php - 计算大型数据库的结果数量

Question

我有两个数据库表，一个有 165 个条目。对于其中的每一个，我都要浏览 100 万个条目的表格，看看这 165 个条目中的每一个被提到了多少次。

“odds_provider”有 165 个条目，“bettingoffer”有 100 万个条目。

    $SQL = "SELECT
    odds_provider.id AS id,
    odds_provider.name AS name,
    COUNT(bettingoffer.odds_providerFK) AS betcount
    FROM odds_provider
    INNER JOIN
    bettingoffer
    ON bettingoffer.odds_providerFK = odds_provider.id
    WHERE 
    odds_provider.active = 'yes'
    GROUP BY
    odds_provider.id,
    odds_provider.name
    ORDER BY betcount DESC";

    $result = mysql_query($SQL);
    while ($db_field = mysql_fetch_assoc($result)) {
echo $db_field['id'] , " " , $db_field['name'] , " " , $db_field['betcount'] , "</BR>";
    }           

它的作用是预期的，但它需要永远。有没有更快的方法？

Answer 1

这个应该更快，尤其是在没有很多非事件 ID 的情况下。

SELECT id, name, betcount
FROM (
  SELECT
    odds_providerFK as id,
    COUNT(*) as betcount
  FROM bettingoffer
  WHERE active = 'yes'
  GROUP BY odds_providerFK
  ORDER BY betcount DESC) as counts
USING (id);