我使用MySQL。我有以下格式的数据 -
DATE , PAGE , PAGE_LOAD_DURATION , VISIT_TYPE
01-01-15 , A , 6 , AM
01-01-15 , B , 4 , PM
01-01-15 , C , 5 , AM
01-01-15 , D , 4 , AM
02-01-15 , B , 13 , PM
03-01-15 , C , 15 , PM
04-01-15 , A , 17 , PM
05-01-15 , A , 34 , PM
06-01-15 , B , 56 , AM
07-01-15 , C , 12 , AM
08-01-15 , D , 89 , AM
09-01-15 , B , 34 , AM
10-01-15 , R , 12 , PM
11-01-15 , T , 56 , AM
12-01-15 , E , 78 , PM
13-01-15 , B , 90 , AM
14-01-15 , A , 34 , PM
15-01-15 , A , 56 , AM
16-01-15 , B , 23 , AM
17-01-15 , A , 15 , PM
18-01-15 , R , 17 , AM
19-01-15 , Y , 37 , PM
20-01-15 , C , 23 , AM
我尝试为给定日期范围生成每次访问类型的每页平均页面加载持续时间
我希望以下列格式生成输出 -
PAGE, VISIT_TYPE, AVERAGE_PAGE_LOAD_DUR
A, AM, average_page_load_dur
A, PM, average_page_load_dur
B, AM, average_page_load_dur
B, PM, average_page_load_dur
C, AM, average_page_load_dur
C, PM, average_page_load_dur
D, AM, average_page_load_dur
D, PM, average_page_load_dur
so on
要计算SUM而不是平均值,我使用以下查询 -
select
PAGE,
SUM(IF(VISIT_TYPE = 'AM',AVERAGE_PAGE_LOAD_DUR,0)) AS SUM_AM,
SUM(IF(VISIT_TYPE = 'PM',AVERAGE_PAGE_LOAD_DUR,0)) AS SUM_PM
from
PAGE_VISITS
where
DATE >= '01-01-15' and DATE <= '09-01-15'
group by
PAGE
为了计算平均值,我用AVG替换了SUM,但是,看起来这不是正确的方法,因为扫描的每行都没有计算平均值(这就是我的方法)认为它有效。如果不这样,请纠正我)
select
PAGE,
AVG(IF(VISIT_TYPE = 'AM',AVERAGE_PAGE_LOAD_DUR,0)) AS SUM_AM,
AVG(IF(VISIT_TYPE = 'PM',AVERAGE_PAGE_LOAD_DUR,0)) AS SUM_PM
from
PAGE_VISITS
where
DATE >= '01-01-15' and DATE <= '09-01-15'
group by
PAGE
我怎样才能让平均值得工作?
答案 0 :(得分:3)
简单地忽略, 0
。我更喜欢case
,所以我会把它写成:
select PAGE,
AVG(CASE WHEN VISIT_TYPE = 'AM' THEN AVERAGE_PAGE_LOAD_DUR END) AS AVG_AM,
AVG(CASE WHEN VISIT_TYPE = 'PM' THEN AVERAGE_PAGE_LOAD_DUR END) AS AVG_PM,
from PAGE_VISITS
where DATE >= '2015-01-01' and DATE <= '2015-09-01'
group by PAGE;
注意:
AVG()
忽略NULL
个值,但计算中包含0
。SUM()
。最后,如果你对每页两行感到满意,那么这就更简单了:
select PAGE, AVG(AVERAGE_PAGE_LOAD_DUR) AS AVG_page_load
from PAGE_VISITS
where DATE >= '2015-01-01' and DATE <= '2015-09-01'
group by PAGE, VISIT_TYPE