Question

我有一个SQL查询，该查询将几个表联接在一起，从而在两行中产生重复。我使用DISTINCT关键字来消除重复项：

SELECT DISTINCT
          o.day as day,
          g.id AS id,
          g.name AS name,
          o.num AS num,
          o.version as version
        FROM
          table_one o
          INNER JOIN table_two t ON
            o.ID = t.ID
          INNER JOIN table_three g ON
            t.ID = g.ID
          INNER JOIN table_four gs ON
            g.ID = gs.ID
            AND
          INNER JOIN table_five s ON
            gs.ID = s.ID
          INNER JOIN table_six z ON
            s.ID = z.ID
          INNER JOIN table_seven bg ON
            bg.ID = g.ID;

这将返回我想要的两行，否则，如果我不使用DISTINCT，我将看到重复的行：

1/2/19, 5, first, 25, 1
1/5/19, 7, second, 20, 1

如果我删除DISTINCT，则两行重复，给了我四行：

1/2/19, 5, first, 25, 1
1/2/19, 5, first, 25, 1
1/5/19, 7, second, 20, 1
1/5/19, 7, second, 20, 1

因此，我的最终目标是使用GROUP BY函数，以便我可以将o.num字段加起来并按其余字段分组。如果我将GROUP BY函数添加到上面的查询中，如下所示：

SELECT DISTINCT
          o.day as day,
          g.id AS id,
          g.name AS name,
          SUM(o.num) AS num,
          o.version as version
        FROM
          table_one o
          INNER JOIN table_two t ON
            o.ID = t.ID
          INNER JOIN table_three g ON
            t.ID = g.ID
          INNER JOIN table_four gs ON
            g.ID = gs.ID
            AND
          INNER JOIN table_five s ON
            gs.ID = s.ID
          INNER JOIN table_six z ON
            s.ID = z.ID
          INNER JOIN table_seven bg ON
            bg.ID = g.ID
        GROUP BY
          o.day as day,
          g.id AS id,
          g.name AS name,
          o.version as version;

我退回了两行，但是o.num的数量增加了一倍（基本上是在没有GROUP BY的情况下执行DISTINCT：

1/2/19, 5, first, 50, 1
1/5/19, 7, second, 40, 1

注意：您可能想知道为什么我在第一次查询中得到想要的结果时尝试使用GROUP BY。我只包括要复制的行。由于某些原因，所有其他行都没有看到此行为。有什么方法可以使GROUP BY和DISTINCT一起工作？

Answer 1

如果您要删除重复项然后求和，请将该查询插入子查询中。

select day, id, name, sum(num) num, version
from (
  -- your query here with DISTINCT clause 
)
group by day, id, name, version

如果重复，则连接条件可能存在一些问题。对我来说不难判断，也不知道数据集。

Answer 2

您可以使用SUM(DISTINCT o.num)：

SELECT o.day as day,
       g.id AS id,
       g.name AS name,
       SUM(DISTINCT o.num) AS num,
       o.version as version
FROM table_one o
INNER JOIN table_two t ON o.ID = t.ID
INNER JOIN table_three g ON t.ID = g.ID
INNER JOIN table_four gs ON g.ID = gs.ID
INNER JOIN table_five s ON gs.ID = s.ID
INNER JOIN table_six z ON s.ID = z.ID
INNER JOIN table_seven bg ON bg.ID = g.ID
GROUP BY o.day as day,
         g.id AS id,
         g.name AS name,
         o.version as version;

GROUP BY函数取消DISTINCT

2 个答案: