JPQL加入一对多数据库

Question

我知道它之前已经在这里提到过几次，但我真的无法让它为我工作，

我有两个实体：配方，成分：

@Entity
@Table(name = "Recipe")
public class Recipe {

    @Id
    @GeneratedValue
    @Column(name = "Recipe_id")
    private Long id;

    private String name;

    private String description;
    @Lob
    @Basic(fetch = FetchType.LAZY)
    @Column(length = 100000)
    private byte[] image;

    @OneToMany(mappedBy = "recipe", fetch = FetchType.EAGER)
    @Cascade({CascadeType.ALL})
    private List<Ingredient> ingredientsList;
....
}

和成分：

@Entity
@Table
public class Ingredient {

    @Id
    @GeneratedValue
    private Long id;

    private String name;

    private int cpt;

    private String cptyType;

    @ManyToOne
    @JoinColumn(name = "Recipe_id")
    private Recipe recipe;

....
}

我还设置了 JPA 存储库，我想创建自定义查询，相当于：

SELECT *
FROM `Recipe`
INNER JOIN `Ingredient` ON Recipe.Recipe_id = Ingredient.Recipe_id
WHERE Ingredient.name = "fancyName"
LIMIT 0 , 30

到目前为止，我已经尝试过这个：

@Query("Select r from Recipe r join r.id i  where i.name = :ingredient")
List<Recipe> findRecipeByIngredient(@Param("ingredient") String ingredient);

以期望结束：

Caused by: java.lang.NullPointerException
        at org.hibernate.hql.internal.ast.HqlSqlWalker.createFromJoinElement(HqlSqlWalker.java:395)
        at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.joinElement(HqlSqlBaseWalker.java:3477)
        at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.fromElement(HqlSqlBaseWalker.java:3263)
        at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.fromElementList(HqlSqlBaseWalker.java:3141)
        at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.fromClause(HqlSqlBaseWalker.java:694)
        at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.query(HqlSqlBaseWalker.java:550)
        at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.selectStatement(HqlSqlBaseWalker.java:287)
        at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.statement(HqlSqlBaseWalker.java:235)
        at org.hibernate.hql.internal.ast.QueryTranslatorImpl.analyze(QueryTranslatorImpl.java:248)
        at org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:183)
        at org.hibernate.hql.internal.ast.QueryTranslatorImpl.compile(QueryTranslatorImpl.java:136)
        at org.hibernate.engine.query.spi.HQLQueryPlan.<init>(HQLQueryPlan.java:101)
        at org.hibernate.engine.query.spi.HQLQueryPlan.<init>(HQLQueryPlan.java:80)
        at org.hibernate.engine.query.spi.QueryPlanCache.getHQLQueryPlan(QueryPlanCache.java:119)
        at org.hibernate.internal.AbstractSessionImpl.getHQLQueryPlan(AbstractSessionImpl.java:214)
        at org.hibernate.internal.AbstractSessionImpl.createQuery(AbstractSessionImpl.java:192)

我尝试过这样的事情：

@Query("Select r from Recipe r join fetch r.ingredientsList where r.name = :ingredient")
    List<Recipe> findRecipeByIngredient(@Param("ingredient") String ingredient);

这不会导致任何错误，但会返回空结果。

这是一个琐碎的问题，但我之前对 jpql 没有任何经验 = /

编辑：

仍然得到空结果：

DEBUG (SqlStatementLogger.java:104) - select recipe0_.Recipe_id as Recipe1_1_0_, ingredient1_.id as id0_1_, recipe0_.description as descript2_1_0_, recipe0_.image as image1_0_, recipe0_.name as name1_0_, ingredient1_.cpt as cpt0_1_, ingredient1_.cptyType as cptyType0_1_, ingredient1_.name as name0_1_, ingredient1_.Recipe_id as Recipe5_0_1_, ingredient1_.Recipe_id as Recipe5_1_0__, ingredient1_.id as id0__ from Recipe recipe0_ inner join Ingredient ingredient1_ on recipe0_.Recipe_id=ingredient1_.Recipe_id where ingredient1_.name=?
DEBUG (CollectionLoadContext.java:224) - No collections were found in result set for role: com.bla.model.Recipe.ingredientsList

编辑 2：

从语句中删除 fetch 后：

DEBUG (SqlStatementLogger.java:104) - select recipe0_.Recipe_id as Recipe1_1_, recipe0_.description as descript2_1_, recipe0_.image as image1_, recipe0_.name as name1_ from Recipe recipe0_ inner join Ingredient ingredient1_ on recipe0_.Recipe_id=ingredient1_.Recipe_id where ingredient1_.name=?
DEBUG (StatefulPersistenceContext.java:899) - Initializing non-lazy collections

Answer 1

您的最后一个查询搜索名称为作为参数传递的成分名称的所有食谱。那不是你想要的。您想要的是所有配方的名称都是作为参数传递的成分名称：

select r from Recipe r
join r.ingredientList i
where i.name = :ingredient

旁注：为什么我不能有两个使用相同成分的食谱？相当有限。关联应该是ManyToMany。

Answer 2

您的 JPQl 查询应如下所示：- （因为 JPA 已经在连接元数据中烘焙，但对于 @OnetoMany，我们可以将其用于 JPQl）

@Query("Select r from Recipe r join Ingredient i on r.id = i.recipe.id where i.name=:ingredient)

List<Recipe> findRecipeByIngredient(@Param("ingredient") String ingredient);