Mysql - 计算日期连续条件并获得范围列表

我需要知道一个团队连续赢了多少个日期（value1> value2），并知道第一个和最后一个匹配获胜的日期。并以韩元和日期显示10个范围的顺序列表。我知道它不容易。它对我来说并不容易..我尝试做一些查询和子查询，但没有好的结果。由于我有一个表Mysql - 计算日期连续条件并获得范围列表

  CREATE TABLE `games` (
       `id` int(11) unsigned NOT NULL AUTO_INCREMENT, 
       `value1` int(11) NOT NULL, 
       `value2` int(11) NOT NULL, 
       `played` date NOT NULL, 
       PRIMARY KEY (`id`), 
       KEY `played` (`played`) 
      ) ENGINE=InnoDB AUTO_INCREMENT=1 DEFAULT CHARSET=latin1; 

      insert into games (value1, value2, played) 
      values 
      ("2", "3", "1943-05-09"), 
      ("4", "3", "1943-08-15"), 
      ("1", "8", "1943-08-22"), 
      ("0", "4", "1943-08-29"), 
      ("1", "0", "1943-09-12"), 
      ("1", "3", "1943-09-26"), 
      ("6", "1", "1943-10-03"), 
      ("3", "2", "1943-10-10"), 
      ("3", "3", "1944-07-16"), 
      ("1", "1", "1944-08-06"), 
      ("4", "1", "1944-09-24"), 
      ("0", "7", "1944-10-08"), 
      ("0", "1", "1945-05-13"), // 1 
      ("4", "2", "1945-11-04"), // 2 
      ("3", "2", "1946-05-12"), // 3 second 3 consecutives win 
      ("4", "2", "1946-11-17"), 
      ("2", "2", "1946-11-24"), 
      ("1", "5", "1946-12-01"), 
      ("1", "0", "1947-05-18"), 
      ("3", "0", "1947-10-05"), 
      ("2", "3", "1948-11-07"), 
      ("0", "1", "1948-11-14"), 
      ("1", "4", "1948-11-21"), 
      ("3", "1", "1949-06-12"), // 1 
      ("4", "0", "1949-06-19"), // 2 
      ("5", "1", "1949-07-24"), // 3 
      ("3", "1", "1949-08-06") // 4 first consecutives win

我需要的结果是这样

  From   To   games_won 

      1949-06-12 1949-08-06   4 
      1945-11-04 1946-11-17   3 
      1943-10-03 1943-10-10   2 
      ...

...

来源

2015-08-15 Diego

你可以指望赢的数量和使用变量的行数。连续胜利的差异是不变的。这给你一个聚合的价值，以得到最终结果：

select min(played), max(played), count(*) 
from (select g.*, 
      (@rn := @rn + 1) as rn, 
      (@rnw := if(value1 > value2, @rnw + 1, @rnw) as rnw 
     from games g cross join 
      (select @rn := 0, @rnw := 0) params 
     order by played 
    ) g 
where value1 > value2 
group by (rn - rnw);

来源

2015-08-15 03:25:51

Gordon Linoff，作品完美.. !!!你很棒..！！：）谢谢..！！ – Diego

Gordon Linoff，我在第4行添加了a）（@rnw：= if（value1> value2，@rnw + 1，@rnw））作为rww 并且将最后一行的订单和限制添加到组中rn - rnw）以count（*）desc命令，播放ASC限制10; 再次感谢。 – Diego

Mysql - 计算日期连续条件并获得范围列表

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