Question

如何提高此查询的效果，同时获取所需的所有信息..

SELECT 
    tr.id, tr.request_status, tr.note, tr.created_date, 
    c.name AS customer_name, c.mobile_phONe, 
    u.full_name AS created_by_name, tt.name AS ticket_type_name 
FROM  
    ticket_request tr 
LEFT JOIN 
    ticket_type tt ON tt.id = tr.ticket_type_id 
LEFT JOIN 
    users u ON u.id = tr.created_by 
LEFT JOIN 
    customer c ON c.id = tr.customer_id 
WHERE 
    tr.is_deleted != 1 
    AND tr.user_id IN (SELECT u.id FROM users u WHERE u.status = '1') 
GROUP BY 
    tr.id  
ORDER BY 
    tr.created_date DESC 
LIMIT 0,20

目前，此查询在7-10秒内运行。

ticket_request表有大约100k行
customers表有大约300k行
users表格和ticket_type没有那么多（约1k行）

Answer 1

下面的加速技术是首先取消LIMIT ，只有在此之后，才能完成所有JOINs。

SELECT  tr3.id, tr3.request_status, tr3.note, tr3.created_date,
        c.name AS customer_name, c.mobile_phONe,
        u2.full_name AS created_by_name,
        tt.name AS ticket_type_name
    FROM  
    (
        SELECT  tr1.id
            FROM  ticket_request tr1
            JOIN  users u1  ON u1.id = tr1.created_by
            WHERE  u1.status = '1'
              AND  tr1.is_deleted != 1
            ORDER BY  tr1.created_date DESC
            LIMIT  0,20 
    ) AS tr2
    JOIN  ticket_request AS tr3  ON tr3.id = tr2.id
    JOIN  user AS u2 ON u2.id = tr3.created_by
    LEFT JOIN  ticket_type tt  ON tt.id = tr3.ticket_type_id
    LEFT JOIN  customer c  ON c.id = tr3.customer_id
    ORDER BY  tr3.created_date

“衍生”表tr2中的JOINs之后仅触及20行;这是加速的主要原因。

这可能同样好：

SELECT  d.id, d.request_status, d.note, d.created_date,
        c.name AS customer_name, c.mobile_phONe, d.created_by_name,
        tt.name AS ticket_type_name
    FROM  
    (
        SELECT  tr.id AS tr_id, tr.request_status, tr.note, tr.created_date,
                tr.ticket_type_id, tr.customer_id
                u.full_name AS created_by_name
            FROM  ticket_request tr
            JOIN  users u  ON u.id = tr.created_by
            WHERE  u.status = '1'
              AND  tr.is_deleted != 1
            ORDER BY  tr.created_date DESC
            LIMIT  0,20 
    ) AS d
    LEFT JOIN  ticket_type tt  ON tt.id = d.ticket_type_id
    LEFT JOIN  customer c  ON c.id = d.customer_id
    ORDER BY  d.created_date

Answer 2

我假设您使用的是MySQL。如果没有，可以稍微修改这个答案以适应另一个数据库，但概念应该保持不变。您可以使用ticket_request列向左连接右侧所涉及的所有ID列添加索引：

ALTER TABLE ticket_type ADD INDEX (id);
ALTER TABLE users ADD INDEX (id);
ALTER TABLE customer ADD INDEX (id);      -- important

要解释索引有用的原因，请考虑LEFT JOIN表和ticket_request表之间的第一个ticket_type。如果没有索引，对于ticket_request中的每个记录，数据库必须扫描整个ticket_type表以查找与连接条件匹配的记录。从性能的角度来看，这是昂贵的。但是使用索引，数据库可以更快地完成此操作，因为它“知道”匹配记录的确切位置（或几乎完全准确）。

虽然您提到只有customer表非常大，但您仍然可以将索引添加到其他表中。将来，它们也可能变大。最有可能涉及customer的联接是您查询的瓶颈。

Answer 3

此处优化的最大机会是使用LIMIT 0,20

GROUP BY tr.id毫无意义，应该删除。
create index ticket_request_ix_is_deleted_created_date on ticket_request (is_deleted,created_date)并将tr.is_deleted != 1更改为tr.is_deleted = 0。

或者

create index ticket_request_ix_created_date on ticket_request (created_date)

Answer 4

SELECT 
    tr.id, tr.request_status, tr.note, tr.created_date, 
    c.name AS customer_name, c.mobile_phONe, 
    u.full_name AS created_by_name, tt.name AS ticket_type_name 
FROM  
    ticket_request tr 
LEFT JOIN 
    ticket_type tt ON tt.id = tr.ticket_type_id and tr.is_deleted != 1 
LEFT JOIN 
    users u ON u.id = tr.created_by 
JOIN 
  users u1 ON u1.id  = tr.user_id and u1.status = '1'
LEFT JOIN 
    customer c ON c.id = tr.customer_id 
GROUP BY 
    tr.id  
ORDER BY 
    tr.created_date DESC 
LIMIT 0,20

尝试此操作可以提高性能并根据您的要求进行调整

Answer 5

除了索引之外，在应用程序级别，您可以使用 Memcached （如果您使用的是php）这样的东西。这也将为您带来出色的表现。

如何在sql查询

5 个答案: