为什么在本机查询中Hibernate惰性负载的子实体？

Question

我不明白，当我使用JPQL与JOIN获取Hibernate应该做一个查询来加入子实体，但是当我想使用本机查询并用一个查询加入所有子实体时，Hibernate仍然懒惰地加载其他查询中的子实体。我正在使用Spring Data 2。

我应该做什么来避免懒惰加载或使用本机查询的n 1查询？

例子：

@Query(value = "SELECT recipe.*, r_ing.*, ing.* FROM recipe recipe join " +
        " on recipe.id = r.recipe_id " +
        " LEFT JOIN recipe_ingredients r_ing on r.recipe_id = r_ing.recipe_id " +
        " LEFT JOIN ingredient ing on r_ing.ingredient_id = ing.id where ing.names in (:ingredientsNames)",
        countQuery = "SELECT count(*) FROM recipe recipe join " +
                " on recipe.id = r.recipe_id " +
                " LEFT JOIN recipe_ingredients r_ing on r.recipe_id = r_ing.recipe_id " +
                " LEFT JOIN ingredient ing on r_ing.ingredient_id = ing.id where ing.names in (:ingredientsNames)",
        nativeQuery = true
)
Page<Recipe> findAllByIngredientsNames(List<String> ingredientsNames, Pageable page);

实体：

@Entity
public class Recipe {
    @OneToMany(mappedBy = "recipe", cascade = CascadeType.ALL, orphanRemoval = true)
    private List<RecipeIngredients> ingredients;
}
@Entity
public class RecipeIngredients implements Serializable {

    @EmbeddedId
    private RecipeIngredientsId recipeIngredientsId;

    @ManyToOne(fetch = FetchType.LAZY)
    @MapsId("recipeId")
    private Recipe recipe;

    @ManyToOne(fetch = FetchType.LAZY, cascade = {CascadeType.PERSIST, CascadeType.MERGE})
    @MapsId("ingredientId")
        private Ingredient ingredient;
}

@Entity
public class Ingredient {

    @NaturalId
    @Column(unique = true)
    private String name;
}

Answer 1

对于本机查询，Hibernate不知道如何映射高级数据。在您的例子中，您有一个获取配方实体的请求，实体映射器知道如何从SELECT*from Recipe中提取结果。但是components属性是反向映射，它被实现为一个延迟的init集合，后面是查询。这就是JPA和Spring数据为您所做的，但它们不够聪明，无法自动理解并进一步映射它，从而急切地将查询结果映射到集合属性。

此外，我猜您在查询结果中看到了多个相同的配方实体。

如果出于某种原因，您真的想要处理本机查询，那么只要正确使用它们：本机查询的结果通常不是JPA管理的实体，而是预测。

因此，为本机查询中的行创建一个特定的投影：

public class FullRecipeProjection {
    private final Integer recipeId; 
    private final Integer recipeIngredientsId;
    private final Integer ingredientId
    private final Integer ingredientName 

    /* Full-arg-constructor */
    public FullRecipeProjection (Integer recipeId, Integer recipeIngredientsId, Integer ingredientId, String ingredientName) {...}

}

然后可以创建查询：

@Query(value = "SELECT new FullRecipeProjection(recipe.recipeId, r_ing.recipeIngredientsId, ing.ingredientId, ing.IngredientName) FROM recipe recipe join " +
        " on recipe.id = r.recipe_id " +
        " LEFT JOIN recipe_ingredients r_ing on r.recipe_id = r_ing.recipe_id " +
        " LEFT JOIN ingredient ing on r_ing.ingredient_id = ing.id where ing.names in (:ingredientsNames)",
        countQuery = "SELECT count(*) FROM recipe recipe join " +
                " on recipe.id = r.recipe_id " +
                " LEFT JOIN recipe_ingredients r_ing on r.recipe_id = r_ing.recipe_id " +
                " LEFT JOIN ingredient ing on r_ing.ingredient_id = ing.id where ing.names in (:ingredientsNames)",
        nativeQuery = true
)
List<FullRecipeProjection> findAllByIngredientsNames(List<String> ingredientsNames);

然后可以将FullRecipeProjection的集合转换为配方的类似对象：

public class FullRecipe {
    private final Integer recipeId;
    private final Set<IngredientProjection> ingredients;
    public FullRecipe(Integer recipeId, Set<IngredientProjection> ingredients) {...}
}

public class IngredientProjection {
    private final Integer ingredientId;
    private final String ingredientName;
    public IngredientProjection(Integer ingredientId, String ingredientName) {...}
}

然后你可以像这样得到你想要的：

final List<FullRecipeProjection> data = repository.findAllByIngredientsNames(ingredientsNames);

final List<FullRecipe> results = data
    .stream()
    // extracting distinct identities of recipes, you have fetched  
    .map(FullRecipeProjection::recipeId)
    .distinct()
    // now we have unique key for the data and can map it
    .map(it -> 
         new FullRecipe(
             it, 
             // extracting all ingredients, which  were fetched in rows with references to recipe.
             data
                 .stream()
                 .filter(o -> o.recipeId.equals(it))
                 .map(ing -> new IngredientProjection(ing.ingredientId, ing.ingredientName))
                 .collect(Collectors.toSet())
    .collect(Collectors.toList()) ; 

很长一段路。但这就是它的工作原理。当您使用JPQL查询时，这种长时间的处理是由Hibernate完成的。

注意：分页对于这种数据提取来说是一个麻烦的操作：按照您指定的方式，分页不是最终结果，而是FullRecipeProjection，这可能会导致不完整的配方获取，当然，在页面不好的数据中（它可能只包含1个FullRecipe，可能无法完全加载！）。