Mysql 选择在同一天多次出现的所有行
我有一个名为“checkins”的MySQL表和4列Mysql 选择在同一天多次出现的所有行,mysql,Mysql,我有一个名为“checkins”的MySQL表和4列 id | userIDFK | checkin_datetime | shopId ------------------------------------------------ 1 | 1 | 2018-01-18 09:44:00 | 3 2 | 2 | 2018-01-18 10:32:00 | 3 3 | 3 | 2018-01-18 11
id | userIDFK | checkin_datetime | shopId
------------------------------------------------
1 | 1 | 2018-01-18 09:44:00 | 3
2 | 2 | 2018-01-18 10:32:00 | 3
3 | 3 | 2018-01-18 11:19:00 | 3
4 | 1 | 2018-01-18 17:57:00 | 3
5 | 1 | 2018-01-18 16:31:00 | 1
6 | 1 | 2018-01-19 08:31:00 | 3
基本上,我希望找到用户在同一天和同一家商店签入多次(>=2)的行。所以,例如,若一个用户在id为1和4的行中签入(同一个用户,同一天,同一个商店),查询应该返回整行的命中率(id,userIDFK,checkin_datetime,shopId)。希望这是有意义的
我已经试过使用
SELECT id, userIDFK, checkin_datetime, shopId
FROM (
SELECT * FROM 'checkins' WHERE COUNT(userIDFK)>=2 AND COUNT(shopId)>=2
)
同一天,我不知道该怎么做,我知道这个查询还远远不够,但这是我能做的最好的查询。你可以尝试按用户ID签入日期和shopID分组
SELECT userIDFK, checkin_datetime, shopId,COUNT(SHOPiD)
FROM checkins
GROUP BY userIDFK, DATE(checkin_datetime), shopId
HAVING COUNT(SHOPID)>1
编辑
可以包含子查询以获取所有行:
select b.id,b.userIDFK, b.checkin_datetime, b.shopId
from checkins b
where (SELECT COUNT(SHOPiD)
FROM checkins a
where a.userIDFK=b.userIDFK and date(a.checkin_datetime)=date(b.checkin_datetime) and a.shopId=b.a.shopId
GROUP BY userIDFK, DATE(checkin_datetime), shopId)>1
您可以尝试按用户ID签入日期和shopID进行分组
SELECT userIDFK, checkin_datetime, shopId,COUNT(SHOPiD)
FROM checkins
GROUP BY userIDFK, DATE(checkin_datetime), shopId
HAVING COUNT(SHOPID)>1
编辑
可以包含子查询以获取所有行:
select b.id,b.userIDFK, b.checkin_datetime, b.shopId
from checkins b
where (SELECT COUNT(SHOPiD)
FROM checkins a
where a.userIDFK=b.userIDFK and date(a.checkin_datetime)=date(b.checkin_datetime) and a.shopId=b.a.shopId
GROUP BY userIDFK, DATE(checkin_datetime), shopId)>1
GROUPBY
可用于获取多个事件
SELECT id, userIDFK, checkin_datetime, shopId
FROM checkins
GROUP BY userIDFK, DATE(checkin_datetime), shopId
HAVING count(id) > 1;
希望有帮助
编辑:
使用内部联接可以实现它。以下是查询:
SELECT c1.* FROM checkins c1 INNER JOIN checkins c2
ON c1.userIDFK = c2.userIDFK
AND date(c1.checkin_datetime) = date(c2.checkin_datetime)
AND c1.shopId = c2.shopId
AND c1.id != c2.id
干杯
GROUPBY
可用于获取多个事件
SELECT id, userIDFK, checkin_datetime, shopId
FROM checkins
GROUP BY userIDFK, DATE(checkin_datetime), shopId
HAVING count(id) > 1;
希望有帮助
编辑:
使用内部联接可以实现它。以下是查询:
SELECT c1.* FROM checkins c1 INNER JOIN checkins c2
ON c1.userIDFK = c2.userIDFK
AND date(c1.checkin_datetime) = date(c2.checkin_datetime)
AND c1.shopId = c2.shopId
AND c1.id != c2.id
干杯 这差不多了,但它只返回1行pr.group,它怎么能返回每个组的所有行呢?添加了查询以返回所有多次出现的行。如果有的话,请告诉我。这差不多了,但它只返回1行pr.group,它如何返回每个组的所有行?添加了查询以返回所有多次出现的行。如果有的话,请告诉我。这差不多了,但它只返回1行pr.group,它怎么能返回每个组的所有行?这差不多了,但它只返回1行pr.group,它怎么能返回每个组的所有行?