mysql - SQL 多重选择和分组依据

Question

这是我的数据:

我想查找满足以下条件的每个日期的唯一 SessionId 的数量:当一个sessionid中“/topic”的数量=1时，同一个sessionid中“/detail”的数量也等于1。所以对于我的情况，sessionid=1(412)是唯一满足要求的sessionid。

这是我使用的代码:(表名称是we)

SELECT count( Distinct sessionid) 
from
    ( Select sessionid, count(search like "/topic%") as TN and Count(search like "/detail%") as DN from we GROUP BY date order by date) as my_table 
WHERE TN ==1 and DN=1

很抱歉代码困惑，但它应该打印出日期 412 和日期 413 的 [1,0]。但它不起作用。有什么建议吗？非常感谢!

Answer 1

SELECT DATE, 
       count(Distinct sessionid) 
FROM (Select sessionid, 
             DATE,
             SUM(CASE WHEN search like '/topic%' THEN 1 ELSE 0 END) as TN, 
             SUM(CASE WHEN search like '/detail%' THEN 1 ELSE 0 END) as DN 
    from we 
    GROUP BY sessionid, DATE
    ) as my_table 
WHERE TN = 1 AND DN = 1
GROUP BY DATE