Question

我知道之前已经提到了几次，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 Reposiotries，我想创建自定义查询，它相当于：

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expcetion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。

这是一个繁琐的问题，但我没有任何jpql之前的经验= /

编辑：

仍然是空的结果：

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

您的上一个查询将搜索其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。你想要的是所有的配方都有一个成分，其名称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：为什么我不能使用相同成分的两种食谱？相当有限。该关联应该是ManyToMany。

JPQL加入一对多数据库

1 个答案: