Sql 编写此脚本以使其不返回聚合错误的更好方法

由于存在t3.Date，此脚本给我一个聚合错误。 请问还有其他的书写方式吗

SELECT t1.ID, t1.Date, t2.Type, t1.Username
FROM t1
INNER JOIN t2
ON t1.CareID = t2.CareID AND 
t1.Date = t2.Date AND
t1.Username = t2.Username
LEFT JOIN t3
ON t1.CareID = t3.CareID AND t1.Username = t3.Username
GROUP BY t1.ID, t1.Date, t2.Type, t1.Username
HAVING COUNT(*) = 1 AND t1.Date <= t3.Date

选择t1.ID、t1.Date、t2.Type、t1.Username
从t1开始
内连接t2
在t1.CareID=t2.CareID和
t1.日期=t2.日期和
t1.Username=t2.Username
左连接t3
在t1.CareID=t3.CareID和t1.Username=t3.Username上
按t1.ID、t1.Date、t2.Type、t1.Username分组
让COUNT（*）=1和t1.Date试试下面的脚本

SELECT t1.ID, t1.Date, t2.Type, t1.Username
FROM t1
INNER JOIN t2
ON t1.CareID = t2.CareID AND 
t1.Date = t2.Date AND
t1.Username = t2.Username
LEFT JOIN t3
ON t1.CareID = t3.CareID AND t1.Username = t3.Username AND t1.Date <= t3.Date
GROUP BY t1.ID, t1.Date, t2.Type, t1.Username
HAVING COUNT(t1.ID) = 1  

选择t1.ID、t1.Date、t2.Type、t1.Username
从t1开始
内连接t2
在t1.CareID=t2.CareID和
t1.日期=t2.日期和
t1.Username=t2.Username
左连接t3
在t1.CareID=t3.CareID和t1.Username=t3.Username和t1.Date上，尝试使用以下脚本

SELECT t1.ID, t1.Date, t2.Type, t1.Username
FROM t1
INNER JOIN t2
ON t1.CareID = t2.CareID AND 
t1.Date = t2.Date AND
t1.Username = t2.Username
LEFT JOIN t3
ON t1.CareID = t3.CareID AND t1.Username = t3.Username AND t1.Date <= t3.Date
GROUP BY t1.ID, t1.Date, t2.Type, t1.Username
HAVING COUNT(t1.ID) = 1  

选择t1.ID、t1.Date、t2.Type、t1.Username
从t1开始
内连接t2
在t1.CareID=t2.CareID和
t1.日期=t2.日期和
t1.Username=t2.Username
左连接t3
在t1.CareID=t3.CareID和t1.Username=t3.Username和t1.Date上删除：

 AND t1.Date <= t3.Date

删除：

 AND t1.Date <= t3.Date

您的语句缺少别名，因此请编写类似于
Count（t1.Id）
的语句，并删除
和t1.Date您的语句缺少别名，因此编写类似于
Count（t1.Id）
的语句，并删除
和t1.Date Having Count（*）的语句。请指定表别名，并且您尚未将group by应用于t3，所以不能用在拥有close@SandipPatel很抱歉，你能不能扩展一下，或者粘贴一个例子？谢谢，这取决于“这”是什么。可能是t1.date