Question

我正在寻找从PostgreSQL中的GROUP BY中选择第一项的方法，直到找到此stackoverflow：Select first row in each GROUP BY group?

在那里，我看到使用了WITH命令。我试图了解更多＆＃34; advanced＆＃34; SQL的命令，如PARTITION，WITH，ROW_NUMBER等。直到两三个月前，我才知道基本命令（SELECT，INNER JOIN， LEFT JOIN，ORDER BY，GROUP BY等）;

我有一点问题（已解决，但我不知道这是否是更好的方法*）。

*更好的方式=我更关心干净的SQL代码而不是性能 - 这仅适用于每天执行一次且不超过5000条记录的报告。

我在PostgreSQL中有两个表：

+----------------------------------------------+
| TABLE NAME: point                            |
+--------+---------------+----------+----------+
|     km |      globalid |      lat |     long |
+--------+---------------+----------+----------+
|  36600 | 1553E2AB-B2F8 | -1774.44 | -5423.58 |
| 364000 | 25EB2465-1B8A | -1773.42 | -5422.03 |
| 362000 | 5FFDE611-88DF | -1771.80 | -5420.37 |
+--------+---------------+----------+----------+


+---------------------------------------------------------+
| TABLE NAME: photo                                       |
+--------------+---------------+------------+-------------+
| attachmentid |  rel_globalid |       date |    filename |
+--------------+---------------+------------+-------------+
|            1 | 1553E2AB-B2F8 | 2015-02-24 | photo01.jpg |
|            2 | 1553E2AB-B2F8 | 2015-02-24 | photo02.jpg |
|          405 | 25EB2465-1B8A | 2015-02-12 | photo03.jpg |
|          406 | 25EB2465-1B8A | 2015-02-12 | photo04.jpg |
|          407 | 25EB2465-1B8A | 2015-02-13 | photo06.jpg |
|            3 | 5FFDE611-88DF | 2015-02-12 | photo07.jpg |
+--------------+---------------+------------+-------------+

所以，对于这个问题：

每个point都有一张或多张照片，但我只需要point数据，第一和最后< / strong> photo。如果point只有一个photo，我只需要第一个photo。如果point有三个photos，我只需要第一个和第三个photo。

所以，我是如何解决的：

首先，我需要每个photo的第一个point，因此，我按rel_globalid分组，并按照组对每张照片进行编号：

WITH photos_numbered AS ( SELECT rel_globalid, date, filename, ROW_NUMBER() OVER ( PARTITION BY rel_globalid ORDER BY date ) AS photo_num FROM photo )

使用此代码，我也可以获得第2个，第3个等等。

好的，现在，我想拍第一张照片（仍然使用上面的WITH）：

SELECT * FROM photos_numbered WHERE photo_num = 1

为了获得最后一张照片，我使用了以下SQL：

SELECT p1.* FROM photos_numbered p1 JOIN ( SELECT rel_globalid, max(photo_num) photo_num FROM photos_numbered GROUP BY rel_globalid ) p2 ON p1.rel_globalid = p2.rel_globalid AND p1.photo_num = p2.photo_num WHERE p1.photo_num > 1

WHERE p1.photo_num > 1是因为如果point只有一个photo，则此photo将显示为第一张照片，最后一张照片将显示为NULL。< / p>
好的，现在我必须＆＃34;转换＆＃34;第一个SELECT的{{1}}和photo的最后一个photo，并为第一个WITH做一个简单SELECT {最后INNER JOIN代表{1}}和photo：

LEFT JOIN

我认为这个SQL对于一个简单的事情来说是巨大的！

有效吗？是的，但是我想要一些建议，一些我可以更好地阅读和理解的文档，一些可以用来制作一个更好的＆＃34; SQL（就像我说的那样，大约两三个月前我甚至不知道photo和WITH photos_numbered AS ( SELECT rel_globalid, date, filename, ROW_NUMBER() OVER ( PARTITION BY rel_globalid ORDER BY date ) AS photo_num FROM photo ), first_photo AS ( SELECT * FROM photos_numbered WHERE photo_num = 1 ), last_photo AS ( SELECT p1.* FROM photos_numbered p1 JOIN ( SELECT rel_globalid, max(photo_num) photo_num FROM photos_numbered GROUP BY rel_globalid ) p2 ON p1.rel_globalid = p2.rel_globalid AND p1.photo_num = p2.photo_num WHERE p1.photo_num > 1 ) SELECT DISTINCT point.km, point.globalid, point.lat, point."long", first_photo.date AS fp_date, first_photo.filename AS fp_filename, last_photo.date AS lp_date, last_photo.filename AS lp_filename FROM point INNER JOIN first_photo ON first_photo.rel_globalid = point.globalid LEFT JOIN last_photo ON last_photo.rel_globalid = point.globalid ORDER BY km命令。

我试图在这里为SQLFiddle添加一个链接，但SQLFiddle从来没有为我工作（总是返回＆＃39; oops＆＃39;消息）。

Answer 1

如果您正在寻找干净的SQL，那么尝试使用lateral_ left和first_value以及last_value窗口函数，而不是公共表表达式（WITH子句）：

select *
from point po
left join lateral 
(
   select first_value( date )     over( order by ph.date) as first_photo_date,
          first_value( filename ) over( order by ph.date) as first_photo_filename,
          last_value( date )      over( order by ph.date) as last_photo_date,
          last_value( filename )  over( order by ph.date) as last_photo_filename    
   from photo ph
   where po.globalid = ph.rel_globalid 
   limit 1
) q on true
;

当只有一条记录时，带有案例表达式的附加count(*) over()可用于“清理”上一张照片的值：

select *
from point po
left join lateral 
(
   select first_value( date )     over( order by ph.date) as first_photo_date,
          first_value( filename ) over( order by ph.date) as first_photo_filename,
          case when count(*) over () > 1 
               then last_value( date )    over( order by ph.date)
          end as last_photo_date,
          case when count(*) over () > 1 
                then last_value( filename )  over( order by ph.date) 
          end as last_photo_filename    
   from photo ph
   where po.globalid = ph.rel_globalid 
   limit 1
) q on true
;

Answer 2

使用krokodilko的答案，我创建了一个没有LEFT JOIN LATERAL的新SQL查询，因为我使用的是PostgreSQL 9.2（没有LEFT JOIN LATERAL）。

SELECT DISTINCT
  po.km,
  po.globalid,
  po.lat,
  po."long",
  ph.fp_date,
  ph.fp_filename,
  ph.lp_date,
  ph.lp_filename
FROM
  point po
INNER JOIN
  (
    SELECT DISTINCT
      rel_globalid,
      first_value(date) OVER (PARTITION BY ph.rel_globalid) AS fp_date,
      first_value(filename) OVER (PARTITION BY ph.rel_globalid) AS fp_filename,
      CASE WHEN count(*) OVER (PARTITION BY ph.rel_globalid) > 1 THEN 
        last_value(date) OVER (PARTITION BY ph.rel_globalid)
      END AS lp_date,
      CASE WHEN count(*) OVER (PARTITION BY ph.rel_globalid) > 1 THEN 
        last_value(filename) OVER (PARTITION BY ph.rel_globalid)
      END AS lp_filename
    FROM
      photo ph
    ORDER BY
      rel_globalid
  ) ph
  ON ph.rel_globalid = po.globalid

OVER (PARTITION)

中几乎每个field中只有我不喜欢的INNER JOIN

需要一些关于SQL的提示/建议（WITH命令，可能还有一些优化）

2 个答案: