Question

对，我不知道为什么但是这个查询执行时间超过6秒，索引的设置都是正确的，如果我单独运行每个查询，它的运行时间不到0.5秒。

这是查询

SELECT c.supplier_id, supplier_name, address1, address2, address3, address4, suppliertype, postcode, contact_name,
(SELECT COUNT(*)
    FROM supplier_questions q1
    WHERE c.supplier_id = q1.supplier_id AND q1.incomplete = '0') AS questions, 
IF (active=1,'Yes', IF (active=2, 'NCR Only','Inactive')) AS rated, 
(SELECT COUNT(*)
    FROM supplier_questions q2
    WHERE c.supplier_id = q2.supplier_id AND q2.reviewed = '1') AS reviewed, 
questapproved, 
ss.supplier_no AS supplier_no
FROM suppliers c
INNER JOIN supplier_site ss ON c.supplier_id = ss.supplier_id
WHERE c.supplier_id != '0' AND ss.site_id = '2'
GROUP BY c.supplier_id
ORDER BY c.supplier_name ASC
LIMIT 0, 20

Explain查询的结果如下

id  select_type table   type    possible_keys   key key_len ref rows    Extra
1   PRIMARY ss  ref site_id,supplier_id site_id 4   const   1287    Using where; Using temporary; Using filesort
1   PRIMARY c   eq_ref  PRIMARY PRIMARY 4   ss.supplier_id  1   
3   DEPENDENT SUBQUERY  q2  ref supplier_id,reviewed    reviewed    4   const   263 Using where
2   DEPENDENT SUBQUERY  q1  ref supplier_id,incomplete  incomplete  4   const   254 Using where

计数查询所在的原因是因为我需要知道这些表中的行数，这不能在另一个查询中完成，因为结果也需要按这些值排序：（

Answer 1

在黑暗中刺伤，这会跑得更快吗？（我没有一个mysql来验证语法，所以原谅任何轻微的错误，但你可能会得到这个想法）

SELECT c.supplier_id, supplier_name, address1, address2, address3, address4, suppliertype, postcode, contact_name, questions, reviewed 
IF (active=1,'Yes', IF (active=2, 'NCR Only','Inactive')) AS rated,  
questapproved,  ss.supplier_no AS supplier_no 
FROM suppliers c 
INNER JOIN supplier_site ss ON c.supplier_id = ss.supplier_id 
inner join 
(SELECT supplier_id, sum(if(incomplete='0',1,0)) as questions,  sum(if(incomplete='1',1,0)) as reviewed FROM supplier_questions q1 group by supplier_id) as tmp
on c.supplier_id = tmp.supplier_id
WHERE c.supplier_id != '0' AND ss.site_id = '2' 
GROUP BY c.supplier_id 
ORDER BY c.supplier_name ASC LIMIT 0, 20

Answer 2

FROM suppliers c
INNER JOIN supplier_site ss ON c.supplier_id = ss.supplier_id
WHERE c.supplier_id != '0' AND ss.site_id = '2'
GROUP BY c.supplier_id 
ORDER BY c.supplier_name ASC

由于自动生成的主键永远不会等于0（除非大数据库设计错误），您可以删除c.supplier_id！='0'子句。

ss.site_id ='2'应该处于JOIN条件以便于阅读。

看起来这应该只匹配每个供应商的表supplier_site中的一行（如果这是您通常的1-N事物 - 地址关系，即您选择每个供应商的第二个地址，也许'2'对应于'帐单地址'或其他东西）所以GROUP BY c.supplier_id是无用的。如果GROUP BY实际上做了什么，那么查询是错误的，因为“地址”列（可能来自supplier_site表）将来自随机行。

所以这里是简化的FROM（WHERE消失了）：

FROM suppliers c
INNER JOIN supplier_site ss ON 
    (c.supplier_id = ss.supplier_id AND ss.site_id = '2')
ORDER BY c.supplier_name ASC

我想你在c.supplier_name上有一个索引，所以这部分查询应该非常快。

现在尝试此查询：

SELECT a.*,
    questapproved, 
    ss.supplier_no AS supplier_no,
    IF (active=1,'Yes', IF (active=2, 'NCR Only','Inactive')) AS rated, 
    sum( q.incomplete = '0') AS questions,
    sum( q.reviewed = '1' ) AS reviewed
FROM
(
    SELECT c.supplier_id, supplier_name, address1, address2, address3, address4, suppliertype, postcode, contact_name
    FROM suppliers c
    INNER JOIN supplier_site ss ON 
        (c.supplier_id = ss.supplier_id AND ss.site_id = '2')
    ORDER BY c.supplier_name ASC
    LIMIT 0, 20
) a
LEFT JOIN supplier_questions q ON (q.supplier_id = c.supplier_id)
GROUP BY c.supplier_id
ORDER BY c.supplier_name;

Answer 3

如果您删除子选项，最终会得到以下内容：

SELECT c.supplier_id, supplier_name, address1, address2, address3, address4, suppliertype, postcode, contact_name,
COUNT(IF (q1.incomplete = '0', '0', null)) AS questions, 
IF (active=1,'Yes', IF (active=2, 'NCR Only','Inactive')) AS rated, 
COUNT(IF (q1.reviewed = '1', '1', null)) AS reviewed,
questapproved, 
ss.supplier_no AS supplier_no
FROM suppliers c
INNER JOIN supplier_site ss ON c.supplier_id = ss.supplier_id
LEFT OUTER JOIN supplier_questions q1 ON c.supplier_id = q1.supplier_id
WHERE c.supplier_id != '0' AND ss.site_id = '2'
GROUP BY c.supplier_id
ORDER BY c.supplier_name ASC
LIMIT 0, 20

我没有可用的MySQL数据库，因此我的SQL可能存在错误。我们的想法是删除子查询并用外连接替换它们并使用IF仅计算相关行。

Answer 4

我首先尝试通过供应商预先查询问题的数量并审核ONCE来进行重组。然后，加入其余的细节。通过使用STRAIGHT_JOIN关键字，它应按显示的顺序进行处理。这将首先预先聚合并使用THAT作为加入供应商和供应商站点的基础。无论如何都不需要外部组，因为它基于供应商ID。但是，对supplier_sites（您的ss.supplier_no）的加入意味着供应商有多个地点。这是否意味着地址和活动状态列来自该表？

问题的加入是否应与特定供应商相关联，是否与相应的网站位置相关？

此外，由于prequery在supplier_id！='0'上具有WHERE子句，因此它不需要下游，因为这将是与其他表正常连接的基础，从而将它们排除在结果集之外。< / p>

SELECT STRAIGHT_JOIN
      PreAggregate.supplier_id, 
      PreAggregate.supplier_name, 
      address1, 
      address2, 
      address3, 
      address4, 
      suppliertype, 
      postcode, 
      contact_name,
      PreAggregate.Questions,
      IF (active=1,'Yes', IF (active=2, 'NCR Only','Inactive')) AS rated, 
      PreAggregate.Reviewed,
      questapproved, 
      ss.supplier_no AS supplier_no
   FROM 
      (select 
             s1.Supplier_ID,
             s1.Supplier_Name,
             SUM( IF( q1.Incomplete = '0', 1, 0 )) Questions,
             SUM( IF( q1.Reviewed = '1', 1, 0 )) Reviewed
          from 
             suppliers s1
                join supplier_questions q1
                   ON s1.supplier_id = q1.supplier_id
          where
             s1.supplier_id != '0'
          group by
             s1.Supplier_ID 
          ORDER BY 
             s1.supplier_name ASC ) PreAggregate

       JOIN suppliers c
          ON PreAggregate.Supplier_ID = c.Supplier_ID

       JOIN supplier_site ss 
          ON PreAggregate.Supplier_ID = ss.supplier_id
          AND ss.Site_ID = '2'
  LIMIT 0, 20

使用COUNT对子查询进行慢速MYSQL查询

4 个答案: