2

我已经成功合并了 3 个表（使用 ID），但是当我尝试添加第 4 个表时 - 输出完全混乱（值变得不切实际/错误）所以我认为我与这个事实有关，即第 4 个表有未分组的 ID，所以我需要在加入这个新表之前对它们进行分组。现在查询如下：

SELECT name, SUM(money) AS MONEY
FROM transactions
JOIN results ON transactions.id = results.id
JOIN more ON results.per_id = more.per_id
GROUP BY name
HAVING SUM(money)>500

当我加入新表时：

SELECT name, SUM(money) AS MONEY, SUM(data_from_NT1), SUM(data_from_NT2)
FROM transactions
JOIN results ON transactions.id = results.id
JOIN more ON results.per_id = more.per_id

JOIN newtable ON results.per_id = newtable.per_id

GROUP BY name
HAVING SUM(money)>500

是否可以执行命令 GROUP BY per_id：

(JOIN newtable ON results.per_id = newtable.per_id GROUP BY per_id)

在将这个新表添加到主表之前？上面的线不起作用。

4

2 回答 2

1

是的，这是可能的，但您必须将其编写为 SELECT。这是可能答案的第一个版本：

SELECT name,
       SUM(money)             AS money,
       SUM(nt2.data_from_NT1) AS data_from_NT1,
       SUM(nt2.data_from_NT2) AS data_from_NT2
  FROM transactions AS t
  JOIN results      AS r ON t.id = r.id
  JOIN more         AS m ON r.per_id = m.per_id
  JOIN (SELECT per_id, SUM(data_from_NT1) AS data_from_NT1, SUM(data_from_NT2) AS data_from_NT2
          FROM newtable GROUP BY per_id
       ) AS nt2 ON results.per_id = nt2.per_id
 GROUP BY name
HAVING SUM(money) > 500;

目前尚不清楚您是否需要主选择列表中的“SUM of SUM”；你可能不知道。也不清楚nameormoney列的来源；我通常也会在它们前面加上适当的表别名。有了这些警告，这可能更接近您所追求的：

SELECT name, SUM(money) AS MONEY, nt2.data_from_NT1, nt2.data_from_NT2
  FROM transactions AS t
  JOIN results      AS r ON t.id = r.id
  JOIN more         AS m ON r.per_id = m.per_id
  JOIN (SELECT per_id, SUM(data_from_NT1) AS data_from_NT1, SUM(data_from_NT2) AS data_from_NT2
          FROM newtable GROUP BY per_id
       ) AS nt2 ON results.per_id = nt2.per_id
 GROUP BY name, nt2.data_from_NT1, nt2.data_from_NT2
HAVING SUM(money) > 500;

毫无疑问，还有其他写法。主查询中的分组最好放在并行子查询中，在主查询中留下简单的直接连接。但是我们没有实际为您执行此操作的信息。

于 2012-04-07T19:22:50.610 回答
1

您可以将 移动group by到子查询：

join   (
       select  per_id
       ,       sum(col1) as col1_avg
       ,       avg(col2) as col2_avg
       from    newtable 
       group by
               per_id
       ) as newtable
on     results.per_id = newtable.per_id

如果您对 进行分组per_id，则必须聚合任何其他列。

于 2012-04-07T19:16:49.973 回答