Question

我想在标题＆amp;的内容表中搜索请求的字词（$ q）。关键字，也适用于模型，它们位于另一个表中，并通过其间的表链接。另外，我需要在另一个表中获取视图的数量。

这是我到目前为止一直在进行的查询，结果很好但是速度太慢（我在PhpMyAdmin中运行它时平均为0.6s ......我们每月有数百万访问者）

SELECT DISTINCT SQL_CALC_FOUND_ROWS
    c.*,
    cv.views,
    (MATCH (c.title) AGAINST ('{$q}') * 3) Relevance1,
    MATCH (c.keywords) AGAINST ('{$q}') Relevance2,
    (MATCH (a.`name`) AGAINST ('{$q}') * 2) Relevance3
FROM
    content AS c
LEFT JOIN
    content_actors AS ca ON ca.content = c.record_num
LEFT JOIN
    actors AS a ON a.record_num = cm.actor
LEFT JOIN
    content_views AS cv ON cv.content = c.record_num
WHERE
    c.enabled = 1
GROUP BY c.title, c.length
HAVING (Relevance1 + Relevance2 + Relevance3) > 0
ORDER BY (Relevance1 + Relevance2 + Relevance3) DESC

表架构如下所示：

content
record_num     title     keywords
1              Video1    Comedy, Action, Supercool
2              Video2    Comet

content_actors
content     model
1           1
1           2
2           1

actors
record_num     name
1              Jennifer Lopez
2              Bruce Willis

content_views
content     views
1           160
2           312

以下是我通过SHOW INDEX FROM tablename：

找到的索引

Table              Column_Name     Seq_in_index     Key_name     Index_type
---------------------------------------------------------------------------
content            record_num      1                PRIMARY      BTREE
content            keywords        1                keywords     FULLTEXT
content            keywords        2                title        FULLTEXT
content            title           1                title        FULLTEXT
content            description     1                description  FULLTEXT
content            keywords        1                keywords_2   FULLTEXT

content_actors     content         1                content      BTREE
content_actors     actor           2                content      BTREE
content_actor      actor           1                actor        BTREE

actors             record_num      1                PRIMARY      BTREE
actors             name            1                name         BTREE
actors             name            1                name_2       FULLTEXT

content_views      content         1                PRIMARY      BTREE
content_views      views           1                views        BTREE

以下是查询的EXPLAIN：

ID     SELECT_TYPE     TABLE     TYPE       POSSIBLE_KEYS          KEY         ROWS      EXTRA
1      SIMPLE          c         ref        enabled_2, enabled     enabled     29210     Using where; Using temporary; Using filesort
1      SIMPLE          ca        ref        content                content     1         Using index
1      SIMPLE          a         eq_ref     PRIMARY                PRIMARY     1
1      SIMPLE          cv        eq_ref     PRIMARY                PRIMARY     1

我正在使用GROUP BY来避免重复内容，但单独使用此组似乎会使处理查询所需的时间加倍。

编辑在稍微玩了一下查询之后，我意识到如果我删除了GROUP BY，我会得到重复项，如果我让GROUP BY在那里，它就不会采取适当的Relevance3值（没有GROUP BY，一个为Relevance3返回值，而另一个不是......）

Answer 1

将MATCHes（或者＆＃39; d加在一起）添加到WHERE - 这将显着减少SQL_CALC_FOUND_ROWS中要处理的行数，并且无需HAVING...。

而不是

cv.views,
...
LEFT JOIN  content_views AS cv ON cv.content = c.record_num

DO

( SELECT views FROM content_views ON content = c.record_num ) AS views,

修改

LEFT和GROUP BY是必需的，因为actors是可选的，可能有多个actors。既然你根本不需要演员姓名，你可以通过
来摆脱它
WHERE ... AND ( EXISTS SELECT * FROM content_actors JOIN actors AS a ON ... WHERE MATCH (a.`name`) AGAINST ('{$q}') AND ca... )

但是这不允许您在ORDER BY中包含相关性。

因此，您需要使用UNION DISTINCT构建子查询。将有2 SELECTs：

SELECT＃1：

SELECT c.id, 3 * MATCH(c.title) AGAINST ('{$q}') + MATCH(c.keywords) AGAINST ('{$q}') AS relevance FROM Content AS c WHERE MATCH(c.title, c.keywords) AGAINST ('{$q}')

（并且FULLTEXT(title, keywords)) This will efficiently fetch the ids for内容行有用。

SELECT＃2：

SELECT c.id, 2*MAX(MATCH(a.actor) AGAINST ('{$q}') AS actor_rel) AS relevance FROM content AS c JOIN content_actors ca ON ca.content = c.record_num JOIN actors a ON a.record_num = ca.actor WHERE MATCH(a.actor) AGAINST ('{$q}') GROUP BY c.id;

请务必拥有content_actors: INDEX(actor)和content: INDEX(record_num)。此SELECT将有效地从actors开始，然后返回content。请注意，当两个演员MATCH时，它会与您的代码有所不同;希望我的MAX是一个更好的解决方案。

现在，让我们把事情放在一起......

SELECT＃3：

SELECT id, SUM(rel) AS relevance FROM ( ... select #1 ... ) UNION ALL ( ... select #2 ... ) GROUP BY id

但这并不是全部...

SELECT＃4：

SELECT c.*, ( ... views ... ) AS views FROM ( ... select #3 ... ) AS u JOIN content c ON c.id = u.id

我建议您手动运行这些步骤以验证它们，逐步将所有部分组合在一起。是的，它很复杂，但应该非常快。

优化跨多个表的全文搜索

1 个答案: