Question

我有一张购买表：

id_purchase, date (timestamp), city ($), price, ...
15484, 1516004158, Phoenix, 147.56, ...
14879, 1516097654, Chicago, 47.99, ...
14788, 1515931918, New Yokr, 87.45, ...

我需要为购买次数最多的十个城市创建每周比较。

WITH 
w1 AS
(SELECT * FROM purchases WHERE relative date limitation),
w2 AS
(SELECT * FROM purchases WHERE relative date limitation),
top_cities
(SELECT city FROM w2 
GROUP BY 1
ORDER BY count(*) DESC)

SELECT w2.city, 
w1.count(*),
w2.count(*),
(w2.count(*)/w1.count(*))-1)*100.00 AS Ratio
FROM w1, w2
WHERE w1.city IN (SELECT * FROM top_cities)
GROUP BY 1
ORDER BY 4 ASC

但查询速度很慢，因此我因超时限制而收到错误。

我想得到的结果如下：

city, w1, w2, Ratio
Chichago, 245, 274, 11.84
Phoenix, 147, 197, 34.01
...

谢谢！

Answer 1

这就是我想到的 - 它会将数据输出为行，这些行更容易从应用程序代码中消耗，您可以进一步过滤和处理它。

架构与您问题的所需输出不符，因为您要求排名前10位的城市＆＃34;但是你的代码只输出数据作为两个明确命名的列，所以它非常不清楚你希望它如何工作。

这是用于MS SQL Server 2000或更高版本的T-SQL编写的。如果用方括号替换方括号[ ]并使用EXTRACT( )而不是YEAR和DATEPART替换大多数其他数据库系统应该能够使用此功能。

SELECT
    city,
    YEAR( [date] ) AS [year],
    DATEPART( wk, [date] ) AS [weekNumber]
    COUNT(*) AS [count]
FROM
    purchases
    INNER JOIN
    (
        SELECT
            TOP 10
            city,
            COUNT(*) AS [count]
        FROM
            purchases
        GROUP BY
            city
        ORDER BY
            [count] DESC
    ) AS top10Cities ON purchases.city = top10Cities.city
GROUP BY
    city,
    YEAR( [date] ),
    DATEPART( wk, [date] )

通过从内到外阅读来更好地理解这个查询：从找到前10个城市的最里面的子查询top10Cities开始，然后它使用这些城市名称来限制返回的行，这些行按周汇总所有购买量 - 数。在我看来，将周数转换为实际日期是一个视图级别的问题，而不是应该在SQL内部完成的事情。

SQL比较查询速度

1 个答案: