我有一个非常简单的 MySQL 表
Name | Time | Count
James | 11:00 | 10
Simon | 11:00 | 5
James | 11:30 | 4
Oliver| 11:30 | 2
James | 12:00 | 1
etc.
我想创建一个 MySQL 查询,对每个人的计数求和并输出最近 1 小时值、最近 3 小时值、最近 6 小时值、最近 12 小时值和最近 24 小时值。
例如如果现在时间是 12:59,上表将输出:
Name | Last-1 | Last-3 | Last-6
James | 1 | 15 | 15
Simon | 0 | 5 | 5
Oliver| 0 | 2 | 2
是否可以在单个查询中完成此操作?到目前为止我已经:
SELECT name, HOUR( NOW( ) ) - HOUR( time ) AS lasthours, SUM( count ) AS c
FROM table
GROUP BY name, lasthours
HAVING lasthours =0
ORDER BY c DESC
这给了我过去一小时的时间,但是我如何获得额外的列?
非常感谢任何帮助。
编辑:
接受的答案需要稍微调整,结果如下:
SELECT name ,
SUM( IF( HOUR( TIMEDIFF( NOW( ) , time ) ) <=0, count , 0 ) ) AS `Last hour` ,
SUM( IF( HOUR( TIMEDIFF( NOW( ) , time ) ) <=3, count , 0 ) ) AS `Last 3 hours` ,
SUM( IF( HOUR( TIMEDIFF( NOW( ) , time ) ) <=6, count , 0 ) ) AS `Last 6 hours` ,
SUM( IF( HOUR( TIMEDIFF( NOW( ) , time ) ) <=12, count , 0 ) ) AS `Last 12 hours` ,
SUM( IF( HOUR( TIMEDIFF( NOW( ) , time ) ) <=24, count , 0 ) ) AS `Last 24 hours` ,
SUM( IF( HOUR( TIMEDIFF( NOW( ) , time ) ) <=48, count , 0 ) ) AS `Last 2 days` ,
SUM( IF( HOUR( TIMEDIFF( NOW( ) , time ) ) <=72, count , 0 ) ) AS `Last 3 days` ,
SUM( IF( HOUR( TIMEDIFF( NOW( ) , time ) ) <=96, count , 0 ) ) AS `Last 4 days` ,
SUM( IF( HOUR( TIMEDIFF( NOW( ) , time ) ) <=120, count , 0 ) ) AS `Last 5 days` ,
SUM( IF( HOUR( TIMEDIFF( NOW( ) , time ) ) <=144, count , 0 ) ) AS `Last 6 days` ,
SUM( IF( HOUR( TIMEDIFF( NOW( ) , time ) ) <=168, count , 0 ) ) AS `Last week`
FROM table
GROUP BY name
ORDER BY `Last hour` DESC
最佳答案
SELECT `name`,
SUM(IF(HOUR(TIMEDIFF(CURTIME(), `Time`)) <= 1, `Count`, 0)) AS `Last-1`,
SUM(IF(HOUR(TIMEDIFF(CURTIME(), `Time`)) <= 3, `Count`, 0)) AS `Last-3`,
SUM(IF(HOUR(TIMEDIFF(CURTIME(), `Time`)) <= 6, `Count`, 0)) AS `Last-6`
FROM `table`
GROUP BY `name`
查看 sqlfiddle .
关于mysql - SQL 查询将结果分组到每小时的存储桶中,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11261060/