Question

我尝试使用以下方法将记录插入到预订表中，

DAO方法

public boolean addBooking(String title,String desc,int slotid,int subscriberid,Session session){
        Booking booking = new Booking();
        booking.setTitle(title);
        booking.setDesc(desc);
        booking.setSlotid(slotid);
        booking.setSubscriberid(subscriberid);
        //booking.setCreated(new Date());
        Integer bookingid = (Integer) session.save(booking);
        if(bookingid > 0)
        return true;
        return false;
    }

错误

 insert 
    into
        BOOKING
        (created, desc, slotid, subscriberid, title) 
    values
        (?, ?, ?, ?, ?)
Hibernate: 
    insert 
    into
        BOOKING
        (created, desc, slotid, subscriberid, title) 
    values
        (?, ?, ?, ?, ?)
2015-11-01 00:43:00 DEBUG SqlExceptionHelper:139 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'desc, slotid, subscriberid, title) values (null, 'hello12', 24, 39, 'hello1')' at line 1 [n/a]

预订域类

import java.util.Date;
import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.OneToOne;
import javax.persistence.Table;
import javax.persistence.Temporal;
import javax.persistence.TemporalType;
import javax.persistence.Version;


@Entity
@Table(name = "BOOKING")
public class Booking {

    public Booking(){

    }

    @Id @GeneratedValue(strategy=GenerationType.IDENTITY)
    @Column(name = "id")
    private Integer id;

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

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


    @OneToOne(fetch=FetchType.EAGER)
    @JoinColumn(name = "slotid",insertable = false, updatable = false)
    private Slot slot;

    private Integer slotid;

    private Integer subscriberid;


    @OneToOne(fetch=FetchType.EAGER)
    @JoinColumn(name = "subscriberid",insertable = false, updatable = false)
    private User subscriber;

    @Column(name = "created")
    @Temporal(TemporalType.TIMESTAMP)
    private Date created;

    public Integer getId() {
        return id;
    }

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

    public String getTitle() {
        return title;
    }

    public void setTitle(String title) {
        this.title = title;
    }

    public Slot getSlot() {
        return slot;
    }

    public void setSlot(Slot slot) {
        this.slot = slot;
    }

    public User getSubscriber() {
        return subscriber;
    }

    public void setSubscriber(User subscriber) {
        this.subscriber = subscriber;
    }

    public Date getCreated() {
        return created;
    }

    public void setCreated(Date created) {
        this.created = created;
    }

    public String getDesc() {
        return desc;
    }

    public void setDesc(String desc) {
        this.desc = desc;
    }

    public Integer getSlotid() {
        return slotid;
    }

    public void setSlotid(Integer slotid) {
        this.slotid = slotid;
    }

    public Integer getSubscriberid() {
        return subscriberid;
    }

    public void setSubscriberid(Integer subscriberid) {
        this.subscriberid = subscriberid;
    }



}

如果我没有将值设置为created字段，则会插入一个空值 - 但是，它默认为mysql表列中的CURRENT_TIMESTAMP。当我设置它的值时，它会抛出另一个语法错误，如下所示

您的SQL语法有错误;检查与MySQL服务器版本对应的手册，以获得正确的语法使用near＆＃39; desc，slotid，subscriberid，title）值（＆＃39; 2015-11-01 00：47：25＆＃39;，＆＃39; hello12＆＃39;，24，＆＃39;在第1行

Answer 1

desc 是关键字。您需要使用反引号或重命名列来引用它：

insert into BOOKING
        (created, `desc`, slotid, subscriberid, title) 
values

<强> 5.4. SQL quoted identifiers ：

您可以强制Hibernate在生成的SQL中引用标识符在映射中的反引号中包含表或列名称文献。 Hibernate将为SQL使用正确的引用样式方言。这通常是双引号，但SQL Server使用括号和MySQL使用反引号。

@Column(name = "`desc`")
private String desc;

插入期间HIbernate语法sql错误

1 个答案: