mysql用子查询更新查询

Question

有人可以看到以下查询有什么问题吗？

当我跑步时，我得到：

  

＃1064 - 您的SQL语法出错;检查与MySQL服务器版本对应的手册，以获得正确的语法   在第8行的'a .CompetitionID = Competition.CompetitionID'附近

Update Competition
Set Competition.NumberOfTeams =
(
SELECT count(*) as NumberOfTeams
FROM PicksPoints
where UserCompetitionID is not NULL
group by CompetitionID
) a
where a.CompetitionID =  Competition.CompetitionID

Answer 1

主要问题是内部查询无法与外部where语句上的update子句相关联，因为where子过滤器首先应用于内部子查询甚至执行之前更新的表。处理这种情况的典型方法是multi-table update。

Update
  Competition as C
  inner join (
    select CompetitionId, count(*) as NumberOfTeams
    from PicksPoints as p
    where UserCompetitionID is not NULL
    group by CompetitionID
  ) as A on C.CompetitionID = A.CompetitionID
set C.NumberOfTeams = A.NumberOfTeams

演示：http://www.sqlfiddle.com/#!2/a74f3/1

Answer 2

谢谢，我没有INNER JOIN UPDATE的想法。

在原始查询中，错误是命名子查询，该子查询必须返回一个值，因此不能为别名。

UPDATE Competition
SET Competition.NumberOfTeams =
(SELECT count(*) -- no column alias
  FROM PicksPoints
  WHERE UserCompetitionID is not NULL
  -- put the join condition INSIDE the subquery :
  AND CompetitionID =  Competition.CompetitionID
  group by CompetitionID
) -- no table alias

应该为每个比赛记录提供技巧。

要注意：

效果与mellamokb提出的查询完全不同，后者不会更新没有相应PickPoints的比赛记录。

由于SELECT id, COUNT(*) GROUP BY id只会计算现有的ID值，

而SELECT COUNT(*)将始终返回一个值，如果没有选择记录则为0。

这可能会或可能不会成为您的问题。

0感知版的mellamokb查询将是：

Update Competition as C
LEFT join (
  select CompetitionId, count(*) as NumberOfTeams
  from PicksPoints as p
  where UserCompetitionID is not NULL
  group by CompetitionID
) as A on C.CompetitionID = A.CompetitionID
set C.NumberOfTeams = IFNULL(A.NumberOfTeams, 0)

换句话说，如果找不到相应的PickPoints，请将Competition.NumberOfTeams设置为零。

Answer 3

对于不耐烦的人：

UPDATE target AS t
INNER JOIN (
  SELECT s.id, COUNT(*) AS count
  FROM source_grouped AS s
  -- WHERE s.custom_condition IS (true)
  GROUP BY s.id
) AS aggregate ON aggregate.id = t.id
SET t.count = aggregate.count

如上所述，@mellamokb的回答减少到最大值

Answer 4

您可以检查您的 eav_attributes 表以查找每个图像角色的相关属性 ID，例如；

然后您可以使用这些将任何角色设置为所有产品的任何其他角色，例如；

UPDATE catalog_product_entity_varchar AS `v` INNER JOIN (SELECT `value`,`entity_id` FROM `catalog_product_entity_varchar` WHERE `attribute_id`=86) AS `j` ON `j`.`entity_id`=`v`.entity_id SET `v`.`value`=j.`value` WHERE `v`.attribute_id = 85 AND `v`.`entity_id`=`j`.`entity_id`

以上内容会将您的所有“基本”角色设置为同一产品的“小”图像。