JPQL 和元组列表作为 SELECT IN 语句的参数

Question

给定以下表格布局：

CREATE TABLE things (
    id      BIGINT PRIMARY KEY NOT NULL,
    foo     BIGINT NOT NULL,
    bar     BIGINT NOT NULL
);

一个实体类（Kotlin）：

@Entity
@Table(name = "things")
class Thing(
        val foo: Long,
        val bar: Long
) : AbstractPersistable<Long>()

还有一个存储库：

interface ThingRepository : JpaRepository<Thing, Long> {
@Query("SELECT t FROM Thing t WHERE t.foo IN ?1")
fun selectByFoos(foos: Iterable<Long>): Iterable<Thing>

@Query("SELECT t FROM Thing t WHERE (t.foo, t.bar) IN ((1, 2), (3, 4))")
fun selectByFoosAndBarsFixed(): Iterable<Thing>

@Query("SELECT t FROM Thing t WHERE (t.foo, t.bar) IN ?1")
fun selectByFoosAndBars(foosAndBars: Iterable<Pair<Long, Long>>): Iterable<Thing>

以下两个调用工作正常：

repo.selectByFoos(listOf(1L, 3L))
repo.selectByFoosAndBarsFixed()

但是这个没有：

repo.selectByFoosAndBars(listOf(Pair(1L, 2L), Pair(3L, 4L)))

它抛出：

org.springframework.dao.DataIntegrityViolationException: could not extract ResultSet; SQL [n/a]; nested exception is org.hibernate.exception.DataException: could not extract ResultSet

Caused by: org.h2.jdbc.JdbcSQLException: Data conversion error converting "aced00057372000b6b6f746c696e2e50616972fa1b06813de78f780200024c000566697273747400124c6a6176612f6c616e672f4f626a6563743b4c00067365636f6e6471007e000178707372000e6a6176612e6c616e672e4c6f6e673b8be490cc8f23df0200014a000576616c7565787200106a6176612e6c616e672e4e756d62657286ac951d0b94e08b020000787000000000000000017371007e00030000000000000002"; SQL statement:
/* SELECT t FROM Thing t WHERE (t.foo, t.bar) IN ?1 */ select thing0_.id as id1_0_, thing0_.bar as bar2_0_, thing0_.foo as foo3_0_ from things thing0_ where (thing0_.foo , thing0_.bar) in (? , ?) [22018-197]

Caused by: java.lang.NumberFormatException: For input string: "aced00057372000b6b6f746c696e2e50616972fa1b06813de78f780200024c000566697273747400124c6a6176612f6c616e672f4f626a6563743b4c00067365636f6e6471007e000178707372000e6a6176612e6c616e672e4c6f6e673b8be490cc8f23df0200014a000576616c7565787200106a6176612e6c616e672e4e756d62657286ac951d0b94e08b020000787000000000000000017371007e00030000000000000002"

我猜作为参数传递的列表元素没有正确插入到查询中。我该如何纠正这个问题？

当然，我可以手动构建查询，如下所示：

@Repository
class SecondThingRepository(private val entityManager: EntityManager) {
    fun selectByFoosAndBars(foosAndBars: Iterable<Pair<Long, Long>>): Iterable<Thing> {
        val pairsRepr = foosAndBars.joinToString(prefix = "(", postfix = ")") { "(${it.first}, '${it.second}')" }
        val query: TypedQuery<Thing> = entityManager.createQuery("SELECT t FROM Thing t WHERE (t.foo, t.bar) IN $pairsRepr", Thing::class.java)
        return query.resultList
    }
}

不过这个好像记的很好。

Answer 1

首先，警告一句：并非所有数据库都支持(t.foo, t.bar) IN ((1, 2), (3, 4)) 语法。使用它会使您的应用程序不可移植。

我假设List 中Pairs 的数量可以是任意的（如果不是，有一个更简单的解决方案，涉及将IN 表达式修改为例如IN (?1, ?2, ?3) 并更新查询方法接受List 类型的三个参数。不过，我想这不是您要的）。

问题在于 Hibernate 不知道如何将 Pair 类映射到数据库类型。似乎集合元素的类型解析逻辑与“外部”类型的解析逻辑不同，因此listOf(listOf(1L, 2L), listOf(3L, 4L)) 也不起作用。

解决方案（请注意，这有点小技巧）是引入 Hibernate 的 UserType 能够映射 Pair 对象并将这个新创建的 PairType 用于 List 的元素。

首先，将以下类添加到您的项目中：

/* It is absolutely crucial that this class extend Pair. If the Pair class you're using
happens to be final, you will have to implement a Pair class yourself. 
For an explanation of why this is required, have a look at SessionFactory.resolveParameterBindType()
and TypeResolver.heuristicType() methods */
public class PairType extends Pair<Long, Long> implements UserType { 

    public PairType(Long first, Long second) {
        super(first, second);
    }

    public PairType() {
        super(null, null);
    }

    @Override
    public int[] sqlTypes() {
        return new int[] {Types.ARRAY};
    }

    @Override
    public Class returnedClass() {
        return Pair.class;
    }

    @Override
    public void nullSafeSet(PreparedStatement st, Object value, int index, SharedSessionContractImplementor session)
            throws HibernateException, SQLException {
        if (Objects.isNull(value)) {
            st.setNull(index, Types.ARRAY);
        } else {
            final Pair pair = (Pair) value;
            st.setArray(index, new Array() {


                @Override
                public Object getArray() throws SQLException {
                    // TODO Auto-generated method stub
                    return new Object[] {pair.getFirst(), pair.getSecond()};
                }

                ...
                //you can leave the rest of the autogenerated method stubs as they are

            });
        }
    }

    @Override
    public Object deepCopy(Object value) throws HibernateException {
        if (Objects.isNull(value)) {
            return null;
        }
        return Pair.of(((Pair) value).getFirst(), ((Pair) value).getSecond());
    }

    @Override
    public boolean isMutable() {
        return false;
    }

    ...
    //you can leave the rest of the autogenerated method stubs as they are here as well

} 

然后，将你的方法签名修改为：

selectByFoosAndBars(foosAndBars: Iterable<PairType>): Iterable<Thing>

注意：上述解决方案对我来说是开箱即用的 H2 数据库。你的旅费可能会改变。