JPQL 加入一對多數據庫

Question

我知道之前在這里已經提到過幾次，但我真的無法讓它為我工作，

我有兩個實體：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;

....
}

我還設置了 JPA Reposiotries，我想創建自定義查詢，它相當於：

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

到目前為止，我已經嘗試過這個：

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

以 expcetion 結束：

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)

我試過這樣的事情：

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

這不會導致任何錯誤，而是返回空結果。

這是一個微不足道的問題，但我之前沒有任何使用 jpql 的經驗 = /

編輯：

仍然得到空結果：

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

編輯2：

從語句中刪除 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

您的最后一個查詢搜索名稱為作為參數傳遞的成分名稱的所有配方。 那不是你想要的。 您想要的是所有具有成分名稱是作為參數傳遞的成分名稱的配方：

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

旁注：為什么我不能使用相同的配料制作兩個食譜？ 相當有限。 該關聯應該是ManyToMany。

Answer 2

您的 JPQl 查詢應如下所示：-（由於 JPA 已在連接元數據中烘焙，但對於@OnetoMany我們可以將其用於 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);