Question

我试图计算所有物品的连续天数。我已经找到了如何连续计算just for one item天数，样本数据

random.uniform(a, b)

会给我这样的结果：

bob 2014-08-10 00:35:22
sue 2014-08-10 00:35:22
bob 2014-08-11 00:35:22
mike 2014-08-11 00:35:22
bob 2014-08-12 00:35:22
mike 2014-08-12 00:35:22

但我需要为所有这样的用户提供：

date_created         | streak|
2014-08-10 00:35:22  |      3|

我一直试图从this response修改sql，但我无法让它工作。我很感激任何建议。

Answer 1

你可以做的一件事是连续几天自己加入桌子并计算它。注意我在计数中添加一个，因为它不计算第一天

SELECT MIN(e.date_created) as date_created, e.username, COUNT(e.username) + 1 AS streak
FROM example e
LEFT JOIN example ee 
  ON e.username = ee.username 
 AND DATE(e.date_created) = DATE(DATE_ADD(ee.date_created, INTERVAL -1 DAY))
WHERE ee.username IS NOT NULL
GROUP BY e.username;

Sql Fiddle

Answer 2

您可以按照以下步骤使用CTE解决此问题，这是sqlfiddle

with ranks as
(
  select
    date,
    name,
    dateadd(day, -row_number() OVER (PARTITION BY name ORDER BY date), date) as rr
  from orders
),
cnt as
(
  select
    name,
    count(*) as ttl
  from ranks
  group by
    name,
    rr
)

select 
  min(date) as date,
  r.name,
  ttl
FROM ranks r
join cnt c
on r.name = c.name
where ttl > 1
group by
  r.name,
  ttl

计算SQL中所有项目的连续天数

2 个答案: