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我正在使用 Postgres 9.1，我有一个查询返回 2 列，即使没有数据存在，我也希望复制这些列。我的查询的重要部分是

SELECT DATE_TRUNC('week', tx_dttm) AS interval_time, institution_id FROM (...)
WHERE institution_id IN (1, 3, 4, 5, 7, 9) GROUP BY interval_time, institution_id

正如我所期望的那样，这将返回给我所有机构和所有 interval_times。现在我也想要没有数据的行。我可以执行以下操作来一次获取任何一个

SELECT DATE_TRUNC('week', tx_dttm) AS interval_time, institution_id FROM (...)
WHERE institution_id IN (1, 3, 4, 5, 7, 9) GROUP BY interval_time, institution_id
RIGHT JOIN generate_series('2012-02-03', '2013-02-18', interval '1 week')
ON DATE_TRUNC('week', generate_series) = interval_time

或者

SELECT DATE_TRUNC('week', tx_dttm) AS interval_time, institution_id FROM (...)
WHERE institution_id IN (1, 3, 4, 5, 7, 9) GROUP BY interval_time, institution_id
RIGHT JOIN unnest(array[1, 3, 4, 5, 7, 9]) AS all_institution_id
ON all_institution_id = institution_id

但是，我想将它们结合起来，以便获得所有机构的所有 interval_times。似乎最好的办法是创建某种中间表，其中包含为每个 interval_time 重复的每个机构 ID，然后在其上进行 RIGHT JOIN，但我不知道该怎么做。任何帮助将非常感激。

编辑

玩了一会儿之后，我想做的似乎是以下，但我的语法是错误的

SELECT DATE_TRUNC('week', tx_dttm) AS interval_time, institution_id FROM (...)
WHERE institution_id IN (1, 3, 4, 5, 7, 9) GROUP BY interval_time, institution_id
RIGHT JOIN unnest(array[1, 3, 4, 5, 7, 9]) AS all_institution_id,
generate_series('2012-02-03', '2013-02-18', interval '1 week')
ON DATE_TRUNC('week', generate_series) = dose_hsts.interval_time
AND all_institution_id = institution_id
4

3 回答 3

1

我认为您CROSS JOIN在执行此操作之前需要一个LEFT JOIN（在此处将其替换RIGHT JOIN为相同的效果）：

SELECT interval_time::date , institution_id, ...
FROM   generate_series(date_trunc('week', timestamp '2012-02-03')
                     , timestamp '2013-02-18'
                     , interval '1 week')   AS d(interval_time)
CROSS  JOIN unnest(array[1, 3, 4, 5, 7, 9]) AS i(institution_id)
LEFT   JOIN (
   SELECT date_trunc('week', tx_dttm) AS interval_time, institution_id
   FROM   ...
   GROUP  BY 1, 2
   ) x USING (interval_time, institution_id);

关于generate_series()：

于 2013-03-27T18:57:23.420 回答
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我认为您想要的是 UNION 两个查询。尝试这个：

SELECT DATE_TRUNC('week', tx_dttm) AS interval_time, institution_id FROM (...)
WHERE institution_id IN (1, 3, 4, 5, 7, 9) GROUP BY interval_time, institution_id
UNION
SELECT DATE_TRUNC('week', tx_dttm) AS interval_time, institution_id FROM (...)
WHERE institution_id IN (1, 3, 4, 5, 7, 9) GROUP BY interval_time, institution_id
RIGHT JOIN generate_series('2012-02-03', '2013-02-18', interval '1 week')
ON DATE_TRUNC('week', generate_series) = interval_time
于 2013-03-27T16:40:03.587 回答
0

我通过使用以下语法弄清楚了

SELECT DATE_TRUNC('week', tx_dttm) AS interval_time, institution_id FROM (...)
WHERE institution_id IN (1, 3, 4, 5, 7, 9) GROUP BY interval_time, institution_id
RIGHT JOIN (SELECT * FROM unnest(array[1, 3, 4, 5, 7, 9]) AS all_institution_id,
generate_series('2012-02-03', '2013-02-18', interval '1 week'))
ON DATE_TRUNC('week', generate_series) = dose_hsts.interval_time
AND all_institution_id = institution_id
于 2013-03-27T16:40:15.047 回答