为什么这个 GROUP BY 子句导致这个错误“42000Expression #1 of SELECT list is not in GROUP BY clause and contains nonaggregated column...”？
Why this GROUP BY clause cause this error "42000Expression #1 of SELECT list is not in GROUP BY clause and contains nonaggregated column..."?
我正在处理 MySql 数据库，并且正在执行此查询：
SELECT
LS.id AS livestock_species_id,
LS.parent_livestock_species_id AS parent_livestock_species_id,
LS.livestock_species_name_en AS livestock_species_name_en,
LSN.livestock_species_name AS livestock_species_name,
LSN.description AS description,
LS.image_link AS image_link
FROM LivestockDetails AS LD
INNER JOIN LivestockSpecies AS LS
ON LD.live_stock_species_id = LS.id
LEFT JOIN LivestockSpeciesName AS LSN
ON LSN.livestock_species_id = LS.id AND LSN.language_id = 3
WHERE
LD.ls_vaccination_id is not null
上次查询return结果集如下：
livestock_species_id parent_livestock_species_id livestock_species_name_en livestock_species_name description image_link
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
1 Cow Inka Inka https://firebasestorage.googleapis.com/v0/b/MY-PROJECT.appspot.com/o/img%2Ficons%2Flivestock%2Fcow.png?alt=media&token=c21866df-448a-4a72-9da2-d55d87f8b31c
1 Cow Inka Inka https://firebasestorage.googleapis.com/v0/b/MY-PROJECT.appspot.com/o/img%2Ficons%2Flivestock%2Fcow.png?alt=media&token=c21866df-448a-4a72-9da2-d55d87f8b31c
1 Cow Inka Inka https://firebasestorage.googleapis.com/v0/b/MY-PROJECT.appspot.com/o/img%2Ficons%2Flivestock%2Fcow.png?alt=media&token=c21866df-448a-4a72-9da2-d55d87f8b31c
1 Cow Inka Inka https://firebasestorage.googleapis.com/v0/b/MY-PROJECT.appspot.com/o/img%2Ficons%2Flivestock%2Fcow.png?alt=media&token=c21866df-448a-4a72-9da2-d55d87f8b31c
2 Chicken Inkoko Inkoko https://firebasestorage.googleapis.com/v0/b/MY-PROJECT.appspot.com/o/img%2Ficons%2Flivestock%2Fchicken.png?alt=media&token=badd0660-48ca-49f8-914a-8a9b8574a221
2 Chicken Inkoko Inkoko https://firebasestorage.googleapis.com/v0/b/MY-PROJECT.appspot.com/o/img%2Ficons%2Flivestock%2Fchicken.png?alt=media&token=badd0660-48ca-49f8-914a-8a9b8574a221
1 Cow Inka Inka https://firebasestorage.googleapis.com/v0/b/MY-PROJECT.appspot.com/o/img%2Ficons%2Flivestock%2Fcow.png?alt=media&token=c21866df-448a-4a72-9da2-d55d87f8b31c
1 Cow Inka Inka https://firebasestorage.googleapis.com/v0/b/MY-PROJECT.appspot.com/o/img%2Ficons%2Flivestock%2Fcow.png?alt=media&token=c21866df-448a-4a72-9da2-d55d87f8b31c
2 Chicken Inkoko Inkoko https://firebasestorage.googleapis.com/v0/b/MY-PROJECT.appspot.com/o/img%2Ficons%2Flivestock%2Fchicken.png?alt=media&token=badd0660-48ca-49f8-914a-8a9b8574a221
2 Chicken Inkoko Inkoko https://firebasestorage.googleapis.com/v0/b/MY-PROJECT.appspot.com/o/img%2Ficons%2Flivestock%2Fchicken.png?alt=media&token=badd0660-48ca-49f8-914a-8a9b8574a221
如您所见，returned 记录是重复的，我不需要这些重复。
所以我的想法是使用 GROUP BY 子句。
我的问题是尝试这样做（按 livestock_species_id 检索字段分组）：
SELECT
LS.id AS livestock_species_id,
LS.parent_livestock_species_id AS parent_livestock_species_id,
LS.livestock_species_name_en AS livestock_species_name_en,
LSN.livestock_species_name AS livestock_species_name,
LSN.description AS description,
LS.image_link AS image_link
FROM LivestockDetails AS LD
INNER JOIN LivestockSpecies AS LS
ON LD.live_stock_species_id = LS.id
LEFT JOIN LivestockSpeciesName AS LSN
ON LSN.livestock_species_id = LS.id AND LSN.language_id = 3
WHERE
LD.ls_vaccination_id is not null
GROUP BY
livestock_species_id
MySql return出现以下错误信息：
42000Expression #1 of SELECT list is not in GROUP BY clause and contains nonaggregated column 'digital_services_DB.LS.id' which is not functionally dependent on columns in GROUP BY clause; this is incompatible with sql_mode=only_full_group_by
为什么？怎么了？我错过了什么？我该如何解决这种情况，避免重复？
您需要指定分组中的所有列（或聚合它们）。在你的情况下，因为你有重复的记录，你可以这样做：
GROUP BY livestock_species_id,parent_livestock_species_id,
livestock_species_name_en, livestock_species_name,
description,image_link
或者你可以使用不同的：
SELECT DISTINCT
LS.id AS livestock_species_id,
LS.parent_livestock_species_id AS parent_livestock_species_id,
LS.livestock_species_name_en AS livestock_species_name_en,
LSN.livestock_species_name AS livestock_species_name,
LSN.description AS description,
LS.image_link AS image_link
关于我在回复中的评论：首先防止重复项出现 - 我猜 LivestockDetails 中有多行与同一物种相关。然而，在您的查询中，您没有从 table.
访问任何数据
如果您只想报告在 LivestockDetails 中至少有一行的物种，请改用 EXISTS 检查：
SELECT
LS.id AS livestock_species_id,
LS.parent_livestock_species_id AS parent_livestock_species_id,
LS.livestock_species_name_en AS livestock_species_name_en,
LSN.livestock_species_name AS livestock_species_name,
LSN.description AS description,
LS.image_link AS image_link
FROM LivestockSpecies AS LS
LEFT JOIN LivestockSpeciesName AS LSN
ON LSN.livestock_species_id = LS.id AND LSN.language_id = 3
WHERE
EXISTS (SELECT * from LivestockDetails LD
WHERE LD.live_stock_species_id = LS.id
and LD.ls_vaccination_id is not null)
这应该会产生更好的结果（如果优化器做得很好），因为我们一开始就不会生成重复项。
（如果存在检查的相关子查询工作不正常，您可能还想尝试：
SELECT
LS.id AS livestock_species_id,
LS.parent_livestock_species_id AS parent_livestock_species_id,
LS.livestock_species_name_en AS livestock_species_name_en,
LSN.livestock_species_name AS livestock_species_name,
LSN.description AS description,
LS.image_link AS image_link
FROM (SELECT DISTINCT live_stock_species_id
FROM LivestockDetails
WHERE ls_vaccination_id is not null) AS LD
INNER JOIN LivestockSpecies AS LS
ON LD.live_stock_species_id = LS.id
LEFT JOIN LivestockSpeciesName AS LSN
ON LSN.livestock_species_id = LS.id AND LSN.language_id = 3
这至少会尽早停止复制）
我正在处理 MySql 数据库，并且正在执行此查询：
SELECT
LS.id AS livestock_species_id,
LS.parent_livestock_species_id AS parent_livestock_species_id,
LS.livestock_species_name_en AS livestock_species_name_en,
LSN.livestock_species_name AS livestock_species_name,
LSN.description AS description,
LS.image_link AS image_link
FROM LivestockDetails AS LD
INNER JOIN LivestockSpecies AS LS
ON LD.live_stock_species_id = LS.id
LEFT JOIN LivestockSpeciesName AS LSN
ON LSN.livestock_species_id = LS.id AND LSN.language_id = 3
WHERE
LD.ls_vaccination_id is not null
上次查询return结果集如下：
livestock_species_id parent_livestock_species_id livestock_species_name_en livestock_species_name description image_link
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
1 Cow Inka Inka https://firebasestorage.googleapis.com/v0/b/MY-PROJECT.appspot.com/o/img%2Ficons%2Flivestock%2Fcow.png?alt=media&token=c21866df-448a-4a72-9da2-d55d87f8b31c
1 Cow Inka Inka https://firebasestorage.googleapis.com/v0/b/MY-PROJECT.appspot.com/o/img%2Ficons%2Flivestock%2Fcow.png?alt=media&token=c21866df-448a-4a72-9da2-d55d87f8b31c
1 Cow Inka Inka https://firebasestorage.googleapis.com/v0/b/MY-PROJECT.appspot.com/o/img%2Ficons%2Flivestock%2Fcow.png?alt=media&token=c21866df-448a-4a72-9da2-d55d87f8b31c
1 Cow Inka Inka https://firebasestorage.googleapis.com/v0/b/MY-PROJECT.appspot.com/o/img%2Ficons%2Flivestock%2Fcow.png?alt=media&token=c21866df-448a-4a72-9da2-d55d87f8b31c
2 Chicken Inkoko Inkoko https://firebasestorage.googleapis.com/v0/b/MY-PROJECT.appspot.com/o/img%2Ficons%2Flivestock%2Fchicken.png?alt=media&token=badd0660-48ca-49f8-914a-8a9b8574a221
2 Chicken Inkoko Inkoko https://firebasestorage.googleapis.com/v0/b/MY-PROJECT.appspot.com/o/img%2Ficons%2Flivestock%2Fchicken.png?alt=media&token=badd0660-48ca-49f8-914a-8a9b8574a221
1 Cow Inka Inka https://firebasestorage.googleapis.com/v0/b/MY-PROJECT.appspot.com/o/img%2Ficons%2Flivestock%2Fcow.png?alt=media&token=c21866df-448a-4a72-9da2-d55d87f8b31c
1 Cow Inka Inka https://firebasestorage.googleapis.com/v0/b/MY-PROJECT.appspot.com/o/img%2Ficons%2Flivestock%2Fcow.png?alt=media&token=c21866df-448a-4a72-9da2-d55d87f8b31c
2 Chicken Inkoko Inkoko https://firebasestorage.googleapis.com/v0/b/MY-PROJECT.appspot.com/o/img%2Ficons%2Flivestock%2Fchicken.png?alt=media&token=badd0660-48ca-49f8-914a-8a9b8574a221
2 Chicken Inkoko Inkoko https://firebasestorage.googleapis.com/v0/b/MY-PROJECT.appspot.com/o/img%2Ficons%2Flivestock%2Fchicken.png?alt=media&token=badd0660-48ca-49f8-914a-8a9b8574a221
如您所见，returned 记录是重复的，我不需要这些重复。
所以我的想法是使用 GROUP BY 子句。
我的问题是尝试这样做（按 livestock_species_id 检索字段分组）：
SELECT
LS.id AS livestock_species_id,
LS.parent_livestock_species_id AS parent_livestock_species_id,
LS.livestock_species_name_en AS livestock_species_name_en,
LSN.livestock_species_name AS livestock_species_name,
LSN.description AS description,
LS.image_link AS image_link
FROM LivestockDetails AS LD
INNER JOIN LivestockSpecies AS LS
ON LD.live_stock_species_id = LS.id
LEFT JOIN LivestockSpeciesName AS LSN
ON LSN.livestock_species_id = LS.id AND LSN.language_id = 3
WHERE
LD.ls_vaccination_id is not null
GROUP BY
livestock_species_id
MySql return出现以下错误信息：
42000Expression #1 of SELECT list is not in GROUP BY clause and contains nonaggregated column 'digital_services_DB.LS.id' which is not functionally dependent on columns in GROUP BY clause; this is incompatible with sql_mode=only_full_group_by
为什么？怎么了？我错过了什么？我该如何解决这种情况，避免重复？
您需要指定分组中的所有列（或聚合它们）。在你的情况下，因为你有重复的记录，你可以这样做：
GROUP BY livestock_species_id,parent_livestock_species_id,
livestock_species_name_en, livestock_species_name,
description,image_link
或者你可以使用不同的：
SELECT DISTINCT
LS.id AS livestock_species_id,
LS.parent_livestock_species_id AS parent_livestock_species_id,
LS.livestock_species_name_en AS livestock_species_name_en,
LSN.livestock_species_name AS livestock_species_name,
LSN.description AS description,
LS.image_link AS image_link
关于我在回复中的评论：首先防止重复项出现 - 我猜 LivestockDetails 中有多行与同一物种相关。然而，在您的查询中，您没有从 table.
如果您只想报告在 LivestockDetails 中至少有一行的物种，请改用 EXISTS 检查：
SELECT
LS.id AS livestock_species_id,
LS.parent_livestock_species_id AS parent_livestock_species_id,
LS.livestock_species_name_en AS livestock_species_name_en,
LSN.livestock_species_name AS livestock_species_name,
LSN.description AS description,
LS.image_link AS image_link
FROM LivestockSpecies AS LS
LEFT JOIN LivestockSpeciesName AS LSN
ON LSN.livestock_species_id = LS.id AND LSN.language_id = 3
WHERE
EXISTS (SELECT * from LivestockDetails LD
WHERE LD.live_stock_species_id = LS.id
and LD.ls_vaccination_id is not null)
这应该会产生更好的结果（如果优化器做得很好），因为我们一开始就不会生成重复项。
（如果存在检查的相关子查询工作不正常，您可能还想尝试：
SELECT
LS.id AS livestock_species_id,
LS.parent_livestock_species_id AS parent_livestock_species_id,
LS.livestock_species_name_en AS livestock_species_name_en,
LSN.livestock_species_name AS livestock_species_name,
LSN.description AS description,
LS.image_link AS image_link
FROM (SELECT DISTINCT live_stock_species_id
FROM LivestockDetails
WHERE ls_vaccination_id is not null) AS LD
INNER JOIN LivestockSpecies AS LS
ON LD.live_stock_species_id = LS.id
LEFT JOIN LivestockSpeciesName AS LSN
ON LSN.livestock_species_id = LS.id AND LSN.language_id = 3
这至少会尽早停止复制）