php - 优化小型 MySQL 连接查询

Question

我在 Venues 表中有一个 field 列表，在 Locations 表中有一个城市/州列表。该地点与我的组织独有的区号相关联，称为 SOYID。 SOYID 由一个地理区域组成 - Locations 表中的每一行都有一个城市、州和相应的 SOYID。 有些场馆行有 SOYID，有些没有；对于那些没有的，我需要找到列出的城市和州的 SOYID。我只想选择特定 SOYID 中的那些场馆。

此查询有效，但是加载需要几秒钟；我认为我没有正确编写查询。目前 Venues 大约有 140 行，Locations 有 40,000 行。

    $sql = "SELECT DISTINCT a.VenueID, a.Name, a.PhotoID, a.City, a.StateAbbr
            FROM Venues AS a LEFT JOIN Locations AS c ON a.City = c.city
            WHERE a.SOYID = '" . mysql_real_escape_string($SOYID) . "'
            OR ((c.city = a.City) AND  (c.state = a.StateAbbr) AND (c.SOYID = '" . mysql_real_escape_string($SOYID) . "'))
            ORDER BY a.Name ASC";

Answer 1

任何时候您在 WHERE 子句中引用 LEFT JOINed 表(c.state 和 c.SOYID)中的列时，您都会强制执行该联接表现得像一个 INNER JOIN。相反，将这些测试作为连接条件的一部分:

"SELECT DISTINCT a.VenueID, a.Name, a.PhotoID, a.City, a.StateAbbr
    FROM Venues AS a 
        LEFT JOIN Locations AS c 
            ON a.City = c.city
                AND a.StateAbbr = c.state
                AND c.SOYID = '" . mysql_real_escape_string($SOYID) . "'
    WHERE a.SOYID = '" . mysql_real_escape_string($SOYID) . "'
        OR c.SOYID IS NOT NULL /* LEFT JOIN found a matching row */
    ORDER BY a.Name ASC"

编辑:根据评论，此版本应该允许您删除DISTINCT 要求:

"SELECT a.VenueID, a.Name, a.PhotoID, a.City, a.StateAbbr
    FROM Venues AS a 
    WHERE a.SOYID = '" . mysql_real_escape_string($SOYID) . "'
        OR EXISTS(SELECT NULL
                      FROM Locations AS c 
                      WHERE a.City = c.city
                          AND a.StateAbbr = c.state
                          AND c.SOYID = '" . mysql_real_escape_string($SOYID) . "') 
    ORDER BY a.Name ASC"