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我有一些名为分类的表，其中包含classification_indicator_id.
我需要总结这一点ID并放入 1 天系列。
我需要添加大约 20 列（与另一列classification_indicator_id）。
我修改了上一个问题的一些答案：

select
data.d::date as "data",
sum(c.classification_indicator_id)::integer as "Segment1",
sum(c4.classification_indicator_id)::integer as "Segment2",
sum(c5.classification_indicator_id)::integer as "Segment3"
from 
  generate_series(
    '2013-03-25'::timestamp without time zone,
    '2013-04-01'::timestamp without time zone,
    '1 day'::interval
) data(d)
left join classifications c on (data.d::date = c.created::date and c.classification_indicator_id = 3)
left join classifications c4 on (data.d::date = c4.created::date and c4.classification_indicator_id = 4)
left join classifications c5 on (data.d::date = c5.created::date and c5.classification_indicator_id = 5)
group by "data"
ORDER BY "data"

但仍然无法正常工作。sum因为每一行都很大，并且当我添加额外的列时会增长。在segment12013-03-26 有 4 列的第二个表中应该与第一个表等中的数量相同。

 With 3 column                      With 4 columns
data       | Segment1 | Segment2   data       | Segment1 | Segment2 | Segment3
--------------------------------   -------------------------------------------
2013-03-25 | 12       | 16         2013-03-25 | 12       | 16       | 20
--------------------------------   -------------------------------------------
2013-03-26 | 18       | 24         2013-03-26 | 108      | 144      | 180
4

2 回答 2

2

正如您之前的回答所评论的那样，您遇到了“代理交叉连接”。
我在这个相关答案中更详细地解释了它：
两个 SQL LEFT JOINS 产生不正确的结果

您的查询应该像这样工作：

SELECT d.created AS data
      ,c3.segment1
      ,c4.segment2
      ,c5.segment3
FROM (
   SELECT generate_series('2013-03-25'::date
                         ,'2013-04-01'::date
                         ,interval '1 day')::date AS created
    ) d
LEFT JOIN (
    SELECT created
          ,sum(classification_indicator_id)::integer AS segment1
    FROM   classifications
    WHERE  classification_indicator_id = 3
    GROUP  BY 1
    ) c3 USING (created)
LEFT JOIN (
    SELECT created
          ,sum(classification_indicator_id)::integer AS segment2
    FROM   classifications
    WHERE  classification_indicator_id = 4
    GROUP  BY 1
    ) c4 USING (created)
LEFT JOIN (
    SELECT created
          ,sum(classification_indicator_id)::integer AS segment3
    FROM   classifications
    WHERE  classification_indicator_id = 5
    GROUP  BY 1
    ) c5 USING (created)
ORDER  BY 1;

假设这created是一个date，而不是一个timestamp。

或者，对于更快的查询，因为这已成为一个主题：

SELECT d.created AS data
      ,count(classification_indicator_id = 3 OR NULL)::int * 3 AS segment1
      ,count(classification_indicator_id = 4 OR NULL)::int * 4 AS segment2
      ,count(classification_indicator_id = 5 OR NULL)::int * 5 AS segment3
FROM (
   SELECT generate_series('2013-03-25'::date
                         ,'2013-04-01'::date
                         ,interval '1 day')::date AS created
    ) d
LEFT   JOIN classifications c USING (created)
GROUP  BY 1
ORDER  BY 1;
于 2013-04-04T12:54:59.727 回答
2

无需连接：

select
    data.d::date as "data",
    sum((classification_indicator_id = 3)::integer * classification_indicator_id)::integer as "Segment1",
    sum((classification_indicator_id = 4)::integer * classification_indicator_id)::integer as "Segment2",
    sum((classification_indicator_id = 5)::integer * classification_indicator_id)::integer as "Segment3",
from 
    generate_series(
        '2013-03-25'::timestamp without time zone,
        '2013-04-01'::timestamp without time zone,
        '1 day'::interval
    ) data(d)
    left join
    classifications c on data.d::date = c.created::date
group by "data"
ORDER BY "data"
于 2013-04-04T13:05:00.017 回答