删除所有值比第二个最高值

Question

我有一个表（价格），有两个字段，代码（字符）和价格（十进制）。我需要查找所有具有相同代码的记录，价格低于或低于两个最高价格。

E.G。在这种情况下，我希望删除id = 1：

id   code   price
1    1001   10
2    1001   101
3    1001   40
4    1001   201
5    1002   122
6    1002   50

Answer 1

DELETE
FROM myTable
WHERE ID IN (
    SELECT *
    FROM (
        SELECT t2.id
        FROM myTable t2
        WHERE EXISTS (
            SELECT 1
            FROM myTable t3
            WHERE t3.code = t2.code
            AND t3.price > t2.price * 5
            HAVING COUNT(*) > 1
            )
    ) t
)
;

Answer 2

我的方法：

DELETE t
FROM t
JOIN (SELECT code, 
         REVERSE(SUBSTRING_INDEX(REVERSE(SUBSTRING_INDEX(
         GROUP_CONCAT(price ORDER BY price DESC SEPARATOR ';'), ';', 2)), ';',1))
         AS second_price
     FROM t
     GROUP BY code) s
  ON t.code = s.code
 AND t.price * 5 < s.second_price;

<强> Rextester Demo

它基于选择第二价格：

SELECT code, 
  REVERSE(SUBSTRING_INDEX(REVERSE(SUBSTRING_INDEX(
  GROUP_CONCAT(price ORDER BY price DESC SEPARATOR ';'), ';', 2)), ';',1))  
  AS second_price
FROM t
GROUP BY code;

<强> Rextester Demo2

修改

实际上它可能更容易：

DELETE t
FROM t
JOIN (SELECT code, 
  SUBSTRING_INDEX(SUBSTRING_INDEX(
                 GROUP_CONCAT(price ORDER BY price DESC SEPARATOR ';'),
                  ';', 2), ';',-1)  AS second_price
     FROM t
     GROUP BY code) s
  ON t.code = s.code
 AND t.price * 5 < s.second_price;

<强> Rextester Demo

  

良好而快速的解决方案，但老实说，我无法清楚地理解它。

很容易理解：

SELECT code, 
  GROUP_CONCAT(price ORDER BY price DESC SEPARATOR ';'),
  SUBSTRING_INDEX(GROUP_CONCAT(price ORDER BY price DESC SEPARATOR ';'), ';', 2),
  SUBSTRING_INDEX(SUBSTRING_INDEX(
                 GROUP_CONCAT(price ORDER BY price DESC SEPARATOR ';'),
                  ';', 2), ';',-1)
FROM t
GROUP BY code;