Question

以下SQL查询：

SELECT DISTINCT b.browserSessionId, v.visitDay, b.sessionTime 
FROM visit AS v 
INNER JOIN bsession AS b 
ON v.browserSessionId=b.browserSessionId

给了我这个结果：

browserSessionId    visitDay    sessionTime     
100                 20150914    150
101                 20150914    100
102                 20150914    50
103                 20150915    75
104                 20150915    150
114                 20150915    225

我想要的结果是按visitDay分组的平均sessionTime：

visitDay    sessionTime     
20150914    50
20150915    75

此查询给出了错误的结果：

SELECT DISTINCT b.browserSessionId, v.visitDay, AVG(b.sessionTime)
FROM visit AS v 
INNER JOIN bsession AS b 
ON v.browserSessionId=b.browserSessionId
GROUp BY v.visitDay

browserSessionId    visitDay    AVG(b.sessionTime)  
100                 20150914    100.0000
103                 20150915    142.5000

而不是每个bowserSessionId的不同值，它计算vist表中browserSession每次出现的平均sessionTime：

SELECT b.browserSessionId, v.visitDay, b.sessionTime 
FROM visit AS v 
INNER JOIN bsession AS b 
ON v.browserSessionId=b.browserSessionId

browserSessionId    visitDay    sessionTime     
100                 20150914    150
100                 20150914    150
100                 20150914    150
101                 20150914    100
101                 20150914    100
101                 20150914    100
101                 20150914    100
102                 20150914    50
102                 20150914    50
102                 20150914    50
103                 20150915    75
103                 20150915    75
103                 20150915    75
103                 20150915    75
104                 20150915    150
104                 20150915    150
104                 20150915    150
114                 20150915    225
114                 20150915    225
114                 20150915    225

如何使SQL仅计算sessionTime的不同值？

Answer 1

使用子查询：

SELECT t.browserSessionId, t.visitDay, AVG(t.sessionTime)
FROM
(
   SELECT DISTINCT b.browserSessionId, v.visitDay, b.sessionTime 
   FROM visit AS v 
   INNER JOIN bsession AS b 
      ON v.browserSessionId=b.browserSessionId 
) AS t
GROUP BY t.browserSessionId, t.visitDay;

或取决于聚合级别

SELECT t.visitDay, AVG(t.sessionTime)
FROM
(
   SELECT DISTINCT b.browserSessionId, v.visitDay, b.sessionTime 
   FROM visit AS v 
   INNER JOIN bsession AS b 
      ON v.browserSessionId=b.browserSessionId 
) AS t
GROUP BY t.visitDay;

使用双GROUP BY计算AVG

1 个答案: