匹配的記錄考慮以下4個表MySQL的次數從多個表

entity table1  table2  table3  
------ ------------- ------------- ------------- 
id  ei(entity.id) ei(entity.id) ei(entity.id) 
name something  somethingelse yetanother

我怎樣才能找出所有三個表在實體的用法，這樣表示

--------------------- 
| id | t1 | t2 | t3 | 
--------------------- 
| 1 | 14 | 23 | 0 | 
| 2 | 66 | 9 | 5 | 
...

我原來的方法是從實體選擇然後離開加入其他表，但MySQL似乎並不喜歡它

SELECT e.id,count(t1.id) FROM entity AS e LEFT JOIN table1 AS t1 on e.id=t1.ei;

編輯：下面是輸出1個表

mysql> explain select e.id,count(o.id) from entity e left join table1 t1 on e.id=o.ei where e.ty=3; 
+----+-------------+-------+------+---------------+------+---------+------+-------+-------------+ 
| id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra  | 
+----+-------------+-------+------+---------------+------+---------+------+-------+-------------+ 
| 1 | SIMPLE  | e  | ALL | NULL   | NULL | NULL | NULL | 1083 | Using where | 
| 1 | SIMPLE  | o  | ALL | NULL   | NULL | NULL | NULL | 90201 |    | 
+----+-------------+-------+------+---------------+------+---------+------+-------+-------------+ 
2 rows in set (0.04 sec)

相反的效果要好得多，但不能擴展到多個表

SELECT e.id,count(t1,id) FROM table1 AS t1 LEFT JOIN entity AS e ON t1.ei=e.id

來源

2012-11-07 MonoThreaded

重寫此查詢的另一種方法。

集團並計算每個單獨的表，然後再加入：

SELECT a.id, 
     COALESCE(b.t1, 0) AS t1, 
     COALESCE(c.t2, 0) AS t2, 
     COALESCE(d.t3, 0) AS t3 
FROM 
     entity a 
    LEFT JOIN 
     (SELECT ei, 
       COUNT(*) AS t1 
      FROM table1 
      GROUP BY ei 
     ) AS b 
      ON a.id = b.ei 
    LEFT JOIN 
     (SELECT ei, 
       COUNT(*) AS t2 
      FROM table2 
      GROUP BY ei 
     ) AS c 
      ON a.id = c.ei 
    LEFT JOIN 
     (SELECT ei, 
       COUNT(*) AS t3 
      FROM table3 
      GROUP BY ei 
     ) AS d 
      ON a.id = d.ei 
    ;

您應該(ei)在每個3代表的絕對增加一個索引，如果你還沒有一個。

來源

2012-11-07 15:19:35

哇。工程奇蹟。您可能希望在每個FROM – MonoThreaded

之前修復額外的'，'順便說一下，如果沒有索引，查詢需要0.5s，這對我來說非常合適。 – MonoThreaded

如果你喜歡0.5超過0.005，對我來說沒問題。我不明白爲什麼人們不想把索引放在表格中，這就是他們的目的，以提高效率。 –

SELECT a.ID, 
     COUNT(b.ei) t1, 
     COUNT(c.ei) t2, 
     COUNT(d.ei) t3 
FROM entity a 
     LEFT JOIN table1 b 
      ON a.id = b.ei 
     LEFT JOIN table2 c 
      ON a.id = c.ei 
     LEFT JOIN table3 d 
      ON a.ID = d.ei 
GROUP BY a.ID

來源

2012-11-07 14:36:04

我試過了，在一分鐘左右之後不得不殺掉查詢，而單獨運行計數需要0.04s ...我預計它會花費0.12s（？） – MonoThreaded

您是否對此列有索引？ 'entity.id'，'table1.ei'，'table2.ei'，'table3.ei'？你可以發佈結果'EXPLAIN pasteyourQueyrHere'？ –

用解釋輸出 – MonoThreaded

select select 
e.id, 
sum(case when t1.name is null then 0 else 1 end) t1, 
sum(case when t2.name is null then 0 else 1 end) t2, 
sum(case when t3.name is null then 0 else 1 end) t3 
from 
entity e left join table1 t1 on e.id=t1.ei left join table2 t2 on e.id=t2.ei left join table3 t3 on e.id=t3.ei 
group by e.id

SQL Fiddle Demo

來源

2012-11-07 14:51:15 Ahmad

這也產生了一個很長的查詢（如在幾分鐘內） – MonoThreaded

MySQL的次數從多個表

回答

SQL Fiddle Demo

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