我在内部Join中使用以下Case子句作为MAX值id上的WHERE条件，Case查询仅根据第一个条件给我结果> lasteditor！=“ abc@example.com”不为空然后lasteditor！=“ abc@example.com”。
我想做的是，如果lasteditor！=“ abc@example.com”为空，则使用lasteditor =“ abc@example.com”否则将abc@example.com忽略为lasteditor并获取下一封lasteditor电子邮件< / p>
WHERE
CASE WHEN
lasteditor != "abc@example.com" IS NOT NULL
THEN
lasteditor != "abc@example.com"
WHEN
lasteditor != "abc@example.com" IS NULL
THEN
lasteditor = "abc@example.com"
ELSE NULL END
我想做的是，在LastEditor不是abc@example.com的任何其他行上进行内部联接，但是，如果LastEditor不是abc@example.com，则根本没有结果，然后获取abc@example.com所在的行
INNER JOIN (SELECT MAX(id) AS maxid,dyna_id,LastEditor FROM t2
WHERE
CASE WHEN t2.lasteditor != "abc@example.com" IS NOT NULL
THEN t2.lasteditor != "abc@example.com"
WHEN t2.lasteditor != "abc@example.com" IS NULL
THEN t2.lasteditor = "abc@example.com"
ELSE NULL END
GROUP BY t2.dyna_id) AS history ON main.id = history.dyna_id
摘要
WHERE t2.lasteditor = (CASE
WHEN t2.lasteditor != 'abc@example.com' --- if rows exists --- THEN 'abc@example.com' (if rows exists after skipping rows where lasteditor is abc@example.com then use this case)
WHEN t2.lasteditor != 'abc@example.com' --- if no rows exists --- THEN pick up the rows where 'abc@example.com' is the last editor
END)
下面的代码做到了，我想它做到了，它正在处理一个巨大的表，希望它显示正确的结果
WHERE
CASE
WHEN
lasteditor <> "abc@example.com" IS NOT NULL
THEN
lasteditor <> "abc@example.com"
ELSE
lasteditor <> "test"
END
GROUP BY
hist.dyna_id
下面的完整查询
SELECT
"Total",
COUNT(IF(main.case_status = "OPEN", 1, NULL)) AS "open_cases",
COUNT(
CASE
WHEN
owner.attr_role = "TM"
AND main.tm_perception > "0 Star"
AND main.tm_perception < "3 Star"
THEN
main.tm_perception
WHEN
owner.attr_role = "TL"
AND main.tl_perception > "0 Star"
AND main.tl_perception < "3 Star"
THEN
main.tl_perception
WHEN
owner.attr_role = "SE"
AND main.owner_perception > "0 Star"
AND main.owner_perception < "3 Star"
THEN
main.owner_perception
WHEN
owner.attr_role = "SE2"
AND main.owner_perception > "0 Star"
AND main.owner_perception < "3 Star"
THEN
main.owner_perception
ELSE
NULL
END
) AS "bb", COUNT(
CASE
WHEN
owner.attr_role = "TM"
AND main.tm_perception > "2 Star"
AND main.tm_perception < "5 Star"
THEN
main.tm_perception
WHEN
owner.attr_role = "TL"
AND main.tl_perception > "2 Star"
AND main.tl_perception < "5 Star"
THEN
main.tl_perception
WHEN
owner.attr_role = "SE"
AND main.owner_perception > "2 Star"
AND main.owner_perception < "5 Star"
THEN
main.owner_perception
WHEN
owner.attr_role = "SE2"
AND main.owner_perception > "2 Star"
AND main.owner_perception < "5 Star"
THEN
main.owner_perception
ELSE
NULL
END
) AS "mb", COUNT(
CASE
WHEN
owner.attr_role = "TM"
AND main.tm_perception > "4 Star"
AND main.tm_perception < "6 Star"
THEN
main.tm_perception
WHEN
owner.attr_role = "TL"
AND main.tl_perception > "4 Star"
AND main.tl_perception < "6 Star"
THEN
main.tl_perception
WHEN
owner.attr_role = "SE"
AND main.owner_perception > "4 Star"
AND main.owner_perception < "6 Star"
THEN
main.owner_perception
WHEN
owner.attr_role = "SE2"
AND main.owner_perception > "4 Star"
AND main.owner_perception < "6 Star"
THEN
main.owner_perception
ELSE
NULL
END
) AS "tb", COUNT(
CASE
WHEN
owner.attr_role = "na"
THEN
main.id
ELSE
NULL
END
) AS "na"
FROM
gt AS main
LEFT JOIN
frontend_forms_users AS caseowner
ON main.se_v_9 = caseowner.alias
INNER JOIN
(
SELECT
MAX(id) AS maxid,
dyna_id AS dn
FROM
history_gt hist
WHERE
CASE
WHEN
lasteditor <> "abc@example.com" IS NOT NULL
THEN
lasteditor <> "abc@example.com"
ELSE
lasteditor <> "test"
END
GROUP BY
hist.dyna_id
)
AS history
ON main.id = dn
LEFT JOIN
history_gt AS owner2
ON maxid = owner2.id
LEFT JOIN
frontend_forms_users AS owner
ON owner2.LastEditor = owner.primary_email
UNION ALL
SELECT
caseowner.attr_lob AS lob,
COUNT(IF(main.case_status = "OPEN", 1, NULL)) AS "open_cases",
COUNT(
CASE
WHEN
owner.attr_role = "TM"
AND main.tm_perception > "0 Star"
AND main.tm_perception < "3 Star"
THEN
main.tm_perception
WHEN
owner.attr_role = "TL"
AND main.tl_perception > "0 Star"
AND main.tl_perception < "3 Star"
THEN
main.tl_perception
WHEN
owner.attr_role = "SE"
AND main.owner_perception > "0 Star"
AND main.owner_perception < "3 Star"
THEN
main.owner_perception
WHEN
owner.attr_role = "SE2"
AND main.owner_perception > "0 Star"
AND main.owner_perception < "3 Star"
THEN
main.owner_perception
ELSE
NULL
END
) AS "bb", COUNT(
CASE
WHEN
owner.attr_role = "TM"
AND main.tm_perception > "2 Star"
AND main.tm_perception < "5 Star"
THEN
main.tm_perception
WHEN
owner.attr_role = "TL"
AND main.tl_perception > "2 Star"
AND main.tl_perception < "5 Star"
THEN
main.tl_perception
WHEN
owner.attr_role = "SE"
AND main.owner_perception > "2 Star"
AND main.owner_perception < "5 Star"
THEN
main.owner_perception
WHEN
owner.attr_role = "SE2"
AND main.owner_perception > "2 Star"
AND main.owner_perception < "5 Star"
THEN
main.owner_perception
ELSE
NULL
END
) AS "mb", COUNT(
CASE
WHEN
owner.attr_role = "TM"
AND main.tm_perception > "4 Star"
AND main.tm_perception < "6 Star"
THEN
main.tm_perception
WHEN
owner.attr_role = "TL"
AND main.tl_perception > "4 Star"
AND main.tl_perception < "6 Star"
THEN
main.tl_perception
WHEN
owner.attr_role = "SE"
AND main.owner_perception > "4 Star"
AND main.owner_perception < "6 Star"
THEN
main.owner_perception
WHEN
owner.attr_role = "SE2"
AND main.owner_perception > "4 Star"
AND main.owner_perception < "6 Star"
THEN
main.owner_perception
ELSE
NULL
END
) AS "tb", COUNT(
CASE
WHEN
owner.attr_role = "na"
THEN
main.id
ELSE
NULL
END
) AS "na"
FROM
gt AS main
LEFT JOIN
frontend_forms_users AS caseowner
ON main.se_v_9 = caseowner.alias
INNER JOIN
(
SELECT
MAX(id) AS maxid,
dyna_id AS dn
FROM
history_gt hist
WHERE
CASE
WHEN
lasteditor <> "abc@example.com" IS NOT NULL
THEN
lasteditor <> "abc@example.com"
ELSE
lasteditor <> "test"
END
GROUP BY
hist.dyna_id
)
AS history
ON main.id = dn
LEFT JOIN
history_gt AS owner2
ON maxid = owner2.id
LEFT JOIN
frontend_forms_users AS owner
ON owner2.LastEditor = owner.primary_email
GROUP BY
lob
ORDER BY
2 DESC
答案 0 :(得分：0)
以下情况可能会有所帮助。主表是带有作者的文章，针对任何文章，作者或其他任何人都可以评论。我只想显示不是作者的任何人的最新评论，除非只有作者发表了评论。
编辑
要显示不是作者的任何人的最新评论-除非只有评论，否则请获取最新的评论。
CREATE TABLE article( id INTEGER NOT NULL PRIMARY KEY ,author VARCHAR(100) NOT NULL ,title VARCHAR(100) NOT NULL );
INSERT INTO article(id,author,title) VALUES (1,'abc@example.com','a first article') , (2,'def@example.com','beware of between') , (3,'ghi@example.com','to be or not whatever');
CREATE TABLE comments( id INTEGER NOT NULL PRIMARY KEY ,article_id INTEGER NOT NULL ,author VARCHAR(100) NOT NULL ,body VARCHAR(200) NOT NULL );
INSERT INTO comments(id,article_id,author,body) VALUES (1,1,'abc@example.com','first comment') , (2,1,'def@example.com','second comment') , (3,1,'ghi@example.com','third comment') , (4,1,'abc@example.com','2nd author comment') , (5,3,'ghi@example.com','only comment') ;
相关查询方法（任何MySQL版本）通过order by case when c.author <> t.author then 1 else 2 end, id DESC和limit
select a.* , c.* from ( select t.* , (select id from comments as c where c.article_id = t.id order by case when c.author <> t.author then 1 else 2 end , id DESC limit 1) AS max_comment_id from article as t ) a left join comments as c on a.max_comment_id = c.id ;id | author | title | max_comment_id | id | article_id | author | body -: | :-------------- | :-------------------- | -------------: | ---: | ---------: | :-------------- | :------------ 1 | abc@example.com | a first article | 3 | 3 | 1 | ghi@example.com | third comment 2 | def@example.com | beware of between | null | null | null | null | null 3 | ghi@example.com | to be or not whatever | 5 | 5 | 3 | ghi@example.com | only comment
分组子查询方法（任何MySQL版本）
select a.* , c.* from article a left join ( select c.article_id , max(coalesce((case when c.author <> a.author then c.id end), (case when c.author = a.author then c.id end))) max_comment_id from comments as c inner join article a on c.article_id = a.id group by c.article_id ) d on a.id = d.article_id left join comments c on d.max_comment_id = c.id ;id | author | title | id | article_id | author | body -: | :-------------- | :-------------------- | ---: | ---------: | :-------------- | :----------------- 1 | abc@example.com | a first article | 4 | 1 | abc@example.com | 2nd author comment 2 | def@example.com | beware of between | null | null | null | null 3 | ghi@example.com | to be or not whatever | 5 | 3 | ghi@example.com | only comment
具有row_number（）的派生表方法（从MySQL 8开始）
select a.* , c.* from article a left join ( select c.* , row_number() over(partition by article_id order by case when c.author <> a.author then 1 else 2 end, id DESC ) as rn from comments as c inner join article a on c.article_id = a.id ) c on a.id = c.article_id and c.rn = 1id | author | title | id | article_id | author | body | rn -: | :-------------- | :-------------------- | ---: | ---------: | :-------------- | :------------ | ---: 1 | abc@example.com | a first article | 3 | 1 | ghi@example.com | third comment | 1 2 | def@example.com | beware of between | null | null | null | null | null 3 | ghi@example.com | to be or not whatever | 5 | 3 | ghi@example.com | only comment | 1
公用表表达式和row_number（）方法（从MySQL 8开始）
with max_comments as ( select c.* , row_number() over(partition by article_id order by case when c.author <> a.author then 1 else 2 end, id DESC ) as rn from comments as c inner join article a on c.article_id = a.id ) select a.* , c.* from article a left join max_comments c on a.id = c.article_id and c.rn = 1id | author | title | id | article_id | author | body | rn -: | :-------------- | :-------------------- | ---: | ---------: | :-------------- | :------------ | ---: 1 | abc@example.com | a first article | 3 | 1 | ghi@example.com | third comment | 1 2 | def@example.com | beware of between | null | null | null | null | null 3 | ghi@example.com | to be or not whatever | 5 | 3 | ghi@example.com | only comment | 1
db <>提琴here
答案 1 :(得分：0)
您的问题，示例联接，子查询没有任何意义...话虽如此，让我尝试以这种方式再次提出您的问题。如果我是正确的话，那么很好，如果不能，那么您的问题需要认真加以澄清。
您有此表“ T2”。在其中，您具有任何给定“ dyna_id”的多个记录。它可能具有1条或更多条记录，并且每个“ ID”都是自动递增的。因此，对于给定的“ Dyna_ID”，您需要该密钥的最新记录。一旦知道每个“ Dyna_ID”实例的最新“ ID”，您就想知道最后一个编辑记录的人是谁。对我来说，这意味着所有记录都将始终有一个编辑帖子的人，除非第一个记录是创建记录并且没有“ addEditor”值并且没有提供“ LastEditor”值（但不要认为这是事实）。是这种情况。）
因此，要获得此结果，它是一个分为两部分的查询。首先，获取给定DYNA_ID的最后一个ID。由此，获取与给定的“ ID”匹配的LastEditor的名称。最后，与其他人加入MAIN的答案相同...
INNER JOIN
-- getting the Details for that last edited per Dyna_ID
(select
T2.Dyna_id,
T2.ID,
coalesce( T2.LastEditor, 'abc@example.com' ) as LastEditorPerDynaID
from
T2
-- getting the last ID for any given Dyna_ID
JOIN ( select
tmp1.dyna_id,
max(tmp1.id) as MaxIdPerDynaID
from
t2 tmp1
group by
tmp1.dyna_id ) PQ
on T2.Dyna_ID = PQ.Dyna_ID
AND T2.ID = PQ.MaxIdPerDynaID ) history
ON main.id = history.dyna_id
在abc@example.com上的澄清
您对abc@example.com的要求。这是“ T2”表的一些示例数据。
ID Dyna_ID LastEditor
1 A ME
2 B abc@example.com
3 A HER
4 C ME
5 B HIM
6 A abc@example.com
7 A HIM
8 C HER
9 B abc@example.com
10 D abc@example.com
根据上述数据，我具有唯一的Dyna_ID值A，B和C，D。因此，原始查询中的max（ID）和最终对应的Last Editor将导致
Dyna_ID HighestID LastEditor
A 7 HIM
B 9 abc@example.com
C 8 HER
D 10 abc@example.com
因此，从此示例数据中，如果您仅查找那些最后一个编辑器不是abc@example.com的那些，则您希望“ B” DynaID提取ID = 5“ HIM”的LastEditor。但是对于“ D” DynaID，他们唯一的记录是“ abc@lexample.com”，没有其他记录，因此可以保留该记录。
如果是这种情况，那么我的内部PQ（预查询）将分别基于“ abc”条目的EITHER条件获得最大ID。因为我们总是会至少有一个条目，但是想要有或没有...的相应最大ID。
INNER JOIN
-- getting the Details for that last edited per Dyna_ID
(select
T2.Dyna_id,
T2.ID,
coalesce( T2.LastEditor, 'abc@example.com' ) as LastEditorPerDynaID
from
T2
-- getting the last ID for any given Dyna_ID
JOIN
( select
PQ1.Dyna_ID,
case when PQ1.MaxIdNotAbcEditor > 0
then PQ1.MaxIdNotAbcEditor
else PQ1.MaxIdAsAbcEditor end as FinalJoinID
from
( select
tmp1.dyna_id,
-- get max ID if NOT the 'abc' editor
max( case when NOT tmp1.LastEditor = 'abc@example.com' then tmp1.id else 0 end ) as MaxIdNotAbcEditor,
-- get max ID if IT IS the 'abc' editor
max( case when tmp1.LastEditor = 'abc@example.com' then tmp1.id else 0 end ) as MaxIdAsAbcEditor
from
t2 tmp1
group by
tmp1.dyna_id ) PQ1 ) PQ
on T2.Dyna_ID = PQ.Dyna_ID
AND T2.ID = PQ.FinalJoinID ) history
ON main.id = history.dyna_id
基于SECOND查询，它分为三部分。对于任何给定的Dyna_ID，最里面的“ PQ1”预查询1都是基于EITHER相应部分得到的最后一个ID，因此
Dyna_ID MaxIdNotAbcEditor MaxIdAsAbcEditor
A 7 6
B 5 9
C 8 0
D 0 10
可以看出，DynaID“ C”没有“ abc @”记录，因此它的最大ID为Abc编辑器= 0 ... DynaID“ D”的逆。它只有一个“ abc @”，因此对于NOT Abc编辑器为0，对于AS Abc编辑器为10。这是正确的假设吗？
好的。因此，现在，对于任何一个Dyna ID，我们都可以为每种分类提供一个最大值。因此，当创建仅动态ID和最大ID的外部“ PQ”结果时，将使用情况进行包装。如果“我不是abc编辑器”列中有一个值，请使用THAT作为ID。否则，抓住唯一的其他选择...是abc编辑器的条目。结果为
Dyna_ID FinalJoinID
A 7
B 5
C 8
D 10