Question

我有一个表记录UNIX时间具有Time字段的事务，因此每行都是一个事务。现在我想找出一天中哪一秒有最大交易？

结果应为：

1-Oct-2013 20:02:34 45 ->(in day 1, at 20:02:34, we have 45 transaction for maximum, so on)
2-Oct-2013 12:34:21 99
3-Oct-2013 15:02:33 70

请提前帮助，谢谢。

Answer 1

一种方法

SELECT FROM_UNIXTIME(time) time, MAX(tcount) tcount
  FROM
(
  SELECT time, COUNT(*) tcount
    FROM transactions
   GROUP BY time
) q
 GROUP BY DATE(FROM_UNIXTIME(time))

这是 SQLFiddle 演示

现在，如果您希望能够通过关系查看每天最高交易的时间戳，那么您需要模拟DENSE_RANK()分析函数。一种方法

SELECT FROM_UNIXTIME(a.time) time, a.tcount
  FROM
(
  SELECT time, COUNT(*) tcount
    FROM transactions
   GROUP BY time
) a JOIN 
(
  SELECT DATE(FROM_UNIXTIME(time)) date, MAX(tcount) tcount
    FROM
  (
    SELECT time, COUNT(*) tcount
      FROM transactions
     GROUP BY time
  ) q
   GROUP BY DATE(FROM_UNIXTIME(time))
) b
    ON DATE(FROM_UNIXTIME(a.time)) = b.date
   AND a.tcount = b.tcount;

这是 SQLFiddle 演示

注意：在此示例中，您有两个时间戳，这些时间戳在第一天具有相同的事务数最大值

Answer 2

尝试以下查询

Select time, Max(count) as max_count from (
SELECT Date_format(from_unixtime(unix_timestamp_column), '%Y-%m-%d') as date, 
       from_unixtime(unix_timestamp_column)  as time, 
       count(*) as count 
from your_table  group by time )  a group by Date

Answer 3

首先，获取独特的时间记录

SELECT DISTINCT your_time_column_name FROM table_name;

然后循环以同时检查记录：

SELECT * FROM table_name WHERE your_time_column_name =“time”

然后计算并找到最大值

Answer 4

SELECT MAX(COUNT_NO), TIME_COLUMN FROM (SELECT COUNT(*) COUNT_NO ,TIME_COLUMN FROM     
TABLE_NAME
GROUP BY TIME_COLUMN)

试试这个。

Answer 5

select yourtimecolumn,max(totalTrans) from (
select yourtimecolumn, count(id) as totalTrans
    from yourtable
    group by yourtimecolumn)a
group by day(yourtimecolumn)

选择每天最多sum（）

5 个答案: