b2b电子商务交易平台有哪些/seo查询是什么意思

今天做练习的遇到了一个问题，下面重现一下错误的场景：

下面是一个Base pojo，里面包含有多个Item pojo。这里我想要在查询base的时候可以把所有的item都查出来。

所以这里使用FetchType=EAGER，会立即引起相关联实体的加载动作，就是为了能够及时抓取所有的Item。

如果使用FetchType=LAZY，就会加载不到Base的所有Iteam。

import org.hibernate.annotations.GenericGenerator;import javax.persistence.*;
import java.util.HashSet;
import java.util.Set;@Entity
@Table(name="Bases")
public class Base {@Id@GenericGenerator(strategy="uuid",name="uuid")@GeneratedValue(generator = "uuid")private String id;private String baseName;@OneToMany(mappedBy = "base",cascade = CascadeType.ALL,fetch = FetchType.EAGER)private Set<Item> items = new HashSet<Item>();public Base(){}public String getId() {return id;}public void setId(String id) {this.id = id;}public Set<Item> getItems() {return items;}public void setItems(Set<Item> items) {this.items = items;}public String getBaseName() {return baseName;}public void setBaseName(String baseName) {this.baseName = baseName;}
}

import org.hibernate.annotations.GenericGenerator;import javax.persistence.*;@Entity
@Table(name = "Items")
public class Item {@Id@GenericGenerator(strategy="uuid",name="uuid")@GeneratedValue(generator = "uuid")private String id;private String itemName;private Integer itemNum;@ManyToOne@JoinColumn(name="b_id")private Base base;public Base getBase() {return base;}public void setBase(Base base) {this.base = base;}public Item(){}public String getId() {return id;}public void setId(String id) {this.id = id;}public String getItemName() {return itemName;}public void setItemName(String itemName) {this.itemName = itemName;}public Integer getItemNum() {return itemNum;}public void setItemNum(Integer itemNum) {this.itemNum = itemNum;}
}

结果在测试删除item的时候，发现无法删除。但是在删除base的时候是可以级联删除掉item的，而在删除item的时候，控制台只打印出两次select语句，并没有执行删除操作。。

  @Testpublic void deleteItem(){ItemServiceImpl itemService = context.getBean(ItemServiceImpl.class);String id = "8a5e9d4464f06f6b0164f06f6d950002";String deleteId = itemService.deleteItem(id);Assert.assertTrue(deleteId.equals(id));}

下面是service层：

 @Transactional@Overridepublic String deleteItem(String id) {return itemDao.deleteItem(id);}

下面是dao层：

    @Overridepublic String deleteItem(String id) {Item item = em.getReference(Item.class, id);em.remove(item);return id;}

结果最后控制台只打印了两次select语句：

Hibernate: select item0_.id as id1_1_0_, item0_.b_id as b_id4_1_0_, item0_.itemName as itemName2_1_0_, item0_.itemNum as itemNum3_1_0_, base1_.id as id1_0_1_, base1_.baseName as baseName2_0_1_ from Items item0_, Bases base1_ where item0_.b_id=base1_.id(+) and item0_.id=?Hibernate: select items0_.b_id as b_id4_1_0_, items0_.id as id1_1_0_, items0_.id as id1_1_1_, items0_.b_id as b_id4_1_1_, items0_.itemName as itemName2_1_1_, items0_.itemNum as itemNum3_1_1_ from Items items0_ where items0_.b_id=?

 之所以打印出两次select语句，是因为@ManyToOne的FetchType默认是FetchType.EAGER

@Target({ElementType.METHOD, ElementType.FIELD})
@Retention(RetentionPolicy.RUNTIME)
public @interface ManyToOne {Class targetEntity() default void.class;CascadeType[] cascade() default {};FetchType fetch() default FetchType.EAGER;boolean optional() default true;
}

所以在dao层中Item item = em.getReference(Item.class, id);，查询item的时候，就会加载关联的base的数据。

而@OneToMany的FetchType默认是FetchType.LAZY

@Target({ElementType.METHOD, ElementType.FIELD})
@Retention(RetentionPolicy.RUNTIME)
public @interface OneToMany {Class targetEntity() default void.class;CascadeType[] cascade() default {};FetchType fetch() default FetchType.LAZY;String mappedBy() default "";boolean orphanRemoval() default false;
}

解决方法：将@oneToMany中的fetch=FetchType.EAGER注释掉就可了。

但是我就是想要及时加载所有的Item数据，这种时候该怎么解决，后来尝试了很多方式，总于找到解决方法

解决方法就是把@ManyToOne的FetchType设置为FetchType.LAZY

@Entity
@Table(name = "Items")
public class Item {@Id@GenericGenerator(strategy="uuid",name="uuid")@GeneratedValue(generator = "uuid")private String id;private String itemName;private Integer itemNum;@ManyToOne(fetch = FetchType.LAZY)//加上LAZY@JoinColumn(name="b_id")private Base base;public Base getBase() {return base;}public void setBase(Base base) {this.base = base;}public Item(){}public String getId() {return id;}public void setId(String id) {this.id = id;}public String getItemName() {return itemName;}public void setItemName(String itemName) {this.itemName = itemName;}public Integer getItemNum() {return itemNum;}public void setItemNum(Integer itemNum) {this.itemNum = itemNum;}
}

测试：

Hibernate: select item0_.id as id1_1_0_, item0_.b_id as b_id4_1_0_, item0_.itemName as itemName2_1_0_, item0_.itemNum as itemNum3_1_0_ from Items item0_ where item0_.id=?
Hibernate: delete from Items where id=?

虽然经过很多尝试找到解决方法，但是还是不太明白为什么会使用FetchType.EAGER就导致无法删除，再慢慢研究吧，研究出来以后会贴出来，也希望有人可以解答我的疑惑。。