Question

数据，如果您无法访问该链接：

create table climate (city varchar(10), status char(1), Curdate date);

insert into climate values ('Chennai', 'S', '2013-08-05');
insert into climate values ('Chennai', 'S', '2013-08-06');
insert into climate values ('Chennai', 'S', '2013-08-07');
insert into climate values ('Chennai', 'S', '2013-08-08');

insert into climate values ('Chennai', 'R', '2013-08-09');
insert into climate values ('Chennai', 'R', '2013-08-10');

insert into climate values ('Chennai', 'S', '2013-08-12');
insert into climate values ('Chennai', 'S', '2013-08-13');

insert into climate values ('Chennai', 'R', '2013-08-14');
insert into climate values ('Chennai', 'R', '2013-08-15');

insert into climate values ('Banglore', 'S', '2013-08-05');
insert into climate values ('Banglore', 'S', '2013-08-06');
insert into climate values ('Banglore', 'R', '2013-08-07');
insert into climate values ('Banglore', 'R', '2013-08-08');

insert into climate values ('Banglore', 'R', '2013-08-09');
insert into climate values ('Banglore', 'S', '2013-08-10');

insert into climate values ('Banglore', 'R', '2013-08-12');
insert into climate values ('Banglore', 'R', '2013-08-13');

insert into climate values ('Banglore', 'R', '2013-08-14');
insert into climate values ('Banglore', 'S', '2013-08-15');

该链接包含大致数据。

从表中我们需要检索城市名称和状态（'R'/'S'）保持相同超过2天的最新最长日期。

即。 R-下雨 S-阳光

当城市连续多雨或晴天超过2天时，我们需要检索城市和最长日期。

例如：来自示例数据，

查询应该检索

City                  Date
Banglore           2013-08-14
Chennai            2013-08-08

提前感谢您的帮助

Answer 1

对于SQL Server 2005/2008：

select city, max(dt) max_dt
from (
    select city
        , dateadd(dd, x, Curdate) dt
        , min(case x when 0 then status end) s0
        , min(case x when 1 then status end) s1
        , min(case x when 2 then status end) s2
    from climate c
    cross join (select 0 x union all select 1 union all select 2)x
    group by city, dateadd(dd, x, Curdate)
) t
where s0 = s1 and s1 = s2
group by city

如果使用SQL Server 2012查询会更加简单。寻找LAG / LEAD功能。

Answer 2

这与Islands和Gaps问题类似，您也可以使用Common Table Expressions来解决它：

;WITH DateIslandByCityStatus_CTE (City, Status, CurDate, Island) AS
(
    SELECT City
         , Status
         , CurDate
         , Island = DATEADD(DAY, -ROW_NUMBER() OVER (PARTITION BY City, Status ORDER BY CurDate), CurDate) 
      FROM Climate
),
DateIslandWithTwoDaysOfWeather (City, Status, MaxDate) AS
(
    SELECT City
         , Status
         , MAX(CurDate)
      FROM DateIslandByCityStatus_CTE
      GROUP BY City, Status, Island
      HAVING COUNT(*) > 2
)
SELECT City
     , Max(MaxDate)
  FROM DateIslandWithTwoDaysOfWeather
 GROUP BY City
 ORDER BY City

另请参阅："The SQL of Gaps and Islands in Sequences - Dwain Camps"

Answer 3

如果使用SQL 2005或更高版本：

select city, status, max(endDate) as Date
from
(
  select city, status, MIN(curdate) as startDate, MAX(curDate) as endDate
  from
  (
    select city, status, curdate,
    DATEADD(dd, - ROW_NUMBER() OVER (PARTITION by city, status ORDER BY city, curdate), curdate)
    from climate
    group by city, status, curdate
  ) consecutiveDates(city, status, curdate, grp)
  group by city, status, grp
  having COUNT(*) > 2
) groupedConsecutiveDates
group by city, status

在复杂的SQL查询中需要帮助

3 个答案: