我正在努力解决这个问题，我需要获得最新的连续输赢或失去成员的结果
tbl_results
result_id member_id result match_date
1 1 W 2014-12-28
2 1 W 2014-12-21
3 1 W 2014-12-14
4 1 L 2014-12-17
5 1 W 2014-11-30
6 2 L 2014-12-28
7 2 L 2014-12-21
8 2 L 2014-12-14
9 2 W 2014-12-17
10 2 L 2014-11-30
11 3 W 2014-12-28
12 3 L 2014-12-21
13 3 W 2014-12-14
14 3 W 2014-12-17
15 3 W 2014-11-30
经过几个小时的互联网研究后，我想出了这个
SET @mID = 0;
SET @num = 1;
//this variable will hold what to count
SET @wtc = 'W';
SET @wl = 1;
SET @consecutives = 'FALSE';
SELECT
tbl_results.member_id,
//I add here a row number for each member_id so that i can determine if the iteration has changed members or not
@num := IF(@mID = tbl_results.member_id, @num + 1, 1) AS row_number,
@mID := tbl_results.member_id AS temp_m_id,
//from here onwards, I started using some programming syntax and logic
//I plan to convert it later to the correct MySQL syntax
//new member detected
IF @num = 1 THEN
//determines what to count (the wins or the loses)
@wtc := IF(tbl_results.remarks = 'W', 'W', 'L') AS what_to_count,
//consecutives started
SET @consecutives = 'TRUE';
DISPLAY 1 in win_lose column
ELSE
//if it is still consecutives
iF @consecutives = 'TRUE' THEN
IF @wtc = 'W' THEN
IF tbl_results.remarks = 'W' THEN
DISPLAY 1 in win_lose column
ELSE
//consecutive wins has been broken
@consecutives = 'FALSE';
DISPLAY 0 in win_lose column
END IF
ELSE
IF tbl_results.remarks = 'L' THEN
DISPLAY 1 in win_lose column
ELSE
//consecutive loses has been broken
consecutives = 'FALSE';
DISPLAY 0 in win_lose column
END IF
END IF
ELSE
DISPLAY 0 in win_lose column here up to the next member
END IF
END IF
FROM
tbl_results
我知道上面的查询肯定是错误的，我没有使用过mysql变量的经验，我刚刚读过它，但我想我会告诉你我遇到了什么，万一你有建议
我的目标是使用上面的查询来输出类似的内容
member_id row_number temp_m_id win_lose what_to_count
1 1 1 1 W
1 2 1 1
1 3 1 1
1 4 1 0
1 5 1 0
2 1 2 1 L
2 2 2 1
2 3 2 1
2 4 2 0
2 5 2 0
3 1 3 1 W
3 2 3 0
3 3 3 0
3 4 3 0
3 5 3 0
然后使用COUNT（）函数的最终输出将是
member_id consecutives what_to_count
1 3 W
2 3 L
3 1 W
what_to_count列确定连续是用于胜利还是用于输掉
离。
1 - 连续3次获胜
2 - 连续3次失败
3 - 获胜1次
请注意，tbl_results是一个由子查询产生的表（它包含按成员ID和日期对原始表进行排序的记录） - 我只是不在此示例中添加它以使其更短
你可以指出我所显示的查询中的错误在哪里，即使我已经知道它确实有很多错误，请原谅我。
您也可以使用正确的mysql语法重写我的查询或使用您自己的
如果你有更好的想法或任何其他方法，只要它会输出连续输赢，请告诉我。请。我真的很感激。非常感谢你
答案 0 :(得分：0)
你的问题真的令人困惑。所以请考虑以下内容......
DROP TABLE IF EXISTS results;
CREATE TABLE results
(result_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,member_id INT NOT NULL
,result CHAR(1) NOT NULL
);
INSERT INTO results VALUES
(1 ,1,'W'),
(2 ,1,'W'),
(3 ,1,'W'),
(4 ,1,'L'),
(5 ,1,'W'),
(6 ,2,'L'),
(7 ,2,'L'),
(8 ,2,'L'),
(9 ,2,'W'),
(10,2,'L'),
(11,3,'W'),
(12,3,'L'),
(13,3,'W'),
(14,3,'W'),
(15,3,'W');
SELECT a.member_id
, a.result_id start
, MIN(c.result_id) end
, a.result
, MIN(c.result_id) - a.result_id +1 total
FROM results a
LEFT
JOIN results b
ON b.member_id = a.member_id
AND b.result = a.result
AND b.result_id + 1 = a.result_id
LEFT
JOIN results c
ON c.member_id = a.member_id
AND c.result = a.result
AND c.result_id >= a.result_id
LEFT
JOIN results d
ON d.member_id = a.member_id
AND d.result = a.result
AND d.result_id - 1 = c.result_id
WHERE b.result_id IS NULL
AND c.result_id IS NOT NULL
AND d.result_id IS NULL
GROUP
BY a.member_id
, a.result_id
, a.result;
+-----------+-------+------+--------+-------+
| member_id | start | end | result | total |
+-----------+-------+------+--------+-------+
| 1 | 1 | 3 | W | 3 |
| 1 | 4 | 4 | L | 1 |
| 1 | 5 | 5 | W | 1 |
| 2 | 6 | 8 | L | 3 |
| 2 | 9 | 9 | W | 1 |
| 2 | 10 | 10 | L | 1 |
| 3 | 11 | 11 | W | 1 |
| 3 | 12 | 12 | L | 1 |
| 3 | 13 | 15 | W | 3 |
+-----------+-------+------+--------+-------+
此解决方案假定result_id是连续的，但如果不是，则会有解决方法。随意挑选骨头。
答案 1 :(得分：0)
我最习惯SQLite，但是这里有一个解决方案：
-- Create a copy from tbl_result appending a column GR_BEGIN
-- GR_BEGIN will be 1 when a different result from previous happen
CREATE TABLE t1 AS SELECT *, COALESCE(prev!=result,1) AS gr_begin FROM (
SELECT *, (
SELECT result FROM tbl_results AS x1 WHERE result_id=(
SELECT MAX(result_id) FROM tbl_results x2 WHERE result_id<tbl_results.result_id AND member_id=tbl_results.member_id
)
) prev FROM tbl_results
) AS x3;
-- Create another table, with another column GR
-- GR will hold sequence number for consecutive results (for each member)
CREATE TABLE t2 AS SELECT *, (
SELECT SUM(gr_begin) FROM t1 x1 WHERE result_id<=t1.result_id AND member_id=t1.member_id
) gr FROM t1;
-- Finally, following query will return last result and count of consecutive
-- results for each member:
SELECT member_id, result last_result, COUNT(*) consecutive
FROM t2
WHERE gr = (SELECT MAX(gr) FROM t2 x1 WHERE member_id=t2.member_id)
GROUP BY member_id
在SQL Fiddle中查看！