Question

我有以下JPA Mapping（出于简洁目的，getter和setter出来，DDL也是从可能/可能不起作用的代码生成的）：

费用

@Entity
public class Expense {

@Id
@GeneratedValue
private Long id;


private String name;
private Long amount;
private Boolean monthly;

@OneToOne
@JoinColumn(name = "category")
@Fetch(FetchMode.JOIN)
private Category category;

@ManyToMany(cascade = {CascadeType.PERSIST, CascadeType.MERGE})
@JsonInclude(JsonInclude.Include.NON_EMPTY)
private List<Label> labels = new ArrayList<>();
//constructor, getters and setters...
}

分类

@Entity
public class Category {

@Id
private String name;

//constructor, getters and setters...
}

标签

@Entity
public class Label {

@Id
private String name;
//constructor, getters and setters...

}

使用JpaRepository

所以我使用的JpaRepository看起来像这样：

public interface ExpensesRepository extends JpaRepository<Expense, Long> {

    @Query("SELECT e FROM Expense e LEFT JOIN FETCH e.category")
    List<Expense> findAllExpensesExploded();
}

当我使用JpaRepository的默认findAll（）方法时，我得到 n + 1选择问题：

2017-01-03 19:35:22.665 DEBUG 26040 --- [nio-8080-exec-1] org.hibernate.SQL                        : select expense0_.id as id1_1_, expense0_.amount as amount2_1_, expense0_.category_name as category5_1_, expense0_.monthly as monthly3_1_, expense0_.name as name4_1_ from expense expense0_
2017-01-03 19:35:22.673 DEBUG 26040 --- [nio-8080-exec-1] org.hibernate.SQL                        : select category0_.name as name1_0_0_ from category category0_ where category0_.name=?
2017-01-03 19:35:22.674 TRACE 26040 --- [nio-8080-exec-1] o.h.type.descriptor.sql.BasicBinder      : binding parameter [1] as [VARCHAR] - [Rent]
2017-01-03 19:35:22.682 DEBUG 26040 --- [nio-8080-exec-1] org.hibernate.SQL                        : select category0_.name as name1_0_0_ from category category0_ where category0_.name=?
2017-01-03 19:35:22.683 TRACE 26040 --- [nio-8080-exec-1] o.h.type.descriptor.sql.BasicBinder      : binding parameter [1] as [VARCHAR] - [Insurance]

但是，当我使用自己的 findAllExpensesExploded（）方法时，我会得到一个SQL查询：

2017-01-03 19:35:22.691 DEBUG 26040 --- [nio-8080-exec-1] org.hibernate.SQL                        : select expense0_.id as id1_1_0_, category1_.name as name1_0_1_, expense0_.amount as amount2_1_0_, expense0_.category_name as category5_1_0_, expense0_.monthly as monthly3_1_0_, expense0_.name as name4_1_0_ from expense expense0_ left outer join category category1_ on expense0_.category_name=category1_.name

我的期望是 findAll（）和 findAllExpensesExploded（），用单个SQL查询执行。

我的查询有效，因为我似乎已正确构建它
但是为什么findAll（）不适用于给定的映射注释？ Spring Data是否可能忽略@Fetch注释？
另一个似乎有道理的问题是，默认 findAll（）是否只应用于简单实体？（其中simple被定义为无关联）。

Answer 1

默认提取模式是懒惰的。使用Spring Data JPA时，使用@NamedEntityGraph和@EntityGraph注释始终是一个好习惯。您可以浏览this

whether the default findAll() should only be used for simple entities? (where simple is defined as no associations).

获取模式 - LAZY只会触发主表。如果在代码中调用任何其他具有父表依赖关系的方法，则它将触发获取模式 - SELECT。

Spring Data JPA是否忽略@Fetch注释？

1 个答案: