Table(name = "player_stat") public class PlayerStat { @Id @GeneratedValue(strategy = GenerationType.IDENTITY) private int id; private int performance; @OneToOne @JoinColumn(name = "squad_player_id") private SquadPlayer squadPlayer; ... }

坚持小队和所有这些关系

记住JPA或Hibernate确定数据库与对象图的关系。这些实体与定义他们的分歧的层级关系持续存在。因此，这是您可以遵循的流程，以便为Squad对象
进行正确的关联

创建new Squad()并使用提供的字段设置所有字段。

当您从表单获取播放器ID列表时，请在同一事务中提取这些player个对象

迭代所有player并为每个new SquadPlayer()创建PlayerStat。设置所有关联字段以及SquadPlayer字段。然后将每个Squad添加到playerSet对象的Squad字段。

最后坚持<input autocomplete="off" class="autocomplete-hint select" data-auto-complete-minlength="1" type="text" onFocus="searchCountry(this)"> <script type="text/javascript"> function searchCountry(a) { $(function() { var cache = {}; $(a).autocomplete({ appendTo: ".countrys", change: function (event, ui) { if (ui.item == null || ui.item == undefined) { $(a).val(""); $(a).empty(); $(a).next("#loading").hide(); } else { var position = $(".countrys").position(), left = position.left, top = position.top; $(".countrys > ul").css({ left: left + 20 + "px", top: top + 4 + "px" }); } }, minLength: 1, select: function (event, ui) { // Set autocomplete element to display the label this.value = ui.item.label; $(this).closest("tr").find(".countryid").val(ui.item.hotelid); $(this).closest("tr").find(".countryid").next(".countryid").val(ui.item.hotelid); // Store value in hidden field $('#hidden_field').val(ui.item.id); $('#hidden_field1').empty().text(ui.item.label); // Prevent default behaviour return false; }, source: function( request, response ) { $(a).next("#loading").show(); var term = request.term; if (term in cache) { response(cache[term]); return; } $.getJSON( "jsonloadi.bc", request, function( data, status, xhr ) { $(a).next("#loading").hide(); cache[ term ] = data; response( data ); }); } }); }); } </script>对象

如果您遵循此流程中的所有内容，则数据库应填充所有具有适当关系的表。

进一步阅读：

了解关系映射，请参阅此OneToMany，ManyToMany，ManyToOne和OneToOne快速指南

以JPA and Hibernate Cascade Types
开头的好文章
关于Cascading Best Practices
的另一篇好文章

在Hibernate中的多个表中插入数据

1 个答案: