Question

我一直在寻找这个时间，我找不到任何相关的东西，所以我会问。我有一个这样的数据库：

id |  date     | name  
 1 |2017-01-01 | sarah  
 2 |2017-01-02 | sarah  
 3 |2017-01-03 | john  
 4 |2017-01-04 | john   
 5 |2017-01-05 | john  
 6 |2017-01-06 | sarah  
 7 |2017-01-07 | sarah  
 8 |2017-01-08 | eddy  
 9 |2017-01-09 | eddy

我想制作如下列表：

start date | end date   | name  
2017-01-01 | 2017-01-02 | sarah  
2017-01-03 | 2017-01-05 | john  
2017-01-06 | 2017-01-07 | sarah  
2017-01-08 | 2017-01-09 | eddy

我怎样才能做到这一点？一直在尝试使用：

SELECT MIN(date) as mindate, MAX(date) as maxdate, name
FROM 'table'
GROUP BY name
ORDER BY date DESC

但我得到了这些：

start date | end date   | name   
2017-01-01 | 2017-01-07 | sarah   
2017-01-03 | 2017-01-05 | john    
2017-01-08 | 2017-01-09 | eddy

请注意，在我想要的列表中，SARAH这个名字出现了两次，而且在GROUP BY中，Sarah只出现过一次......

Answer 1

一种选择是使用行号变量来标识每个岛的开始和结束日期。最终结果与您最初的预期结果类似，不同之处在于现在我们将每个岛的开始和结束日期，而不是整个表中的每个名称。

SET @island_number = 0;
SET @name = '';

SELECT MIN(t.date) AS start_date,
       MAX(t.date) AS end_date,
       t.name
FROM
(
    SELECT @island_number:=CASE WHEN @name = name
                                THEN @island_number
                                ELSE @island_number + 1
                           END AS gn,
           @name:=name AS name,
           date
    FROM yourTable
    ORDER BY id
) t
GROUP BY t.gn, t.name

Answer 2

尝试以下查询。

SELECT
MIN(_date) as start_date
,MAX(_date) as end_date
,name_new as name
FROM
(
Select
   _date
   ,if(name = @prev, @seq, @seq := @seq + 1) as seq
   ,if(name = @prev, name, @prev := name) as name_new
From test, (SELECT @seq := 0, @prev := '') AS init
Order by `_date`
  ) as der
  group by name_new, seq
  order by seq

显示具有相同组的SQL数据库出现两次或更多次

2 个答案: