Question

这里有一个问题，我需要计算每个日期30天的unque用户总数。我的意思是今天它是今天30天和昨天之间的独特之和 - 昨天 - 30天等等。表结构是：

date        uid   visits
27-06-2013   11     40
27-06-2013   14     40
26-06-2013   13     45
25-06-2013   11     20
24-06-2013   12     40

它继续这样下去。我需要的是每天的总和。

Answer 1

以下简单的事情应该可以正常工作：

SELECT
    `date`,
    COUNT(DISTINCT `uid`) as `unique_visits`
FROM
    `foo`
WHERE
    `date` >= DATE_SUB(NOW(), INTERVAL 30 DAY)
GROUP BY
    `date`

SQLFiddle DEMO

Answer 2

假设我已正确阅读您的问题并且您的日期字段属于数据类型日期，那么这样的事情可能是： -

SELECT Sub1.EndDate, COUNT(DISTINCT a.uid)
FROM SomeTable a
INNER JOIN
(
    SELECT DISTINCT date AS EndDate, DATE_SUB(date, INTERVAL 30 DAY) AS StartDate
    FROM SomeTable
) Sub1
ON a.date BETWEEN Sub1.StartDate AND Sub1.EndDate

这是使用子选择来获取每个30天的范围，然后将其加入到该表中以获取该日期范围内的所有记录并进行不同的计数以查找所有不同的用户

Answer 3

Consider this example. It provides a rolling sum on values within 3 points of the current value

SELECT * FROM ints;
+---+
| i |
+---+
| 0 |
| 1 |
| 2 |
| 3 |
| 4 |
| 5 |
| 6 |
| 7 |
| 8 |
| 9 |
+---+

SELECT x.i, SUM(y.i) FROM ints x JOIN ints y ON y.i BETWEEN x.i-3 AND x.i GROUP BY x.i;
+---+----------+
| i | SUM(y.i) |
+---+----------+
| 0 |        0 |
| 1 |        1 |
| 2 |        3 |
| 3 |        6 |
| 4 |       10 |
| 5 |       14 |
| 6 |       18 |
| 7 |       22 |
| 8 |       26 |
| 9 |       30 |
+---+----------+

30天的唯一身份用户总数

3 个答案: