Question

我有一个包含2列，日期和金额的表格。我希望能够找到数量上的x天峰值。因此，如果我有下表，并且我正在寻找说3天的峰值，那意味着当天的数量高于当天之前或之后3天的任何一天。

date         amount
2012-09-04 | 53137.47
2012-09-05 | 53137.2
2012-09-06 | 53137.54
2012-09-07 | 53138.58
2012-09-10 | 53138.73
2012-09-11 | 53138.28
2012-09-12 | 53138.22
2012-09-13 | 53138.48
2012-09-14 | 53140.14
2012-09-17 | 53139.82
2012-09-18 | 53139.86
2012-09-19 | 53140.01
2012-09-20 | 53139.75
2012-09-21 | 53139.82
2012-09-24 | 53139.01
2012-09-25 | 53138.93
2012-09-26 | 53138.48
2012-09-27 | 53138.83
2012-09-28 | 53138.62

应选择2012-09-10,2012-09-14

我尝试过使用连接和派生表，但我似乎无法让它工作。

尝试了以下内容：

    select * from a WHERE  
amount=(select MAX(amount) from a where 
date<date_add(date,interval 3 day) and 
date>date_sub(date,interval 3 day));

和

    SELECT a1.*
    FROM a AS a1
    JOIN a AS a2 ON 
(select * from a1 where 
a2.DATE < a1.DATE + 3 and 
a2.DATE > a1.DATE - 3) myalias
    WHERE a1.amount > a2.amount;

Answer 1

为了在日期间隔中找到最大值，您必须将中心日期的记录加入间隔中其他日期的记录，然后按中心日期分组。

尝试此查询：

SELECT a1.date, a1.amount, max(a2.amount) AS highest
FROM tbl a1 LEFT JOIN tbl a2
ON a2.date BETWEEN date_sub(a1.date, interval 3 DAY)
           AND date_add(a1.date, interval 3 DAY)
GROUP BY a1.date, a1.amount
HAVING a1.amount = highest
ORDER BY a1.date;

要根据连续记录而不是日历天来获得答案，您必须使用行号（通过将表自身连接到自身）来扩展表，然后将此扩展表自行连接到在比较金额时，我们自己选择每个日期的+/- 3行内的行：

SELECT t3.date FROM
  (SELECT t1.date AS date, t1.amount AS amount, count(t2.date) AS rownum
   FROM tbl t1
   INNER JOIN tbl t2
   ON t2.date <= t1.date
   GROUP BY t1.date, t1.amount) AS t3
LEFT JOIN
  (SELECT t1.date AS date, t1.amount AS amount, count(t2.date) AS rownum
   FROM tbl t1
   INNER JOIN tbl t2
   ON t2.date <= t1.date
   GROUP BY t1.date, t1.amount) AS t4
ON t3.rownum <> t4.rownum
AND t3.rownum - t4.rownum  BETWEEN -3 AND 3
AND t4.amount >= t3.amount
WHERE t4.date IS NULL
ORDER BY t3.date

这适用于MySQL和PostgreSQL，也适用于其他数据库引擎。在MySQL之外的其他数据库中，可以通过对两个子查询使用CTE来简化它。

请使用此SQLfiddle进行测试。

如何从包含日期和金额列的表中选择峰值金额日期

1 个答案: