Question

我正在尝试使用重复记录清理数据库。我需要将引用移动到单个记录并删除另一个记录。

我有两个表：促销员和场地，每个表都有一个名为cities的表的引用。问题是有些城市名称相同，ID不同，与场地和推广人有关系。

通过此查询，我可以将所有发起人和场地分组为一个城市记录：

SELECT c.id as id, c.name as name, GROUP_CONCAT( DISTINCT p.id ) as promoters_ids, GROUP_CONCAT( DISTINCT v.id ) as venues_ids
FROM cities as c
LEFT JOIN promoters as p ON p.city_id = c.id
LEFT JOIN venues as v ON v.city_id = c.id
WHERE c.name IN ( SELECT name from cities group by name having count(cities.name) > 1 )
GROUP BY c.name

现在我想在启动器上运行UPDATE查询，将city_id设置为上面查询的结果。

这样的事情：

    UPDATE promoters AS pr SET pr.city_id = (
        SELECT ID
        FROM (
            SELECT c.id as id, c.name as name, GROUP_CONCAT( DISTINCT p.id ) as promoters_ids
            FROM cities as c
            LEFT JOIN promoters as p ON p.city_id = c.id

            WHERE c.name IN ( SELECT name from cities group by name having count(cities.name) > 1 ) AND pr.id IN promoters_ids
            GROUP BY c.name
            ) AS T1 

    )

我该怎么做？

由于

Answer 1

如果我理解正确，您希望删除重复的城市（最后），因此您需要更新链接到您要在该过程中删除的任何城市的推广者。

我认为使用具有相同名称的任何城市的最低ID是有意义的（也可能是最高的，但我想至少指定它，并且不要留给我。

因此，为了获得推广者的正确ID，我需要：选择与已经链接到推广者的城市同名的所有城市的最低ID。

幸运的是，这种需求非常适合查询：

UPDATE promoters AS pr 
SET pr.city_id = (
  SELECT 
    -- Select the lowest ID ..
    Min(c.id)
  FROM
    -- .. of all cities ..
    Cities c
    -- .. that have the same name ..
    INNER JOIN Cities pc on pc.Name = c.Name
  WHERE
    .. as the city already linked to the promoter being updated
    pc.id = pr.city_id
  GROUP BY
    c.name)

诀窍是按名称加入城市，这样您就可以轻松获得所有同名城市。我认为你对IN子句做了同样的尝试，但这比它需要的要复杂一些。

我认为你根本不需要group_concat，除了检查inned查询是否确实返回了正确的城市，尽管它没有意义，因为你已经在对名称进行分组。如果这样写，你可以说这不会出错：

  SELECT 
    -- Select the lowest ID ..
    MIN(c.id) AS id,
    GROUP_CONCAT(c.name) AS names --< already grouped by this, so why...
  FROM
    -- .. of all cities ..
    Cities c
    -- .. that have the same name.
    INNER JOIN Cities pc on pc.Name = c.Name
  GROUP BY
    c.name

我希望我能正确理解这个问题。

使用IN与group_concat结果的Mysql查询

1 个答案: