Mysql

Question

我需要知道一支球队连续赢了多少天 (value1 > value2)，并知道第一场和最后一场比赛获胜的日期。并显示按获胜日期和日期排序的 10 个范围的列表。我知道这并不容易。这对我来说并不容易。我尝试进行一些查询和子查询，但没有很好的结果。谢谢我有一个table和

            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

我需要这样的结果

            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
            ...

...

Answer 1

您可以使用变量计算获胜次数和行数。连续获胜的差异是恒定的。这为您提供了聚合值以获得最终结果：

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);

Mysql - 计算条件连续的日期并获取范围列表

Mysql - Count date consecutive with condition and get range list

date

rows

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