MariaDB 按键选择值的计数

Question

我有这张表，我想通过 2 个键获取常用值的计数... 例如，如果我输入键 1981 和 1975，我得到的计数是 3（它们都有 2255、2228、2232）等等。我只知道如何用 PHP 解决它，但它比用 SQL 慢得多。谢谢

全表转储（mysql 命令）：DUMP

PHP-ish 解决方案将是（从头开始，没有优化）： （对于伪代码式的文本，我们深表歉意）

$1 = Select * From tbl Where Key = key1;
$2 = Select * From tbl Where Key = key2;

$counter = 0;
foreach($1 as $item)
{
    foreach($2 as $item2)
    {
        if($item->value == $item2->value)
        {
            $counter++;
        }
    }
}

return $counter;

Answer 1

您需要 2 个级别的聚合。

首先，使用这个查询：

SELECT value
FROM stats
WHERE `key` IN (1981, 1975)
GROUP BY value
HAVING COUNT(*) = 2 // the number of values that you search for

你得到满足keys都属于它们的条件的values，然后将其用作子查询来计算values：

SELECT COUNT(*) counter
FROM (
  SELECT value
  FROM stats
  WHERE `key` IN (1981, 1975)
  GROUP BY value
  HAVING COUNT(*) = 2
) t 

请参阅demo。

Answer 2

你可以试试：

CREATE TABLE `keyvalue` (
  `v` int DEFAULT NULL,
  `k` int DEFAULT NULL
);

insert into keyvalue values 
   (1,1982),(2,1982),(3,1982),
   (1,1975),(2,1975),(3,1975),(4,1975),
   (3,1983);

select k1, k2
from (
   select 
      k1.k as k1,
      k1.v as v1, 
      k2.k as k2, 
      k2.v as v2, 
      row_number() over(partition by k1.k,k2.k) as R 
   from keyvalue k1 
   inner join keyvalue k2 on k1.k!=k2.k and k1.v=k2.v 
   ) x
where k2>k1
group by k1,k2
having max(R)=3
;

见DBFIDDLE