Java Hibernate中缺少列错误?
我在Hibernate中遇到缺少列错误。整个错误如下:org.hibernate.hibernateeexception:vaccum.inquiry中缺少列:inquiry\u type\u inquiry\u type\u id。这是我的密码Java Hibernate中缺少列错误?,java,hibernate,hibernate-mapping,hibernate-criteria,Java,Hibernate,Hibernate Mapping,Hibernate Criteria,我在Hibernate中遇到缺少列错误。整个错误如下:org.hibernate.hibernateeexception:vaccum.inquiry中缺少列:inquiry\u type\u inquiry\u type\u id。这是我的密码 package com.beans; import java.sql.Timestamp; import javax.management.loading.PrivateClassLoader; import javax.persistence.C
package com.beans;
import java.sql.Timestamp;
import javax.management.loading.PrivateClassLoader;
import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.ManyToOne;
import javax.persistence.OneToOne;
import javax.persistence.PrimaryKeyJoinColumn;
import javax.persistence.Table;
import org.hibernate.annotations.Cascade;
@Entity
@Table(name="enquiry")
public class Enquiry {
@Id@GeneratedValue
@Column(name="enquiry_id")
int enquiry_id;
@Column(name="text")
String text;
@Column(name="location")
String location;
@Column(name="name")
String name;
@Column(name="mobile")
String mobile;
@Column(name="date_time")
Timestamp date_time;
/////////////////////
@ManyToOne(cascade = CascadeType.ALL)
private Enquiry_Type enquiry_type;
public Enquiry_Type getEnquiry_type() {
return enquiry_type;
}
public void setEnquiry_type(Enquiry_Type enquiry_type) {
this.enquiry_type = enquiry_type;
}
//////////////////////////
public Enquiry() {
// TODO Auto-generated constructor stub
}
public Enquiry(String text,String location,String name,String mobile,Timestamp date_time) {
this.text = text;
this.location = location;
this.name = name;
this.mobile = mobile;
this.date_time = date_time;
}
public int getEnquiry_id() {
return enquiry_id;
}
public void setEnquiry_id(int enquiry_id) {
this.enquiry_id = enquiry_id;
}
public String getText() {
return text;
}
public void setText(String text) {
this.text = text;
}
public String getLocation() {
return location;
}
public void setLocation(String location) {
this.location = location;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getMobile() {
return mobile;
}
public void setMobile(String mobile) {
this.mobile = mobile;
}
public Timestamp getDate_time() {
return date_time;
}
public void setDate_time(Timestamp date_time) {
this.date_time = date_time;
}
}
查询_Type.java在这里
package com.beans;
import java.sql.Timestamp;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.Table;
@Entity
@Table(name="enquiry_type")
public class Enquiry_Type {
@Id@GeneratedValue
@Column(name="enquiry_type_id")
int enquiry_id;
@Column(name="enquiry_type")
String enquiry_type;
public Enquiry_Type() {
}
public Enquiry_Type(String enquiry_type) {
this.enquiry_type = enquiry_type;
}
public String getEnquiry_type() {
return enquiry_type;
}
public void setEnquiry_type(String enquiry_type) {
this.enquiry_type = enquiry_type;
}
}
我从SessionFactory=cfg.buildSessionFactory()获取错误代码>。下面给出了代码
//creating configuration object
Configuration cfg=new Configuration();
cfg.configure("hibernate.cfg.xml");//populates the data of the configuration file
//creating seession factory object
SessionFactory factory=cfg.buildSessionFactory();
//creating session object
Session session=factory.openSession();
//creating transaction object
Transaction t=session.beginTransaction();
表1:查询
查询| id |文本|位置|姓名|手机|类型|日期|时间|
表2:查询类型
查询类型(id |查询类型)|
这里,Enquiry.type=Enquiry\u type.Enquiry\u type\u id您是否可以交叉检查表“Enquiry”中是否有一个名为“Enquiry\u type\u Enquiry\u type\u id”的列
您必须确保表中的所有列都应该存在于实体类中如何生成数据库架构?如果使用hibernate执行此操作,则应将hibernate.hbm2ddl.auto属性的值更改为“创建”或“创建删除”。首先确保您的要求,如果要将父元素及其新的OneToMany children by cascade属性插入,请使用下面的一个
@OneToMany(cascade = CascadeType.ALL)
private Enquiry_Type enquiry_type;
注释@JoinColumn表示该实体是关系的所有者,即对应的表有一列,该列带有引用表的外键)
你能把你的表结构也添加到questionWell中吗?)。你已经把表注释为“@table(name=“enquiry\u type”)。所以你的类应该在enquiry\u type表中有列。例如,查询类型和查询类型id应该出现在您的查询类型类中。试试看我已经更新了我的答案,并显示了另一个具有查询类型和查询类型id的类。@ThomasNo。很多事情我已经申请了,但还没有解决@Thomashey您遗漏了查询类型字段的@column name注释。这可能是因为它是外键约束。因此,我尝试在@ManyToOne(cascade=CascadeType.ALL)
注释上编写此字段,但它表明在这种关系上无法应用此字段@war_Heroyes这是正确的,但是,如果您不编写@column name注释,那么在我使用的“validate”数据库中永远不会创建feild@通过使用“创建”或“创建放置”将9kb以上的数据放入板条箱中,每次都会将整个表重新装入。所以,这不是我想要的解决方案。因为,我想要数据库中以前的数据。是的,它将表1中的字段名“type”更改为“inquiry\u type\u inquiry\u type\u id”,其中包含外键@超过9K
@ManyToOne
@JoinColumn(name = "enquiryTypeID")
private Enquiry_Type enquiry_type;