使用 COUNT 进行子查询的慢 MYSQL 查询

Question

是的，我不知道为什么，但是这个查询需要超过 6 秒才能执行，索引都设置正确，如果我单独运行每个查询，它在不到 0.5 秒的时间内就可以很好地执行。

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

解释查询的结果如下

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

计数查询存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

作为在黑暗中的刺，这跑得更快吗？ （我还没有 mysql 来验证语法，所以请原谅任何轻微的错误，但你可能会明白）

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' 应该在 JOIN 条件下以提高可读性。

看起来这应该只匹配表供应商站点中每个供应商的一行（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 是没用的。 如果 GROUP BY 确实做了某事，那么查询是错误的，因为可能来自供应商站点表的“地址”列将来自随机行。

所以这是简化的 FROM（WHERE 不见了）：

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。

现在试试这个查询：

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

如果您删除子选择，您最终会得到如下结果：

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 中可能存在错误。 这个想法是删除子查询并用外连接替换它们，并使用 IF 只计算相关行。

Answer 4

我将首先尝试通过供应商预先查询聚合问题的数量并审查一次来进行重组。 然后，加入rest的详细信息。 通过使用 STRAIGHT_JOIN 关键字，它应该按照显示的顺序进行处理。 这将首先进行预聚合，并以此为基础重新加入供应商，然后再加入供应商站点。 不需要外部组，因为它无论如何都基于供应商 ID。 但是，加入供应商站点（您的 ss.supplier_no）意味着供应商拥有多个位置。 这是否意味着地址和活动状态列来自该表？

问题的加入是否应该与特定供应商及其相应的站点位置相关联？

此外，由于预查询在supplier_id = '0' 上有WHERE 子句，因此stream 不需要它，因为这将是正常连接到其他表的基础。 从而将它们从结果集中消除。

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