Question

有一个列出健康训练的MySQL表。

recordId (primary key, ,integer, auto incrementing)
workoutNumber (integer)
date (date, ex- "2014-07-29")

我需要知道用户最近连续几天的工作时间。我们可以在MySQL查询中执行此操作吗？我使用PHP作为应用程序语言。

Answer 1

我可以给你一个纯粹的SQL＆＃34;解决方案，使用临时变量：

此查询将创建连续几天1 s的列，如果日期不连续则创建0。

select a.*
    , coalesce(date_diff(a.date, @prevDate), 0) = 1 as consecutiveDay -- '1' if the days are consecutive,
                                                                      -- '0' otherwise
                                                                      -- (The first row will be 0)
    , @prevDate := a.date as this_date -- You need to store the date of the current record
                                       -- to compare it with the next one
from
    (select @prevDate := null) as init  -- This is where you initialize the temp 
                                        -- variable that will track the previous date
    , yourTable as a
-- WHERE -- (Put any where conditions here)
order by a.date;

现在，您可以将使用上述查询作为第二个查询的行源进行求和：

select sum(consecutiveDays) as consecutiveDays
from 
    ( select a.*
           , coalesce(date_diff(a.date, @prevDate), 0) = 1 as consecutiveDay
           , @prevDate := a.date as this_date
      from (select @prevDate := null) as init
         , yourTable as a
      -- WHERE -- (add where conditions here)
      order by a.date
    ) as b

希望这有帮助

Answer 2

你可以这样做。

将前一天与当天和昨天与前一天相结合。此连接看起来像这样：

SELECT *
FROM table AS t1 INNER JOIN table AS t2 ON t1.date = SUBDATE(t2.date, 1) --or whatever you need to get the previous day

您现在会注意到未配对的天数在边界上，而配对天数则在连续的间隔内。要获得边界，必须使用外连接。

SELECT *
FROM table AS t1 FULL OUTER JOIN table AS t2 ON t1.date = SUBDATE(t2.date, 1)

现在您需要的是选择那些未配对的日期：

SELECT *
FROM table AS t1 FULL OUTER JOIN table AS t2 ON t1.date = SUBDATE(t2.date, 1)
WHERE t1.date IS NULL OR t2.date IS NULL;

当t1.date IS NULL为开始日时。当t2.date IS NULL为结束日时。

只需获取所需内容的MAX()：

SELECT MAX(t2.date)
FROM table AS t1 FULL OUTER JOIN table AS t2 ON t1.date = SUBDATE(t2.date, 1)
WHERE t1.date IS NULL;

免责声明我即时记下了这个答案，-1，IS NULL或MIN() vs MAX()可能是错的，但基本的想法应该清楚。

作为解释，你想要的是：

t1          t2
NULL        16/07/1994  <-- interesting join (start of an interval)
16/07/1994  17/07/1994
17/07/1994  18/07/1994
18/07/1994  NULL        <-- interesting join (end of an interval)

如何检查mysql日期列中的连续天数

2 个答案: