SQL中同一行的列值中第二高的值

Question

样本数据：

id score1 score2 score3 score4
1  10     05      30    50
2  05     15      10    00
3  25     10      05    15

预期结果集：

id col_value
1    30
2    10
3    15

Answer 1

使用CASE表达式来告诉您在GREATEST()调用中省略哪个得分。

SELECT id,
    CASE GREATEST(score1, score2, score3, score4)
        WHEN score1 THEN GREATEST(score2, score3, score4)
        WHEN score2 THEN GREATEST(score1, score3, score4)
        WHEN score3 THEN GREATEST(score1, score2, score4)
        ELSE GREATEST(score1, score2, score3)
    END AS col_value
FROM your_table ;

该解决方案可轻松推广到任意数量的列。

---

和一个没有CASE的变体，同时使用GREATEST()和LEAST() ：

SELECT id,
    LEAST(
        GREATEST(score1, score2, score3),
        GREATEST(score2, score3, score4),
        GREATEST(score3, score4, score1),
        GREATEST(score4, score1, score2)
    ) AS col_value
FROM your_table ;

Answer 2

请考虑以下内容，它们更容易概括：

DROP TABLE IF EXISTS my_table;

CREATE TABLE my_table 
(id INT NOT NULL
,score_no INT NOT NULL
,score INT NOT NULL
,PRIMARY KEY(id,score_no)
);

INSERT INTO my_table VALUES
(1, 1 ,10),
(1 ,2 ,05),
(1 ,3 ,30),
(1 ,4 ,50),
(2 ,1 ,05),
(2 ,2 ,15),
(2 ,3 ,10),
(2 ,4 ,00),
(3 ,1 ,25),
(3 ,2 ,10),
(3 ,3 ,05),
(3 ,4 ,15);

SELECT id
     , score_no
     , score 
  FROM 
     ( SELECT x.*
            , CASE WHEN @prev=id THEN @i:=@i+1 ELSE @i:=1 END rank
            , @prev:=id 
         FROM my_table x
            , (SELECT @prev:=null,@i:=0) vars 
        ORDER 
           BY id
            , score DESC
            , score_no 
     ) a
 WHERE rank = 2;
 +----+----------+-------+
 | id | score_no | score |
 +----+----------+-------+
 |  1 |        3 |    30 |
 |  2 |        3 |    10 |
 |  3 |        4 |    15 |
 +----+----------+-------+

如果绑定分数，此解决方案将选择具有较低“score_no”的分数。

Answer 3

假设没有关系，您可以使用大case表达式：

select t.*,
       (case when score1 > score2 and score1 > score3 and score1 < score 4 then score1
             when score1 > score2 and score1 < score3 and score1 > score 4 then score1
             when score1 < score2 and score1 > score3 and score1 > score 4 then score1
             when score2 > score1 and score2 > score3 and score2 < score 4 then score2
             when score2 > score1 and score2 < score3 and score2 > score 4 then score2
             when score2 < score1 and score2 > score3 and score2 > score 4 then score2
             . . .
        end) as second_score          
from t;

但是，一般而言，这类问题表明数据结构存在问题。 我怀疑你应该有一个每个id和score一行的表（也许是一个score号）。 这通常更容易在SQL中操作。

Answer 4

使用UNPIVOT尝试此查询

    CREATE TABLE #my_table
(id INT NOT NULL, score1 INT NOT NULL, score2 INT NOT NULL, score3 INT NOT NULL, score4 INT NOT NULL) 

INSERT INTO #my_table VALUES(1,  10,     05,      30,    50)
INSERT INTO #my_table VALUES(2,  05,     15,      10,    00)
INSERT INTO #my_table VALUES(3,  25,     10,      05,    15)

;WITH getHighestValue as (
SELECT id, Scores, ScoreText, ROW_NUMBER() OVER(PARTITION BY id ORDER BY Scores DESC) AS Ranks
FROM #my_table
UNPIVOT(
Scores for ScoreText in (score1,score2,score3,score4)
) unpiv
)
SELECT id, Scores as col_value 
FROM getHighestValue
WHERE Ranks = 2

结果：