Question

我有这样的查询：

SELECT
  SUM(c.cantitate) AS num,
  p.id             AS pid,
  p.titlu          AS titlu,
  p.alias          AS alias,
  p.gramaj         AS gramaj,
  p.prettotal      AS prettotal,
  p.pretunitar     AS pretunitar,
  p.pretredus      AS pretredus,
  p.stoc           AS stoc,
  p.cant_variabila AS cant_variabila,
  p.nou            AS nou,
  p.congelat       AS congelat,
  p.cod            AS cod,
  p.poza           AS poza,
  cc.seo           AS seo
FROM produse p
  LEFT JOIN (SELECT  produs, cantitate, COS FROM comenzi) c ON p.id = c.produs
  LEFT JOIN (SELECT STATUS, id FROM cosuri) cs ON c.cos = cs.id
  LEFT JOIN (SELECT id, seo FROM categorii) cc ON p.categorie = cc.id
WHERE cs.status = 'closed' AND p.vizibil = '1'
GROUP BY pid
ORDER BY num DESC
LIMIT 0, 14

查询正常，但1个查询的持续时间：2.922秒。如何改进查询？

密钥如下：

comenzi：cos，produs as unique key

cosuri：id为唯一键

产品：titlu，类别，别名为关键

Answer 1

由于您有很多sub-queries，所以花费在完整数据读取上的时间。如果您直接在桌面上应用联接而不是sub query，则会提高效果。

删除所有子查询后尝试一下。像：

SELECT
  SUM(c.cantitate) AS num,
  p.id             AS pid,
  p.titlu          AS titlu,
  p.alias          AS alias,
  p.gramaj         AS gramaj,
  p.prettotal      AS prettotal,
  p.pretunitar     AS pretunitar,
  p.pretredus      AS pretredus,
  p.stoc           AS stoc,
  p.cant_variabila AS cant_variabila,
  p.nou            AS nou,
  p.congelat       AS congelat,
  p.cod            AS cod,
  p.poza           AS poza,
  cc.seo           AS seo
FROM produse p
  LEFT JOIN comenzi c
    ON p.id = c.produs
  LEFT JOIN cosuri cs
    ON c.cos = cs.id
  LEFT JOIN categorii cc
    ON p.categorie = cc.id
WHERE cs.status = 'closed'
    AND p.vizibil = '1'
GROUP BY pid
ORDER BY num DESC
LIMIT 0, 14

你也应该有以下索引：

+---------+------------+
|  TABLE  |   COLUMN   |
+---------+------------+
| produse | categorie  |
| produse | vizibil    |
| comenzi | produs     |
| comenzi | cos        |
| cosuri  | status     |
+---------+------------+

假设所有表中的id列表都是PK，否则这些列也需要index。

Answer 2

你可以这样做：花费更少的时间

SELECT
  SUM(c.cantitate) AS num,
  p.id             AS pid,
  p.titlu          AS titlu,
  p.alias          AS alias,
  p.gramaj         AS gramaj,
  p.prettotal      AS prettotal,
  p.pretunitar     AS pretunitar,
  p.pretredus      AS pretredus,
  p.stoc           AS stoc,
  p.cant_variabila AS cant_variabila,
  p.nou            AS nou,
  p.congelat       AS congelat,
  p.cod            AS cod,
  p.poza           AS poza,
  cc.seo           AS seo
FROM produse p
  LEFT JOIN comenzi c ON p.id = c.produs
  LEFT JOIN cosuri cs ON c.cos = cs.id
  LEFT JOIN categorii cc ON p.categorie = cc.id
WHERE cs.status = 'closed' AND p.vizibil = '1'
GROUP BY pid
ORDER BY num DESC
LIMIT 0, 14

Answer 3

会很快

SELECT
  SUM(c.cantitate) AS num,
  p.id             AS pid,
  p.titlu          AS titlu,
  p.alias          AS alias,
  p.gramaj         AS gramaj,
  p.prettotal      AS prettotal,
  p.pretunitar     AS pretunitar,
  p.pretredus      AS pretredus,
  p.stoc           AS stoc,
  p.cant_variabila AS cant_variabila,
  p.nou            AS nou,
  p.congelat       AS congelat,
  p.cod            AS cod,
  p.poza           AS poza,
  cc.seo           AS seo
FROM produse p
  LEFT JOIN comenzi c ON p.id = c.produs
  LEFT JOIN cosuri cs ON c.cos = cs.id
  LEFT JOIN categorii cc ON p.categorie = cc.id
WHERE cs.status = 'closed' AND p.vizibil = '1'
GROUP BY pid
ORDER BY num DESC
LIMIT 0, 14

Answer 4

只需删除子查询。没有帮助你。测试一下，告诉我们需要多少。

SELECT
  SUM(c.cantitate) AS num,
  p.id             AS pid,
  p.titlu          AS titlu,
  p.alias          AS alias,
  p.gramaj         AS gramaj,
  p.prettotal      AS prettotal,
  p.pretunitar     AS pretunitar,
  p.pretredus      AS pretredus,
  p.stoc           AS stoc,
  p.cant_variabila AS cant_variabila,
  p.nou            AS nou,
  p.congelat       AS congelat,
  p.cod            AS cod,
  p.poza           AS poza,
  cc.seo           AS seo
FROM produse p
  LEFT JOIN comenzi   c  ON p.id = c.produs
  LEFT JOIN cosuri    cs ON c.cos = cs.id
  LEFT JOIN categorii cc ON p.categorie = cc.id
WHERE cs.status = 'closed' AND p.vizibil = '1'
GROUP BY pid
ORDER BY num DESC
LIMIT 0, 14

当然，如果有以下字段的索引会有所帮助：

p.id（主键已经编入索引并且不需要创建新主键）
cc.id（主键已经编入索引并且不需要创建新主键）
cs.id（主键已经编入索引并且不需要创建新主键）
p.categorie
p.vizibil（此处潜在收益较小）
c.produs
c.cos
cs.status（此处潜在收益较小）
p.pid（根据需要订购记录而忘记该组的典型错误）

优化mysql多个左连接

4 个答案: