postgresql 如何将group by语句限制为1列[重复]

y53ybaqx  于 2023-08-04  发布在  PostgreSQL
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PostgreSQL: Select only the first record per id based on sort order(1个答案)
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(抱歉,如果这听起来很熟悉,我删除并重新上传了我的问题,因为我标记了错误的SQL版本)
我有一个包含约会数据的表,当关联数据更新时,每个约会可以有多行数据。我想选择每个约会的最后一条记录,以了解每个约会的最新快照。
在附加的代码中,我被迫按close_pallets和close_units分组,这影响了我看到的内容(即它返回多行每个约会)。我只想按appointment_id分组,以便每个约会只得到一行。我该怎么做?

SELECT
MAX(appointment_record_version_number),
appointment_id,
appointment_pallets AS close_pallets,
appointment_units AS close_units
FROM
b.dh
WHERE
last_updated_datetime BETWEEN '2023-06-01' AND '2023-06-30'
AND warehouse_id = 'xxx'
GROUP BY
appointment_id,
close_pallets,
close_units

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lymnna71

lymnna711#

您可以使用brilliant distinct on with custom ordering而不是group by来实现您的结果。

SELECT DISTINCT ON (appointment_id)
  appointment_record_version_number,
  appointment_id,
  appointment_pallets AS close_pallets,
  appointment_units AS close_units
FROM b.dh
WHERE last_updated_datetime BETWEEN '2023-06-01' AND '2023-06-30'
  AND warehouse_id = 'xxx'
ORDER BY appointment_id,
         appointment_record_version_number desc; -- pick the biggest one

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gdrx4gfi

gdrx4gfi2#

这将给予group by提供所需的输出。

SELECT MAX(appointment_record_version_number),
appointment_id,
MAX(appointment_pallets) AS close_pallets,
MAX(appointment_units) AS close_units FROM b.dh WHERE
last_updated_datetime BETWEEN '2023-06-01' AND '2023-06-30'
AND warehouse_id = 'xxx'
GROUP BY
appointment_id

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