Question

我正在尝试在具有ManyToOne关系的实体上在Hibernate中创建映射。我正在尝试这个：

CampaignItemSlot类：

 package models;

 import java.sql.Timestamp;

 import javax.persistence.Column;
 import javax.persistence.Entity;
 import javax.persistence.GeneratedValue;
 import javax.persistence.GenerationType;
 import javax.persistence.Id;
 import javax.persistence.JoinColumn;
 import javax.persistence.JoinTable;
 import javax.persistence.ManyToOne;
 import javax.persistence.Table;

@Entity
@Table(name = "campaign_item_slots")
public class CampaignItemSlot {

@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
private long id;

@JoinColumn(name = "advert_slot_id")
@ManyToOne
private AdvertSlot advertSlot;

private boolean active;

private Timestamp date_created;

 private Timestamp date_updated;

 public CampaignItemSlot() {
    super();
    // TODO Auto-generated constructor stub
 }
}

但是我在日志文件中得到了这个：

  Caused by: org.hibernate.HibernateException: Missing column: advertSlot_id in text_advertising.campaign_item_slots

这是我的表SQL：

  CREATE TABLE IF NOT EXISTS `text_advertising`.`campaign_item_slots` (
  `id` BIGINT NOT NULL AUTO_INCREMENT,   
  `campaign_item_id` BIGINT NOT NULL,
  `advert_slot_id` BIGINT NOT NULL,
  `active` TINYINT(1) NOT NULL DEFAULT TRUE,
  `date_created` DATETIME NOT NULL,
  `date_updated` DATETIME NOT NULL,
  PRIMARY KEY (`id`),
  INDEX `fk_campaignitems_1_idx` (`campaign_item_id` ASC),
  INDEX `fk_campaignitems_2_idx` (`advert_slot_id` ASC),
  CONSTRAINT `fk_campaign_item_slots_1`
  FOREIGN KEY (`campaign_item_id`)
  REFERENCES `text_advertising`.`campaignitems` (`id`)
  ON DELETE NO ACTION
  ON UPDATE NO ACTION,
  CONSTRAINT `fk_campaign_item_slots_2`
  FOREIGN KEY (`advert_slot_id`)
  REFERENCES `text_advertising`.`advert_slots` (`id`)
  ON DELETE NO ACTION
  ON UPDATE NO ACTION)
  ENGINE = InnoDB
  DEFAULT CHARACTER SET = utf8
  COLLATE = utf8_general_ci;

AdvertSlot类：

  package models;

  import java.sql.Timestamp;

  import javax.persistence.Entity;
  import javax.persistence.EnumType;
  import javax.persistence.Enumerated;
  import javax.persistence.GeneratedValue;
  import javax.persistence.GenerationType;
  import javax.persistence.Id;
  import javax.persistence.ManyToOne;
  import javax.persistence.OneToMany;
  import javax.persistence.Table;

  @Entity
  @Table(name = "advert_slots")
  public class AdvertSlot {

@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
private long id;

@ManyToOne
private Publication publication;

private String name;

private String description;

private boolean active;

private Timestamp date_created;

private Timestamp date_updated;

public AdvertSlot() {
    super();
    // TODO Auto-generated constructor stub
}

public long getId() {
    return id;
}

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

public Publication getPublication() {
    return publication;
}

public void setPublication(Publication publication) {
    this.publication = publication;
}

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 boolean isActive() {
    return active;
}

public void setActive(boolean active) {
    this.active = active;
}

public Timestamp getDate_created() {
    return date_created;
}

public void setDate_created(Timestamp date_created) {
    this.date_created = date_created;
}

public Timestamp getDate_updated() {
    return date_updated;
}

public void setDate_updated(Timestamp date_updated) {
    this.date_updated = date_updated;
}   
 }

不知怎的，Hibernate没有看到我的advert_slot_id，请帮忙吗？

Answer 1

这里的答案是通过扩展org.hibernate.cfg.DefaultNamingStrategy然后通过hibernate config引用它来创建自定义命名策略：hibernate.ejb.naming_strategy

以下是一个例子：

＆＃13;

    @Override
    public  String foreignKeyColumnName(String propertyName, String propertyEntityName, String propertyTableName, String referencedColumnName) {
    	String changed = CaseFormat.UPPER_CAMEL.to(CaseFormat.LOWER_UNDERSCORE, propertyName) + "_id";
    	return changed;
    }

＆＃13;

Answer 2

它将起作用。尝试这种方式：

第一张桌子：

words = zenPython.split()
for i in [',', '.', '-', '*', '!']:
    words = [x.strip(i) for x in words]    #list of words with ", . - * !" characters removed

第二张表：

@Table(name = "buyer_city")
public class BuyerCity{

    //other fields
    .....

    @OneToMany(mappedBy="buyerCity", fetch = FetchType.LAZY)
    @JsonManagedReference
    private Set<BuyerCityMapping> buyerCityMapping = new HashSet<>();
}

Hibernate manyToOne映射，多个下划线不起作用

2 个答案: