Question

这是我的问题：

SELECT 
    h.id,
    h.subject,
    h.body matnF,
    h.amount,
    h.keywords tags,
    h.closed,
    h.author_id author,
    h.AcceptedAnswer,
    h.type,
    h.visibility,
    h.date_time,
    v.value AS vote_value,
    u.reputation,
    u.user_fname,
    u.user_lname,
    u.avatar,
    (h.author_id = user_id1) as hasUserId,
    (select COALESCE(sum(vv.value), 0)
     from votes vv
     where h.id = vv.post_id and vv.table_code = '$this->table_code'
    ) as total_votes,
    (select count(1)
     from favorites ff
     where h.type = 0 and h.id = ff.post_id and ff.table_code = '$this- >table_code'
    ) as total_favorites,
    CASE WHEN h.type = 1 THEN '0'
         WHEN h.amount IS NULL OR h.author_id = :user_id2 THEN '1'
    ELSE EXISTS( select 1
            from money m
            where m.user_id = :user_id3 and m.post_id = h.id
           )END paid,
    CASE WHEN h.type = 0 AND f.id IS NOT NULL THEN '2'
    ELSE '3'
    END AS favorite
FROM qanda h
LEFT JOIN votes v ON h.id = v.post_id AND v.user_id = :user_id4 AND   v.table_code = '$this->table_code'
LEFT JOIN favorites f ON h.type = 0 AND h.id = f.post_id AND f.user_id = :user_id5 AND f.table_code = '$this->table_code'
LEFT JOIN users u ON h.author_id = u.id and h.visibility = 1
WHERE h.id = :id1 OR h.related = :id2
ORDER BY h.type , h.AcceptedAnswer DESC , h.date_time
LIMIT 20;

请关注上面的这一行查询：

WHERE h.id = :id1 OR h.related = :id2

如您所见，这两个条件之间存在OR逻辑运算符。如您所知，OR通常会阻止索引的有效使用。因此，当存在大量数据时，我的查询的性能确实很弱。

我该如何改进？实际上我正试图用OR替换UNION，但是当你看到我的查询太长时......那么有什么想法吗？

编辑：以下是EXPLAIN：的结果（对于非常短的数据集 - 总共20行）

Answer 1

我要尝试的第一件事是子查询：

from ((select q.* from quanda q where q.id = :id1) union
      (select q.* from quanda q where q.related = :id2)
     ) left join
     . . .

注意：这确实需要quanda(id)和quanda(related)上的索引来提高性能。

如果选择了很少的行，那么这可能会快得多。

Answer 2

你可以尝试一下：

SELECT h.id, h.subject, h.body matnF, h.amount, h.keywords tags, h.closed, h.author_id author, h.AcceptedAnswer, h.type, h.visibility, h.date_time, v.value AS vote_value, u.reputation, u.user_fname, u.user_lname, u.avatar, (h.author_id = :user_id1) as hasUserId,
(select COALESCE(sum(vv.value),0) from votes vv where h.id = vv.post_id and vv.table_code = '$this->table_code') as total_votes,
(select count(1) from favorites ff where h.type = 0 and h.id = ff.post_id and ff.table_code = '$this->table_code') as total_favorites,
CASE WHEN h.type = 1 THEN '0'
WHEN h.amount IS NULL OR h.author_id = :user_id2 THEN '1'
ELSE EXISTS (select 1 from money m where m.user_id = :user_id3 and m.post_id = h.id) END paid,
CASE WHEN h.type = 0 AND f.id IS NOT NULL THEN '2' ELSE '3' END AS favorite
FROM qanda h
LEFT JOIN votes v ON h.id = v.post_id AND v.user_id = :user_id4 AND v.table_code = '$this->table_code'
LEFT JOIN favorites f ON h.type = 0 AND h.id = f.post_id AND f.user_id = :user_id5 AND f.table_code = '$this->table_code'
LEFT JOIN users u ON h.author_id = u.id and h.visibility = 1
WHERE h.id = :id1 
ORDER BY h.type, /*(tans.id IS NOT NULL) DESC,*/ h.AcceptedAnswer DESC, h.date_time
LIMIT 20

UNION 

SELECT 
*
FROM 
(SELECT h.id, h.subject, h.body matnF, h.amount, h.keywords tags, h.closed, h.author_id author, h.AcceptedAnswer, h.type, h.visibility, h.date_time, v.value AS vote_value, u.reputation, u.user_fname, u.user_lname, u.avatar, (h.author_id = :user_id1) as hasUserId,
(select COALESCE(sum(vv.value),0) from votes vv where h.id = vv.post_id and vv.table_code = '$this->table_code') as total_votes,
(select count(1) from favorites ff where h.type = 0 and h.id = ff.post_id and ff.table_code = '$this->table_code') as total_favorites,
CASE WHEN h.type = 1 THEN '0'
WHEN h.amount IS NULL OR h.author_id = :user_id2 THEN '1'
ELSE EXISTS (select 1 from money m where m.user_id = :user_id3 and m.post_id = h.id) END paid,
CASE WHEN h.type = 0 AND f.id IS NOT NULL THEN '2' ELSE '3' END AS favorite
FROM qanda h
LEFT JOIN votes v ON h.id = v.post_id AND v.user_id = :user_id4 AND v.table_code = '$this->table_code'
LEFT JOIN favorites f ON h.type = 0 AND h.id = f.post_id AND f.user_id = :user_id5 AND f.table_code = '$this->table_code'
LEFT JOIN users u ON h.author_id = u.id and h.visibility = 1
WHERE h.id = :id2 
ORDER BY h.type, /*(tans.id IS NOT NULL) DESC,*/ h.AcceptedAnswer DESC, h.date_time
LIMIT 20) t;

注意：如果您希望在最终结果集中允许重复，请将UNION替换为UNION ALL。

Answer 3

如果要从使用UNION而不是OR获得任何优势，那么使查询的那部分尽可能小，特别是将列数保持最小。在任何其他联接之前做联盟。我建议这个：

SELECT
      h.id
    , h.subject
    , h.body      AS matnF
    , h.amount
    , h.keywords  AS tags
    , h.closed
    , h.author_id AS author
    , h.AcceptedAnswer
    , h.type
    , h.visibility
    , h.date_time
    , v.value     AS vote_value
    , u.reputation
    , u.user_fname
    , u.user_lname
    , u.avatar
    , h.author_id AS hasUserId
    , (     SELECT COALESCE(SUM(vv.value), 0)
            FROM votes vv
            WHERE h.id = vv.post_id
                  AND vv.table_code = '$this->table_code'
      )  AS total_votes
    , (     SELECT COUNT(1)
            FROM favorites ff
            WHERE h.type = 0 AND h.id = ff.post_id
                  AND ff.table_code = '$this- >table_code'
      ) AS total_favorites
    , CASE
            WHEN h.type = 0 AND f.id IS NOT NULL THEN '2'
            ELSE '3'
      END  AS favorite
    , CASE
            WHEN h.type = 1 THEN '0'
            WHEN h.amount IS NULL OR h.author_id = :user_id2 THEN '1'
            ELSE EXISTS( select 1
                from money m
                where m.user_id = :user_id3 and m.post_id = h.id
               )
      END paid
FROM (
      SELECT q.id , q.author_id , q.visibility , q.type FROM quanda q WHERE q.id = :id1
      UNION
      SELECT q.id , q.author_id , q.visibility , q.type FROM quanda q WHERE q.related = :id2
      ) h
      LEFT JOIN votes v ON h.id = v.post_id
                  AND v.user_id = :user_id4
                  AND v.table_code = '$this->table_code'
      LEFT JOIN favorites f ON h.type = 0
                  AND h.id = f.post_id
                  AND f.user_id = :user_id5
                  AND f.table_code = '$this->table_code'
      LEFT JOIN users u ON h.author_id = u.id
                  AND h.visibility = 1

如何用UNION运算符替换OR运算符？

3 个答案: