与QueryDSL和JPASQLQuery建立多对多关系
我有以下实体:
@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;
}
我使用 JPASQLQuery 代码>通过提供的对象的 QueryDSL 强>(4.2.2)中的PostgreSQL创建一些原生查询:
I'm using JPASQLQuery
object provided by QueryDSL (4.2.2) to create some native queries in PostgreSQL:
public JPASQLQuery<T> getJPASQLQuery() {
return new JPASQLQuery<>(
entityManager,
PostgreSQLTemplates.builder().printSchema().build()
);
}
尝试使用 join
函数出现问题,例如:
The problem comes trying to use join
functions, for example:
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));
如果我保持当前的 innerJoin((SubQueryExpression<>)pizza.ingredients,成分?)代码>我接收:
If I keep the current innerJoin((SubQueryExpression<?>) pizza.ingredients, ingredient)
I receive:
类com.querydsl.core.types.dsl.SetPath不能转换到类com.querydsl.core.types.SubQueryExpression
我不能删除当前的(SubQueryExpression<?>)
,因为 innerJoin
不接受 SetPath
作为参数.
I cannot remove current (SubQueryExpression<?>)
because innerJoin
doesn't accept SetPath
as parameter.
另一方面,以下内容:
.from(pizza)
.innerJoin(ingredient)
由于 pizza_ingredient
未包含在生成的查询中而无法使用.
Doesn't work due to pizza_ingredient
is not included in the generated query.
如何使用 innerJoin 代码>在 JPASQLQuery 代码>有许多以像上面许多关系?
How can I use innerJoin
in JPASQLQuery
with a many to many relationship like above?
基本上,有试图解决它的两种主要方法:
Basically, there are two main approaches trying to solve it:
建议一位QueryDSL开发人员此处,替换为 JPASQLQuery
>通过JPA替代方案.
As suggest one QueryDSL developer here, replacing JPASQLQuery
by JPA alternatives.
首先重要的一点是将 name
属性添加到每个 @Table
批注中,因为内部是QueryDSL NativeSQLSerializer
类用于生成 from
和 join
子句.
First is important to add name
property into every @Table
annotation because internally is the one used by QueryDSL NativeSQLSerializer
class to generate from
and join
clauses.
例如,
@Table(schema = "eat")
public class Pizza ...
应替换为:
@Table(name = "pizza", schema = "eat")
public class Pizza ...
接下来,为多对多表创建自定义的 Path
:
Next, create for custom Path
for the many to many table:
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");
因此完整的代码应为:
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();