Question

所以我想在Spring内存中有两个表要查询。我已经成功地在名为Medicine的实体中对“ Drugs”表进行了建模，但是现在需要对“ drugInteraction”表进行建模-该表将具有drug_id（在medicine表中称为id的PK）和drugInteraction中的drug_name的组合表，作为组合的主键。

在python中使用的架构：

cursor.execute("CREATE TABLE drugs(id INT AUTO_INCREMENT PRIMARY KEY, name VARCHAR(255), description TEXT, toxicity TEXT)")


cursor.execute("CREATE TABLE drugInteractions (drug_id INT NOT NULL, name VARCHAR(90), description TEXT, PRIMARY KEY(drug_id, name), FOREIGN KEY (drug_id) REFERENCES drugs (id))")

drugInteractions表的一些示例数据：


drug_id     name       description

1       "Abciximab"         "The risk or severity of bleeding can be increased when Abciximab is combined with Lepirudin."
1       "Aceclofenac"        "The risk or severity of bleeding and hemorrhage can be increased when Aceclofenac is combined with Lepirudin."
1.      "Acemetacin"         "The risk or severity of bleeding and hemorrhage can be increased when Lepirudin is combined with Acemetacin."

药物表的一些示例数据：

id.       name.       description

1       "Lepirudin"      "Lepirudin is identical to...."
2       "Cetuximab"      "Cetuximab is an epidermal growth..."

这是我对Medicine.java的要求：

package com.example.configbackendspring;

import net.minidev.json.JSONObject;

import javax.persistence.*;

@Entity
@Table(name = "drugs")
public class Medicine {
    @Id
    @GeneratedValue
    @Column(name = "id")
    private Integer id;
    @Column(name = "name")
    private String name;
    @Column(name = "description")
    private String description;
    @Column(name = "toxicity")
    private String toxicity;

    public Medicine(int id, String name, String description, String toxicity) {
        this.id=id;
        this.name=name;
        this.description=description;
        this.toxicity=toxicity;
    }


    public Medicine(){}


    public int getId() {
        return id;
    }

    public void setId(Integer id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public String getDescription() {
        return description;
    }

    public void setDescription(String description) {
        this.description = description;
    }

    public String getToxicity() {
        return toxicity;
    }

    public void setToxicity(String toxicity) {
        this.toxicity = toxicity;
    }

    public JSONObject toJSONObject(){
        JSONObject object = new JSONObject();
        JSONObject medicineObject = new JSONObject();
        medicineObject.appendField("name", this.name);
        medicineObject.appendField("description", this.description);
        medicineObject.appendField("toxicity", this.toxicity);
        medicineObject.appendField("id", this.id);
        object.appendField("medicine", medicineObject);
        return object;

    }
}

这就是我对drugInteraction.java所拥有的...不起作用

package com.example.configbackendspring;

import net.minidev.json.JSONObject;

import javax.persistence.*;
import javax.resource.cci.Interaction;
import java.io.Serializable;


@Entity
@Table(name = "drugInteractions")
public class DrugInteraction {


    @EmbeddedId
    private InteractionId interactionId;

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


    public DrugInteraction(int drug_id, String name, String description) {
        this.interactionId.drug_name = name;
        this.interactionId.drug_id = drug_id;
        this.description=description;

    }


    public DrugInteraction(){}


    public Integer getId() {
        return interactionId.drug_id;
    }


    public String getName() {
        return interactionId.drug_name;
    }


    public String getDescription() {
        return description;
    }

    public void setDescription(String description) {
        this.description = description;
    }


    public JSONObject toJSONObject(){
        JSONObject object = new JSONObject();
        JSONObject interactionObject = new JSONObject();
        interactionObject.appendField("name", interactionId.drug_name);
        interactionObject.appendField("description", this.description);
        interactionObject.appendField("drug_id", interactionId.drug_id);
        object.appendField("drugInteraction", interactionObject);
        return object;

    }
}

这是InteractionId.java

package com.example.configbackendspring;

import lombok.*;

import java.io.Serializable;

@RequiredArgsConstructor
@NoArgsConstructor
@Getter
@Setter
@ToString
@EqualsAndHashCode
public class InteractionId implements Serializable
{

//    public InteractionId(int drug_id, String drug_name){
//        this.drug_id=drug_id;
//        this.drug_name=drug_name;
//
//    }


    @NonNull
    public int drug_id;

    @NonNull
    public String drug_name;
}

我当前的问题是我不知道如何将药物中的外键ID与复合键关联起来。上面的代码正在编译，但是数据库为空，因此导入必定会在某处失败

请，有人可以建议我如何更改文件以模拟上述行为吗？我在弄清楚如何使用外键对复合ID建模时遇到了麻烦

Answer 1

您有很多错误。如果您使用的是@EmbeddedId，则您的ID类必须是可嵌入的。

@Embeddable
public class InteractionId {
    @Column(name="name")
    String name;
    Long drugId; //type should be same as for ID field on Medicine

    //equals and hashcode etc.
}

您还需要从DrugInteraction到Medicine并用MapsId注释的关系：

@Entity
@Table(name = "drugInteractions")
public class DrugInteraction {

    @EmbeddedId
    private InteractionId interactionId;

    @MapsId("drugId")//value corresponds to property in the ID class
    @ManyToOne
    @JoinColumn(name = "drug_id")
    private Medicine medicine;
}

要保存新实例：

DrugInteraction di = new DrugInteraction();
Medicine medicine = //an existing medicine
di.setName("Some Name");
di.setMedicine(medicine);
//save

作为替代方案，也可以使用IDClass而不是EmbeddedId来做到这一点：

//not an embeddable
public class InteractionId {
    String name;
    Long drugId; //type should be same as for ID field on Medicine

    //equals and hashcode etc.
}

并更改映射：

@Entity
@Table(name = "drugInteractions")
@IdClass(InteractionId.class) //specify the ID class
public class DrugInteraction {

    @Id
    private String name;

    @Id
    @ManyToOne
    @JoinColumn(name = "drug_id")
    private Medicine medicine;
}

JPA和Hibernate-具有外键的复合主键

1 个答案: