在 SQL 查询中计算随后两个月的不同行程

Question

我有一个简单的 table，其中包含不同日期的旅行。

trip_id	start_date	end_date
160320	2017-12-31 20:40:25 UTC	2017-12-31 20:45:25 UTC
160321	2018-01-12 21:01:51 UTC	2018-01-12 22:01:51 UTC

我只想创建一个显示这些字段的 SQL 查询。

年
月
trips_this_month,
trips_previous_month
difference_from_previous_month (count_this_month - count_previous_month)
is_increased（是一个布尔值列，如果我们看到增加则为真，为假除此以外）更新：我可以集中精力编写一个简单的查询来获取它们，但我仍然觉得我可以优化这个查询。任何帮助将不胜感激。

    SELECT
    year,
    month,
    trips_this_month,
    trips_previous_month,
    case when difference_from_previous_month < 0 then false else true end as is_increased
FROM    
(SELECT    
    year,
    month,
    number_of_trips AS trips_this_month,
    LAG(number_of_trips,1,0) over (order by year,month) AS trips_previous_month,    
    number_of_trips - LAG(number_of_trips,1,0) OVER(order by year,month) AS difference_from_previous_month,    
    FROM(
        SELECT EXTRACT(Month FROM start_date) AS month,
               EXTRACT(Year FROM start_date) AS year,
               COUNT(*) as number_of_trips
               FROM a_table
    group by  month ,year
    )
    order by year, month
    limit 100
)

但我还是忍不住想做更多。感谢您进一步帮助完成它。

Answer 1

这是一个 MySQL 答案，是在 OP 更改标签之前发布的。
我不使用 Bigquery，所以我不确定我的答案在用于 Bigquery 之前需要调整多少。我所知道的是我已经测试了 OP 的原始查询以及 MySQL 服务器上接受的答案的查询并且它有效所以我假设我建议的 (MySQL) 答案不需要很多调整它以在 Bigquery 中工作。

试试这个：

WITH RECURSIVE cte AS (
SELECT MIN(start_date) minstdt, MAX(start_date) maxstdt FROM mytable
UNION ALL
SELECT minstdt+INTERVAL 1 MONTH, maxstdt FROM cte 
       WHERE minstdt+INTERVAL 1 MONTH <= maxstdt )
       
SELECT year,
       month,
       number_of_trips,
       number_of_trips-IFNULL(prev_month_number_of_trips,0) AS This_month_vs_prev_month,
       IF(number_of_trips > prev_month_number_of_trips,1,0) AS Is_increased
FROM
(SELECT
     YEAR(cte.minstdt) AS year,
     MONTH(cte.minstdt) AS month,
     SUM(CASE WHEN start_date IS NULL THEN 0 ELSE 1 END) AS number_of_trips,
     LAG(SUM(CASE WHEN start_date IS NULL THEN 0 ELSE 1 END)) 
         OVER (ORDER BY YEAR(cte.minstdt), MONTH(cte.minstdt)) AS prev_month_number_of_trips
    FROM cte 
LEFT JOIN mytable 
 ON YEAR(cte.minstdt)=YEAR(start_date) 
  AND MONTH(cte.minstdt)=MONTH(start_date) 
GROUP BY year, month) V
ORDER BY year, month;

我使用递归通用 table 表达式 (cte) 根据 table 的 start_date 中出现的最小和最大日期生成日期。
我已经用 YEAR() 替换了 EXTRACT() 并且 MONTH() 函数稍微短了一些。
我LEFT JOIN cte 数据table.

Demo fiddle

看看你是否可以使用它。

Answer 2

考虑使用标准化的每月第一天比较当前和之前月份的聚合：

WITH sub AS (
    SELECT 
       DATE_SUB(
           DATE_ADD(LAST_DAY(start_date), INTERVAL 1 DAY), 
           INTERVAL 1 MONTH
       ) AS month_year,
       COUNT(*) AS number_of_trips
    FROM a_table
    GROUP BY month_year
), calc AS (
    SELECT
       EXTRACT(YEAR FROM curr.month_year) AS year,
       EXTRACT(MONTH FROM curr.month_year) AS month,
       COALESCE(curr.number_of_trips, 0) AS trips_this_month,
       COALESCE(prev.number_of_trips, 0) AS trips_previous_month
    FROM sub AS curr
    LEFT JOIN sub AS prev
        ON prev.month_year = DATE_SUB(curr.month_year, INTERVAL 1 MONTH) 
)

SELECT
    year,
    month,
    trips_this_month,
    trips_previous_month,
    trips_this_month - trips_previous_month AS difference_from_previous_month,
    (trips_this_month - trips_previous_month) > 0 AS is_increased
FROM calc

在 SQL 查询中计算随后两个月的不同行程

calculate different trips over two subsequent month in a SQL query

google-bigquery