为什么Find All方法选择了错误的字段

Question

public class Userr extends Model{
    @Id
    public Long id;
    @OneToMany
    List<Badge> badges;
}

public class Badge extends Model{
    @Id
    public Long id;
    @ManyToOne
    Userr user;
}

这些是我的2个型号，这是产生的错误：

  

[RuntimeException ：java.lang.IllegalArgumentException：查询抛出SQLException：未找到列"T1.USERR_ID"; SQL语句：select t0.id c0 , t1.id_ c1, t1.name_ c2, t1.desc_ c3, t1.img_url_ c4, t1.effect_ c5, t1.user_id c6 from userr t0 left outer join badge t1 on t1.userr_id = t0.id where t0.id = ? order by t0.id [42122-168]绑定值：[null]查询为：select t0.id c0 , t1.id_ c1, t1.name_ c2, t1.desc_ c3, t1.img_url_ c4, t1.effect_ c5, t1.user_id c6 from userr t0 left outer join badge t1 on t1.userr_id = t0.id where t0.id = ? order by t0.id（通过参考链：com.avaje.ebean.common.BeanList[0]->models.Userr["created_badges"]）]

我想知道为什么要使用findAll代替userr_id来翻译user_id查询 虽然我的模型文件中没有任何内容，但该字段称为user_id。

我尝试将user更改为userr，因为它将其翻译为<classname>_id

这是一个错误吗？或者我错了吗？

Answer 1

这不是一个BUG！它是DDL生成模式的默认行为。我将向您解释通常发生的可能情况：

1st case：您之前没有在数据库上创建表

当您第一次构建应用程序而不使用现有表时，我认为您之前遇到的问题不会发生（只要您使用默认配置值并将应用程序设置为DEV模式）。直截了当，假设你有两个像以前一样的模型：

@Entity
public class Userr extends Model {
    @Id
    public Long id;
    public String realName;
    @OneToMany
    List<Badge> badges;
}

@Entity
public class Badge extends Model {
    @Id
    public Long id;
    @ManyToOne
    Userr user;
}

在运行应用程序时，简单框架应该告诉DDL生成模式查看连接的数据库并查找是否存在名为userr和badge的表。因为您之前没有创建过这些表，所以框架将自动在您的数据库上生成这两个表。根据您的模型，框架将使用类似于（PostgreSQL脚本）的脚本生成您的表：

-- the java class name is 'Userr' generated as 'userr'
CREATE TABLE userr {
    -- long on java generated as bigint
    id        bigint,
    -- 'realName' on java generated as 'realname' and string on java generated as varchar whose length 255 
    realname  character varying(255),
    -- one to many 'badges' field is not generated
    CONSTRAINT pk_userr PRIMARY KEY (id), -- id as primary key
}
-- the java class name is 'Badge' generated as 'badge'
CREATE TABLE badge {
    id        bigint,
    -- 'Userr' Object with MayToOne relation generated as the foreign key to the id on 'userr' table
    userr_id  bigint,
    CONSTRAINT pk_badge PRIMARY KEY (id),                      
    CONSTRAINT fk_badge_userr FOREIGN KEY (userr_id)
       REFERENCES userr (id) MATCH SIMPLE
       ON UPDATE NO ACTION ON DELETE NO ACTION,

}

2nd case：您在数据库

上有现有表

假设您之前有2个名为user_account和badge的表格。 user_account表看起来像这样：

CREATE TABLE user_account {
    id        bigint,
    real_name character varying(100),
    CONSTRAINT pk_userr PRIMARY KEY (id),
}
CREATE TABLE badge {
    id        bigint,
    user_id   bigint,
    CONSTRAINT pk_badge PRIMARY KEY (id),                      
    CONSTRAINT fk_badge_user_account FOREIGN KEY (user_id)
       REFERENCES user_account (id) MATCH SIMPLE
       ON UPDATE NO ACTION ON DELETE NO ACTION,
}

您可以使用以下代码正确使用以前的型号（进行一些修改）来处理当前表格：

@javax.persistence.Entity
// define this model as table named 'user_account'
@javax.persistence.Table(name = "user_account")
public class Userr extends Model {
    @Id
    public Long id;
    // define this field as column named 'real_name' whose length is 100
    @javax.persistence.Column(name = "real_name", length = 100)
    public String realName;
    @OneToMany
    List<Badge> badges;
}

@javax.persistence.Entity
public class Badge extends Model {
    @Id
    public Long id;
    @ManyToOne
    // define this field as column named 'user_id' and 
    // this act as reference to model 'Userr' using field named 'id' on it
    @javax.persistence.JoinColumn(name = "user_id", referencedColumnName = "id")
    Userr user;
}

希望这可以帮助您更好地理解DDL生成模式..：）