緩慢的MYSQL查詢與使用COUNT的子查詢

Question

正確我不知道爲什麼，但這個查詢需要超過6秒的執行，索引的所有設置都正確，如果我分別運行每個查詢它執行很好，執行時間少於0.5秒。

下面是該查詢

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 

的解釋查詢的結果是如下

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 

原因計數查詢是在那裏，是因爲我需要知道那些行數表，這是不能在另一個查詢中完成，因爲結果還需要按這些值排序：（

Answer 1

作爲在黑暗中的刺，它運行得更快嗎？（我沒有得到一個MySQL來驗證Y開，所以請原諒任何輕微的錯誤，但你可能會得到的想法）

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

如果刪除的語法子選擇你最終的東西是這樣的：

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 

我不沒有MySQL數據庫可用，所以可能在我的SQL中有錯誤。 這個想法是刪除子查詢並將其替換爲外連接 並使用IF來只計數相關行。

Answer 3

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 

由於自動生成的主鍵永遠不會等於0（除非大數據庫設計錯誤），您可以刪除c.supplier_id！='0'子句。

爲了提高可讀性，ss.site_id ='2'應該處於連接條件。

它看起來應該只匹配每個供應商的表supplier_site中的一行（如果這是您通常的1-N事物地址關係，即您選擇每個供應商的第二個地址，也許'2'對應於'帳單地址'或其他），所以GROUP BY c.supplier_id是無用的。如果GROUP BY實際上做了某事，那麼查詢是錯誤的，因爲「地址」列可能來自supplier_site表，它來自隨機行。

因此，這裏的簡化FROM（在WHERE消失）：

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 

我想你上了車c.supplier_name一個索引，查詢的這部分應該是非常快的。

現在試試這個查詢：

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 4

我會先通過查詢預聚集體通過供應商的問題數和經審查後嘗試重組。然後，加入其餘的細節。通過使用STRAIGHT_JOIN關鍵字，它應該按照顯示的順序進行處理。這將首先進行預先彙總，然後使用THAT作爲參加回供應商和供應商網站的依據。無論是否需要外部組織，因爲它基於供應商ID。然而，加入supplier_sites（您的ss.supplier_no）意味着供應商擁有多個位置。這是否意味着地址和活動狀態列是源自該表的？

問題連接是否應與特定供應商相關聯，並且是否與相應的網站位置相關？

此外，由於prequery在supplier_id！='0'上有WHERE子句，因此它不需要向下流，因爲這將成爲正常連接到其他表的基礎，從而將它們排除在結果集之外。

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