CREATE TABLE `t_student` (
  `id` int NOT NULL AUTO_INCREMENT,
  `t_id` int DEFAULT NULL COMMENT '學(xué)科id',
  `score` int DEFAULT NULL COMMENT '分?jǐn)?shù)',
  PRIMARY KEY (`id`)
);

數(shù)據(jù)如下：

題目一：獲取每個科目下前五成績排名（允許并列）

允許并列情況可能存在如4、5名成績并列情況，會導(dǎo)致取前4名得出5條數(shù)據(jù)，取前5名也是5條數(shù)據(jù)。

SELECT
	s1.* 
FROM
	student s1
	LEFT JOIN student s2 ON s1.t_id = s2.t_id 
	AND s1.score < s2.score 
GROUP BY
	s1.id
HAVING
	COUNT( s2.id ) < 5 
ORDER BY
	s1.t_id,
	s1.score DESC

ps：取前4名時

分析：

1.自身左外連接，得到所有的左邊值小于右邊值的集合。以t_id=1時舉例，24有5個成績大于他的（74、64、54、44、34），是第6名，34只有4個成績大于他的，是第5名......74沒有大于他的，是第一名。

SELECT
	* 
FROM
	student s1
	LEFT JOIN student s2 ON s1.t_id = s2.t_id 
	AND s1.score < s2.score

2. 把總結(jié)的規(guī)律轉(zhuǎn)換成SQL表示出來，就是group by 每個student 的 id(s1.id)，Having統(tǒng)計這個id下面有多少個比他大的值(s2.id)

SELECT
	s1.* 
FROM
	student s1
	LEFT JOIN student s2 ON s1.t_id = s2.t_id 
	AND s1.score < s2.score 
GROUP BY
	s1.id
HAVING
	COUNT( s2.id ) < 5

3. 最后根據(jù) t_id 分類，score 倒序排序即可。

題目二：獲取每個科目下最后兩名學(xué)生的成績平均值

取最后兩名成績

SELECT
	s1.* 
FROM
	student s1
	LEFT JOIN student s2 ON s1.t_id = s2.t_id 
	AND s1.score > s2.score 
GROUP BY
	s1.id 
HAVING
	COUNT( s1.id )< 2 
ORDER BY
	s1.t_id,
	s1.score

并列存在情況下可能導(dǎo)致篩選出的同一t_id 下結(jié)果條數(shù)大于2條，但題目要求是取最后兩名的平均值，多條平均后還是本身，故不必再對其處理，可以滿足題目要求。

分組求平均值：

SELECT
	t_id,AVG(score)
FROM
	(
	SELECT
		s1.*
	FROM
		student s1
		LEFT JOIN student s2 ON s1.t_id = s2.t_id 
		AND s1.score > s2.score
	GROUP BY
		s1.id 
	HAVING
		COUNT( s1.id )< 2 
	ORDER BY
		s1.t_id,
		s1.score 
	) tt 
GROUP BY
	t_id

結(jié)果：

分析：

1. 查詢出所有t1.score>t2.score 的記錄

SELECT
		s1.*,s2.*
	FROM
		student s1
		LEFT JOIN student s2 ON s1.t_id = s2.t_id 
		AND s1.score > s2.score

2. group by s.id 去重，having 計數(shù)取2條

3. group by t_id 分別取各自學(xué)科的然后avg取均值

題目三：獲取每個科目下前五成績排名（不允許并列）

SELECT
	* 
FROM
	(
	SELECT
		s1.*,
		@rownum := @rownum + 1 AS num_tmp,
		@incrnum :=
	CASE
			
			WHEN @rowtotal = s1.score THEN
			@incrnum 
			WHEN @rowtotal := s1.score THEN
			@rownum 
		END AS rownum 
	FROM
		student s1
		LEFT JOIN student s2 ON s1.t_id = s2.t_id 
		AND s1.score > s2.score,
		( SELECT @rownum := 0, @rowtotal := NULL, @incrnum := 0 ) AS it 
	GROUP BY
		s1.id 
	ORDER BY
		s1.t_id,
		s1.score DESC 
	) tt 
GROUP BY
	t_id,
	score,
	rownum 
HAVING
	COUNT( rownum )< 5

分析：

1.引入輔助參數(shù)

SELECT
	s1.*,
	@rownum := @rownum + 1 AS num_tmp,
	@incrnum :=
CASE
		
		WHEN @rowtotal = s1.score THEN
		@incrnum 
		WHEN @rowtotal := s1.score THEN
		@rownum 
	END AS rownum 
FROM
	student s1
	LEFT JOIN student s2 ON s1.t_id = s2.t_id 
	AND s1.score > s2.score,
	( SELECT @rownum := 0, @rowtotal := NULL, @incrnum := 0 ) AS it

2.去除重復(fù)s1.id，分組排序

SELECT
		s1.*,
		@rownum := @rownum + 1 AS num_tmp,
		@incrnum :=
	CASE
			
			WHEN @rowtotal = s1.score THEN
			@incrnum 
			WHEN @rowtotal := s1.score THEN
			@rownum 
		END AS rownum 
	FROM
		student s1
		LEFT JOIN student s2 ON s1.t_id = s2.t_id 
		AND s1.score > s2.score,
		( SELECT @rownum := 0, @rowtotal := NULL, @incrnum := 0 ) AS it 
	GROUP BY
		s1.id 
	ORDER BY
		s1.t_id,
		s1.score DESC