如何在 JPA CriteriaQuery 的 ORDER BY 子句中重写子查询
How to rewrite subquery in ORDER BY clause in JPA CriteriaQuery
我正在尝试使用 CriteriaQuery 编写 SQL 查询，但我很难做到。这个查询基本上得到一个发货清单，并按授权日期对它们进行排序。此授权日期表示为状态转换消息 table 中第一条记录的日期属性，初始状态为 3，最终状态为 4。这是我的查询：
SELECT s.id
FROM shipment s
ORDER BY (SELECT min(stm.date)
FROM status_transition_message stm
WHERE stm.initial_status = 1 AND stm.final_status = 3 AND stm.shipment_id = s.id) desc;
我尝试了多种不同的解决方案，但 none 到目前为止都有效。
我目前的迭代如下：
private void sortByAuthDate(Root<ShipmentTbl> root, CriteriaQuery<?> query, CriteriaBuilder builder, ListSort sort) {
Subquery<Timestamp> authDateQuery = query.subquery(Timestamp.class);
Root<StatusTransitionMessageTbl> stmRoot = authDateQuery.from(StatusTransitionMessageTbl.class);
Predicate shipmentId = builder.equal(stmRoot.<ShipmentTbl>get("shipment").<String>get("id"), root.<String>get("id"));
Predicate initialStatus = builder.equal(stmRoot.<Integer>get("initialStatus"), 3);
Predicate finalStatus = builder.equal(stmRoot.<Integer>get("finalStatus"), 4);
// returns the authorization date for each queried shipment
authDateQuery.select(builder.least(stmRoot.<Timestamp>get("date")))
.where(builder.and(shipmentId, initialStatus, finalStatus));
Expression<Timestamp> authDate = authDateQuery.getSelection();
Order o = sort.getSortDirection() == ListSort.SortDirection.ASC ? builder.asc(authDate) : builder.desc(authDate);
query.multiselect(authDate).orderBy(o);
}
该方案的问题是CriteriaQuery生成的SQL查询不支持ORDER BY子句中的子查询，导致解析异常。
我的 CriteriaQuery-fu 不足以帮助您完成该部分，但您可以将 SQL 查询重写为：
SELECT s.id
FROM shipment s
LEFT JOIN status_transition_message stm
ON stm.initial_status = 1 AND stm.final_status = 3 AND stm.shipment_id = s.id
GROUP BY s.id
ORDER BY min(stm.date) DESC;
对我来说，这很可能是比 运行 ORDER BY 子句中的相关子查询更快的解决方案，尤其是在优化器不太复杂的 RDBMS 上。
所以我尝试遵循@Lukas Eder 的解决方案并得出了这个解决方案：
private void sortByAuthDate(Root<ShipmentTbl> root, CriteriaQuery<?> query, CriteriaBuilder builder, ShipmentListSort sort) {
Join<ShipmentTbl, StatusTransitionMessageTbl> shipmentStatuses = root.join("shipmentStatus", JoinType.LEFT);
Predicate initialStatus = builder.equal(shipmentStatuses.<Integer>get("initialStatus"), 1);
Predicate finalStatus = builder.equal(shipmentStatuses.<Integer>get("finalStatus"), 3);
Expression<Timestamp> authDate = builder.least(shipmentStatuses.<Timestamp>get("date"));
Order o = sort.getSortDirection() == ShipmentListSort.SortDirection.ASC ? builder.asc(authDate) : builder.desc(authDate);
shipmentStatuses.on(builder.and(initialStatus, finalStatus));
query.multiselect(authDate).groupBy(root.<String>get("id")).orderBy(o);
}
}
但是现在它抛出这个异常：
ERROR o.h.e.jdbc.spi.SqlExceptionHelper - ERROR: for SELECT DISTINCT, ORDER BY expressions must appear in select list
发生这种情况是因为查询稍后只会获得不同的发货，并且它要求排序列也出现在 select 中。问题是我不知道如何强制 CriteriaQuery 将该列保留在 SELECT 语句中。它会自动只放入 ORDER BY。
这是它在我的测试中执行的 JPQL 查询：
select
distinct generatedAlias0
from
ShipmentTbl as generatedAlias0
left join
generatedAlias0.shipmentStatus as generatedAlias1 with ( generatedAlias1.initialStatus=:param0 )
and (
generatedAlias1.finalStatus=:param1
)
where
lower(generatedAlias0.shipmentName) like :param2
group by
generatedAlias0.id
order by
min(generatedAlias1.date) desc
和生成的 SQL 查询：
select
distinct shipmenttb0_.id as id1_13_,
shipmenttb0_.archived_date as archived2_13_,
shipmenttb0_.auth_code as auth_cod3_13_,
shipmenttb0_.authorization_date as authoriz4_13_,
shipmenttb0_.booked_in_by_user as booked_i5_13_,
shipmenttb0_.business_channel as business6_13_,
shipmenttb0_.courier as courier7_13_,
shipmenttb0_.courier_amount as courier_8_13_,
shipmenttb0_.courier_currency as courier_9_13_,
shipmenttb0_.ship_to as ship_to39_13_,
shipmenttb0_.estimated_shipment_date as estimat10_13_,
shipmenttb0_.last_updated_date as last_up11_13_,
shipmenttb0_.measurement_unit as measure12_13_,
shipmenttb0_.original_submitted_date as origina13_13_,
shipmenttb0_.packaging_type as packagi14_13_,
shipmenttb0_.placeholder_message as placeho15_13_,
shipmenttb0_.scheduled_period_of_day as schedul16_13_,
shipmenttb0_.scheduled_shipment_date as schedul17_13_,
shipmenttb0_.ship_from as ship_fr40_13_,
shipmenttb0_.ship_origin as ship_or41_13_,
shipmenttb0_.shipment_name as shipmen18_13_,
shipmenttb0_.status as status19_13_,
shipmenttb0_.submitted_date as submitt20_13_,
shipmenttb0_.supplier_contact_email as supplie21_13_,
shipmenttb0_.supplier_contact_name as supplie22_13_,
shipmenttb0_.supplier_contact_phone_number as supplie23_13_,
shipmenttb0_.supplier_email as supplie24_13_,
shipmenttb0_.supplier_secondary_contact_email as supplie25_13_,
shipmenttb0_.supplier_secondary_contact_name as supplie26_13_,
shipmenttb0_.supplier_secondary_contact_phone_number as supplie27_13_,
shipmenttb0_.tenant as tenant28_13_,
shipmenttb0_.total_received_boxes as total_r29_13_,
shipmenttb0_.total_units as total_u30_13_,
shipmenttb0_.total_value as total_v31_13_,
shipmenttb0_.total_volume as total_v32_13_,
shipmenttb0_.total_weight as total_w33_13_,
shipmenttb0_.tracking_number as trackin34_13_,
shipmenttb0_.tt_note as tt_note35_13_,
shipmenttb0_.tt_priority as tt_prio36_13_,
shipmenttb0_.updated_by_user as updated37_13_,
shipmenttb0_.weight_unit as weight_38_13_
from
shipment shipmenttb0_
left outer join
status_transition_message shipmentst1_
on shipmenttb0_.id=shipmentst1_.shipment_id
and (
shipmentst1_.initial_status=?
and shipmentst1_.final_status=?
)
where
lower(shipmenttb0_.shipment_name) like ?
group by
shipmenttb0_.id
order by
min(shipmentst1_.date) desc limit ?
我正在尝试使用 CriteriaQuery 编写 SQL 查询，但我很难做到。这个查询基本上得到一个发货清单，并按授权日期对它们进行排序。此授权日期表示为状态转换消息 table 中第一条记录的日期属性，初始状态为 3，最终状态为 4。这是我的查询：
SELECT s.id
FROM shipment s
ORDER BY (SELECT min(stm.date)
FROM status_transition_message stm
WHERE stm.initial_status = 1 AND stm.final_status = 3 AND stm.shipment_id = s.id) desc;
我尝试了多种不同的解决方案，但 none 到目前为止都有效。
我目前的迭代如下：
private void sortByAuthDate(Root<ShipmentTbl> root, CriteriaQuery<?> query, CriteriaBuilder builder, ListSort sort) {
Subquery<Timestamp> authDateQuery = query.subquery(Timestamp.class);
Root<StatusTransitionMessageTbl> stmRoot = authDateQuery.from(StatusTransitionMessageTbl.class);
Predicate shipmentId = builder.equal(stmRoot.<ShipmentTbl>get("shipment").<String>get("id"), root.<String>get("id"));
Predicate initialStatus = builder.equal(stmRoot.<Integer>get("initialStatus"), 3);
Predicate finalStatus = builder.equal(stmRoot.<Integer>get("finalStatus"), 4);
// returns the authorization date for each queried shipment
authDateQuery.select(builder.least(stmRoot.<Timestamp>get("date")))
.where(builder.and(shipmentId, initialStatus, finalStatus));
Expression<Timestamp> authDate = authDateQuery.getSelection();
Order o = sort.getSortDirection() == ListSort.SortDirection.ASC ? builder.asc(authDate) : builder.desc(authDate);
query.multiselect(authDate).orderBy(o);
}
该方案的问题是CriteriaQuery生成的SQL查询不支持ORDER BY子句中的子查询，导致解析异常。
我的 CriteriaQuery-fu 不足以帮助您完成该部分，但您可以将 SQL 查询重写为：
SELECT s.id
FROM shipment s
LEFT JOIN status_transition_message stm
ON stm.initial_status = 1 AND stm.final_status = 3 AND stm.shipment_id = s.id
GROUP BY s.id
ORDER BY min(stm.date) DESC;
对我来说，这很可能是比 运行 ORDER BY 子句中的相关子查询更快的解决方案，尤其是在优化器不太复杂的 RDBMS 上。
所以我尝试遵循@Lukas Eder 的解决方案并得出了这个解决方案：
private void sortByAuthDate(Root<ShipmentTbl> root, CriteriaQuery<?> query, CriteriaBuilder builder, ShipmentListSort sort) {
Join<ShipmentTbl, StatusTransitionMessageTbl> shipmentStatuses = root.join("shipmentStatus", JoinType.LEFT);
Predicate initialStatus = builder.equal(shipmentStatuses.<Integer>get("initialStatus"), 1);
Predicate finalStatus = builder.equal(shipmentStatuses.<Integer>get("finalStatus"), 3);
Expression<Timestamp> authDate = builder.least(shipmentStatuses.<Timestamp>get("date"));
Order o = sort.getSortDirection() == ShipmentListSort.SortDirection.ASC ? builder.asc(authDate) : builder.desc(authDate);
shipmentStatuses.on(builder.and(initialStatus, finalStatus));
query.multiselect(authDate).groupBy(root.<String>get("id")).orderBy(o);
}
}
但是现在它抛出这个异常：
ERROR o.h.e.jdbc.spi.SqlExceptionHelper - ERROR: for SELECT DISTINCT, ORDER BY expressions must appear in select list
发生这种情况是因为查询稍后只会获得不同的发货，并且它要求排序列也出现在 select 中。问题是我不知道如何强制 CriteriaQuery 将该列保留在 SELECT 语句中。它会自动只放入 ORDER BY。
这是它在我的测试中执行的 JPQL 查询：
select
distinct generatedAlias0
from
ShipmentTbl as generatedAlias0
left join
generatedAlias0.shipmentStatus as generatedAlias1 with ( generatedAlias1.initialStatus=:param0 )
and (
generatedAlias1.finalStatus=:param1
)
where
lower(generatedAlias0.shipmentName) like :param2
group by
generatedAlias0.id
order by
min(generatedAlias1.date) desc
和生成的 SQL 查询：
select
distinct shipmenttb0_.id as id1_13_,
shipmenttb0_.archived_date as archived2_13_,
shipmenttb0_.auth_code as auth_cod3_13_,
shipmenttb0_.authorization_date as authoriz4_13_,
shipmenttb0_.booked_in_by_user as booked_i5_13_,
shipmenttb0_.business_channel as business6_13_,
shipmenttb0_.courier as courier7_13_,
shipmenttb0_.courier_amount as courier_8_13_,
shipmenttb0_.courier_currency as courier_9_13_,
shipmenttb0_.ship_to as ship_to39_13_,
shipmenttb0_.estimated_shipment_date as estimat10_13_,
shipmenttb0_.last_updated_date as last_up11_13_,
shipmenttb0_.measurement_unit as measure12_13_,
shipmenttb0_.original_submitted_date as origina13_13_,
shipmenttb0_.packaging_type as packagi14_13_,
shipmenttb0_.placeholder_message as placeho15_13_,
shipmenttb0_.scheduled_period_of_day as schedul16_13_,
shipmenttb0_.scheduled_shipment_date as schedul17_13_,
shipmenttb0_.ship_from as ship_fr40_13_,
shipmenttb0_.ship_origin as ship_or41_13_,
shipmenttb0_.shipment_name as shipmen18_13_,
shipmenttb0_.status as status19_13_,
shipmenttb0_.submitted_date as submitt20_13_,
shipmenttb0_.supplier_contact_email as supplie21_13_,
shipmenttb0_.supplier_contact_name as supplie22_13_,
shipmenttb0_.supplier_contact_phone_number as supplie23_13_,
shipmenttb0_.supplier_email as supplie24_13_,
shipmenttb0_.supplier_secondary_contact_email as supplie25_13_,
shipmenttb0_.supplier_secondary_contact_name as supplie26_13_,
shipmenttb0_.supplier_secondary_contact_phone_number as supplie27_13_,
shipmenttb0_.tenant as tenant28_13_,
shipmenttb0_.total_received_boxes as total_r29_13_,
shipmenttb0_.total_units as total_u30_13_,
shipmenttb0_.total_value as total_v31_13_,
shipmenttb0_.total_volume as total_v32_13_,
shipmenttb0_.total_weight as total_w33_13_,
shipmenttb0_.tracking_number as trackin34_13_,
shipmenttb0_.tt_note as tt_note35_13_,
shipmenttb0_.tt_priority as tt_prio36_13_,
shipmenttb0_.updated_by_user as updated37_13_,
shipmenttb0_.weight_unit as weight_38_13_
from
shipment shipmenttb0_
left outer join
status_transition_message shipmentst1_
on shipmenttb0_.id=shipmentst1_.shipment_id
and (
shipmentst1_.initial_status=?
and shipmentst1_.final_status=?
)
where
lower(shipmenttb0_.shipment_name) like ?
group by
shipmenttb0_.id
order by
min(shipmentst1_.date) desc limit ?