mysql - 如何获取mysql每5分钟查询数据的总和(列)和最后一个索引值

Question

该图像文件是示例数据

第一张图片是您的查询

第二张图片是我想要的 View

时间值正好是00秒~00秒

我的意思是110000 ~110450 喜欢这个值“04m:50s”

这是考试数据

例如，这是示例数据

每 10 秒插入一次数据。

## time    money amount  date
HHmmss
100000 500 30 2016-07-28
------------------------

100010 800 740 2016-07-28
-------------------------

100450 300 80 2016-07-28
------------------------

100500 200 5 2016-07-28
-----------------------

100510 900 9144 2016-07-28
--------------------------

 ~~~~~~~~~~~~~~~~~~~~~~~~ 

110000 500 5 2016-07-28
-----------------------

110500 233 5 2016-07-28
=======================

我想获取每5分钟的数据总和和 View

此查询是每1分钟的数据

如何更改方便使用

select sum(amount),
       lpad(cast(time - 5  as char),6,'0') AS dateVal,
       substr(lpad(cast(time - 5  as char),6,'0'),1,4),
       date,
       max(time)
from example
where date = '2016-07-28'
group by substr(dateVal, 1, 4)
order by dateVal asc;

我想要得到这样的数据

amountsum   time(range)         money  
888         100000 ~100450      i want last insert money value 44200
95          100500 ~100950      i want last insert money value 44200
22          101000 ~101450      i want last insert money value 44200
6843        101500 ~101950      i want last insert money value 44200  
2213        102000 ~102450      i want last insert money value 44200


sequence time    money  mpare sel    buy   amount  date
1875    082100  710,000 0   710,000 702,000 27  2016-07-29
1874    082710  710,000 0   710,000 702,000 22  2016-07-29
1873    082730  710,000 0   710,000 702,000 1   2016-07-29
1877    090030  710,000 0   711,000 710,000 848 2016-07-29
1876    090100  711,000 1,000   711,000 710,000 2   2016-07-29
1884    090110  711,000 1,000   711,000 710,000 66  2016-07-29
1883    090120  710,000 0   711,000 710,000 5   2016-07-29
1882    090130  711,000 1,000   711,000 710,000 53  2016-07-29
1881    090140  710,000 0   711,000 710,000 11  2016-07-29
1880    090150  710,000 0   711,000 710,000 2   2016-07-29
1879    090200  710,000 0   711,000 710,000 10  2016-07-29
1878    090210  710,000 0   711,000 710,000 10  2016-07-29
1889    090220  712,000 2,000   712,000 711,000 313 2016-07-29
1888    090230  712,000 2,000   712,000 711,000 21  2016-07-29
1887    090240  712,000 2,000   712,000 711,000 4   2016-07-29
1886    090250  712,000 2,000   712,000 711,000 2   2016-07-29
1885    090300  713,000 3,000   713,000 712,000 25  2016-07-29
1894    090310  713,000 3,000   714,000 713,000 13  2016-07-29
1893    090330  714,000 4,000   714,000 713,000 13  2016-07-29
1892    090350  713,000 3,000   714,000 713,000 3   2016-07-29
1891    090400  714,000 4,000   714,000 713,000 1   2016-07-29
1890    090410  714,000 4,000   714,000 713,000 49  2016-07-29
1900    090420  715,000 5,000   715,000 714,000 27  2016-07-29
1899    090430  716,000 6,000   716,000 715,000 46  2016-07-29
1898    090440  716,000 6,000   716,000 715,000 86  2016-07-29
1897    090450  716,000 6,000   716,000 715,000 59  2016-07-29
1896    090500  716,000 6,000   716,000 715,000 3   2016-07-29
1895    090510  716,000 6,000   716,000 715,000 2   2016-07-29
1906    090520  717,000 7,000   717,000 716,000 15  2016-07-29
1905    090530  717,000 7,000   717,000 716,000 31  2016-07-29
1904    090540  717,000 7,000   717,000 716,000 3   2016-07-29
1903    090550  716,000 6,000   717,000 716,000 77  2016-07-29

Answer 1

不确定您到底想要什么结果，但试试这个:

select 
    sum(amount) as amountsum,
    concat(min(`time`), '~', max(`time`)) as `time(range)`,
    `date`
from example
cross join (
    select min(`time`) as minTime from example where `date` = '2016-07-28'
) t
where `date` = '2016-07-28'
group by truncate((time_to_sec(`time`) - time_to_sec(`minTime`)) / 60 / 5, 0)
order by max(`time`) asc;

Demo Here