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编辑最终：修改后的代码适用于 5.5，但在 5.0 上，我得到的行集略有不同，任何人都有任何想法......

以下查询

SELECT DISTINCT
  sql_no_cache as job.id,
  job.id as job_id,
  job.job_status,
  job.servee_name,
  kase.plaintiff,
  index_number,
  lawfirm_filenumber,
  DATE_FORMAT(job.datetime_received, '%m/%d/%Y') as date_received,
  DATE_FORMAT(job.date_sent_to_court, '%m/%d/%Y') as date_sent_to_court,
  DATE_FORMAT(kase.date_kase_filed, '%m/%d/%Y') as date_kase_filed,
  TO_DAYS(NOW())-TO_DAYS(kase.date_kase_filed) as thedays,
  DATE_FORMAT(kase.court_datetime, '%m/%d/%Y %h:%i %p') as court_date,
  kase.court_room,
  CONCAT( LUcourt.court_county," - ",LUcourt.court_type) as court_name,
  DATE_FORMAT( job.datetime_served, '%m/%d/%Y') as datetime_served,
  CONCAT( server.namefirst,' ', server.namelast) as server_name

FROM      job 
LEFT JOIN kase        ON job.kase_id = kase.id
LEFT JOIN kasebilling ON kasebilling.kase_id = kase.id
LEFT JOIN server      ON job.server_id = server.id
LEFT JOIN client      ON client.id     = kase.client_id
LEFT JOIN LUcourt     ON LUcourt.id    = kase.court_id

   WHERE job.subscriber_id = 1
ORDER BY job_id + 0
   LIMIT 0,500;

</pre>
i modified the query but it wont run ...

    SELECT DISTINCT sql_no_cache  job.id, job.id as job_id, job.job_status, job.servee_name, kase.plaintiff, index_number, lawfirm_filenumber, DATE_FORMAT(job.datetime_received, '%m/%d/%Y') as date_received, DATE_FORMAT(job.date_sent_to_court, '%m/%d/%Y') as date_sent_to_court, DATE_FORMAT(kase.date_kase_filed, '%m/%d/%Y') as date_kase_filed, TO_DAYS(NOW())-TO_DAYS(kase.date_kase_filed) as thedays, DATE_FORMAT(kase.court_datetime, '%m/%d/%Y %h:%i %p') as court_date, kase.court_room, CONCAT(LUcourt.court_county," - ",LUcourt.court_type) as court_name, DATE_FORMAT(job.datetime_served, '%m/%d/%Y') as datetime_served, concat(server.namefirst,' ', server.namelast) as server_name FROM job 
    LEFT JOIN kase ON job.kase_id = kase.id
    LEFT JOIN kasebilling  ON kasebilling.kase_id = kase.id
    LEFT JOIN server ON job.server_id = server.id
    left join client on client.id=kase.client_id
    left join LUcourt on LUcourt.id=kase.court_id  WHERE job.subscriber_id = 1 and job_id in (select distinct job.id from job where job.subscriber_id = '1' order by id+0) LIMIT 0,500
</pre>

我已经修改了上面的查询以删除一些表名......但是发生的情况是，如果我通过查询删除订单会非常快......我认为它的排序不仅仅是限制 500，而是所有行...我怎样才能通过 order by 更快地完成这项工作。任何帮助将不胜感激谢谢....

PS：它在 order_by 上使用文件排序...

解释：

+----+-------------+-------------+--------+------- -------------------------------------------------- -----------------------------------------+-------- -------------------------+---------+-------------- ---------+--------+------------------------------- ---------------+
| 编号 | 选择类型 | 表| 类型 | 可能的键 | 关键 | key_len | 参考 | 行 | 额外 |
+----+-------------+-------------+--------+------- -------------------------------------------------- -----------------------------------------+-------- -------------------------+---------+-------------- ---------+--------+------------------------------- ---------------+
| 1 | 简单 | 工作 | 参考 | 订阅者 ID，订阅者 ID-job_loc-client_loc-server_loc，sub_id_idx_servee_name_idx | subscriber_id-job_loc-client_loc-server_loc | 4 | 常量 | 362381 | 使用哪里；使用临时的；使用文件排序 |
| 1 | 简单 | 加濑 | eq_ref | 初级 | 初级 | 4 | pserve.job.kase_id | 1 | |
| 1 | 简单 | 卡西比林 | 参考 | kase_id | kase_id | 4 | pserve.kase.id | 2 | 使用索引 |
| 1 | 简单 | 服务器 | eq_ref | 初级 | 初级 | 4 | pserve.job.server_id | 1 | |
| 1 | 简单 | 客户 | eq_ref | 初级 | 初级 | 4 | pserve.kase.client_id | 1 | 使用索引 |
| 1 | 简单 | 卢考特 | eq_ref | 初级 | 初级 | 4 | pserve.kase.court_id | 1 | |
+----+-------------+-------------+--------+------- -------------------------------------------------- -----------------------------------------+-------- -------------------------+---------+-------------- ---------+--------+------------------------------- ---------------+

我修改了查询，但它不会运行...

SELECT DISTINCT sql_no_cache  job.id, job.id as job_id, job.job_status, job.servee_name, kase.plaintiff, index_number, lawfirm_filenumber, DATE_FORMAT(job.datetime_received, '%m/%d/%Y') as date_received, DATE_FORMAT(job.date_sent_to_court, '%m/%d/%Y') as date_sent_to_court, DATE_FORMAT(kase.date_kase_filed, '%m/%d/%Y') as date_kase_filed, TO_DAYS(NOW())-TO_DAYS(kase.date_kase_filed) as thedays, DATE_FORMAT(kase.court_datetime, '%m/%d/%Y %h:%i %p') as court_date, kase.court_room, CONCAT(LUcourt.court_county," - ",LUcourt.court_type) as court_name, DATE_FORMAT(job.datetime_served, '%m/%d/%Y') as datetime_served, concat(server.namefirst,' ', server.namelast) as server_name FROM job 
LEFT JOIN kase ON job.kase_id = kase.id
LEFT JOIN kasebilling  ON kasebilling.kase_id = kase.id
LEFT JOIN server ON job.server_id = server.id
left join client on client.id=kase.client_id
left join LUcourt on LUcourt.id=kase.court_id  WHERE job.subscriber_id = 1 and job_id in (select distinct job.id from job where job.subscriber_id = '1' order by id+0) LIMIT 0,500
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1 回答 1

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第一个丑陋的尝试：添加条件job.subscriber_id = '1'以LEFT JOIN job减少要排序和限制的总行数，但也将其保留在 WHERE 部分：

FROM      kase
LEFT JOIN job     ON k.kase_id   = kase.id
                 AND job.subscriber_id = '1' /* condition copied here ... */
LEFT JOIN kb      ON kb.kase_id  = kase.id
LEFT JOIN server  ON job.server_id = server.id
LEFT JOIN client  ON client.id   = kase.client_id
LEFT JOIN LUcourt ON LUcourt.id  = kase.court_id
    WHERE job.subscriber_id = '1'            /* ... but also kept here */
 ORDER BY job.id
    LIMIT 0,500;

第二次尝试，更丑陋：子选择前 500 kase*job组合，并针对子选择的行左连接其他表：

FROM
( SELECT * /* you probably have to list & alias column names if necessary */
  FROM      kase
  LEFT JOIN job   ON k.kase_id   = kase.id
      WHERE job.subscriber_id = '1'
   ORDER BY job.id
      LIMIT 0,500
) AS kase_job
/* Now we have 500 rows*/

LEFT JOIN kb      ON kb.kase_id  = kase_job.id
LEFT JOIN server  ON kase_job.server_id = server.id
LEFT JOIN client  ON client.id   = kase_job.client_id
LEFT JOIN LUcourt ON LUcourt.id  = kase_job.court_id

/* If you do `(500 rows) LEFT JOIN whatever` you can be sure you have
   at least 500 rows (never less) here, so you have to ORDER/LIMIT then again
*/

 ORDER BY kase_job.id
    LIMIT 0,500;
于 2012-09-10T19:56:59.203 回答