使用 QueryDSL 和 JPASQLQuery 加入多對多關系

Question

我有以下實體：

@AllArgsConstructor
@EqualsAndHashCode(of = {"name"})
@Data
@NoArgsConstructor
@Entity
@Table(schema = "eat")
public class Pizza {

   @Id
   @GeneratedValue(strategy = GenerationType.SEQUENCE, generator="pizza_id_seq")
   private Integer id;

   @NotNull       
   private String name;

   @NotNull
   @Positive
   private Double cost;

   @ManyToMany
   @JoinTable(schema = "eat",
              name = "pizza_ingredient",
              inverseJoinColumns = { @JoinColumn(name = "ingredient_id") })
   private Set<Ingredient> ingredients;

}


@AllArgsConstructor
@EqualsAndHashCode(of = {"name"})
@Data
@NoArgsConstructor
@Entity
@Table(schema = "eat")
public class Ingredient {

   @Id
   @GeneratedValue(strategy = GenerationType.SEQUENCE, generator="ingredient_id_seq")
   private Integer id;

   @NotNull
   @Size(min=1, max=64)
   private String name;

}

我正在使用JPASQLQuery (4.2.2) 提供的JPASQLQuery object在 PostgreSQL 中創建一些本機查詢：

public JPASQLQuery<T> getJPASQLQuery() {
   return new JPASQLQuery<>(
      entityManager,
      PostgreSQLTemplates.builder().printSchema().build()
   );
}

問題來自嘗試使用join函數，例如：

QIngredient ingredient = QIngredient.ingredient;
QPizza pizza = QPizza.pizza;

StringPath ingredientPath = Expressions.stringPath("ingredient");
StringPath pizzaPath = Expressions.stringPath("pizza");
NumberPath<Double> costPath = Expressions.numberPath(Double.class, "cost");
Expression rowNumber = SQLExpressions.rowNumber().over().partitionBy(ingredientPath).orderBy(costPath.desc()).as("rnk");

JPASQLQuery subQuery = getJPASQLQuery()
   .select(ingredient.name.as(ingredientPath), pizza.name.as(pizzaPath), pizza.cost.as(costPath), rowNumber)
   .from(pizza)
   // The error is in next innerJoin
   .innerJoin((SubQueryExpression<?>) pizza.ingredients, ingredient)
   .where(ingredient.name.in(ingredientNames));

如果我保留當前的innerJoin((SubQueryExpression<?>) pizza.ingredients, ingredient)我會收到：

class com.querydsl.core.types.dsl.SetPath cannot be cast to class com.querydsl.core.types.SubQueryExpression

我無法刪除當前(SubQueryExpression<?>)因為innerJoin不接受SetPath作為參數。

另一方面，以下內容：

.from(pizza)               
.innerJoin(ingredient)

由於pizza_ingredient不包含在生成的查詢中，因此不起作用。

如何在innerJoin中使用具有上述多對多關系的JPASQLQuery ？

Answer 1

基本上，有兩種主要方法試圖解決它：

---

包括所需的本機功能

正如這里的一位 QueryDSL 開發人員所建議的那樣，用 JPA 替代品替換JPASQLQuery 。

---

為多對多表創建所需的Path

首先將name屬性添加到每個@Table注釋中很重要，因為內部是 QueryDSL NativeSQLSerializer class 用來生成from和join子句的屬性。

因此，例如：

@Table(schema = "eat")
public class Pizza ...

應替換為：

@Table(name = "pizza", schema = "eat")
public class Pizza ...

接下來，為多對多表創建自定義Path ：

RelationalPathBase<Object> pizzaIngredient = new RelationalPathBase<>(Object.class, "pi", "eat", "pizza_ingredient");
NumberPath<Integer> pizzaIngredient_PizzaId = Expressions.numberPath(Integer.class, pizzaIngredient, "pizza_id");
NumberPath<Integer> pizzaIngredient_IngredientId = Expressions.numberPath(Integer.class, pizzaIngredient, "ingredient_id");

所以完整的代碼是：

QIngredient ingredient = QIngredient.ingredient;
QPizza pizza = QPizza.pizza;

RelationalPathBase<Object> pizzaIngredient = new RelationalPathBase<>(Object.class, "pi", "eat", "pizza_ingredient");
NumberPath<Integer> pizzaIngredient_PizzaId = Expressions.numberPath(Integer.class, pizzaIngredient, "pizza_id");
NumberPath<Integer> pizzaIngredient_IngredientId = Expressions.numberPath(Integer.class, pizzaIngredient, "ingredient_id");

StringPath ingredientPath = Expressions.stringPath("ingredient");
StringPath pizzaPath = Expressions.stringPath( "pizza");
NumberPath<Double> costPath = Expressions.numberPath(Double.class, "cost");

Expression rowNumber = SQLExpressions.rowNumber().over().partitionBy(ingredientPath).orderBy(costPath.desc()).as("rnk");
NumberPath<Long> rnk = Expressions.numberPath(Long.class, "rnk");

SubQueryExpression subQuery = getJPASQLQuery()
   .select(ingredient.name.as(ingredientPath), pizza.name.as(pizzaPath), pizza.cost.as(costPath), rowNumber)
   .from(pizza)
   .innerJoin(pizzaIngredient).on(pizzaIngredient_PizzaId.eq(pizza.id))
   .innerJoin(ingredient).on(ingredient.id.eq(pizzaIngredient_IngredientId))
   .where(ingredient.name.in(ingredientNames));

return getJPASQLQuery()
          .select(ingredientPath, pizzaPath, costPath)
          .from(
              subQuery,
              Expressions.stringPath("temp")
          )
          .where(rnk.eq(1l))
          .fetch();