SQL 左链接(left join) 排序 分页 中遇到的未按理想状态排序分页的解决方案

SELECT a.id AS "id", a.code AS "code", a.name AS "name", a.type AS "type", a.importance_degree AS "importanceDegree", a.tech_state AS "techState", a.customerid AS "customerid", a.departmentid AS "departmentid", a.project_managerid AS "projectManagerid", a.plan_starttime AS "planStarttime", a.plan_endtime AS "planEndtime", a.state AS "state", a.act_starttime AS "actStarttime", a.act_endtime AS "actEndtime", a.sending_time AS "sendingTime", a.old_plan_starttime AS "oldPlanStarttime", a.old_plan_endtime AS "oldPlanEndtime", a.delivery_time AS "deliveryTime", a.complete_status AS "completeStatus", a.marketerid AS "marketerid", a.create_by AS "createBy.id", a.create_date AS "createDate.id", a.update_by AS "updateBy.id", a.update_date AS "updateDate.id", a.remarks AS "remarks", a.del_flag AS "delFlag", department.id AS "department.id", department.name AS "department.name", projectManager.id AS "projectManager.id", projectManager.name AS "projectManager.name", projectManager.mobile AS "projectManager.mobile", marketer.id AS "marketer.id", marketer.name AS "marketer.name", marketer.mobile AS "marketer.mobile", customer.id AS "customer.id", customer.code AS "customer.code", customer.name AS "customer.name", push.id AS "push.id", push.projectid AS "push.project.id", push.userid AS "push.user.id", push.is_readed AS "push.isReaded"
FROM (SELECT * FROM ps_project WHERE del_flag = 0 ORDER BY code DESC limit 2,2) a
LEFT JOIN ps_customer customer ON a.customerid = customer.id
LEFT JOIN sys_office department ON a.departmentid = department.id
LEFT JOIN sys_user projectManager ON a.project_managerid = projectManager.id
LEFT JOIN sys_user marketer ON a.marketerid = marketer.id
LEFT JOIN ps_project_push push ON a.id = push.projectid 
AND push.del_flag = 0 
AND customer.del_flag= 0 
AND department.del_flag= 0 
AND projectManager.del_flag= 0 
AND marketer.del_flag= 0

语句目标：

    以主表排序后并进行分页，然后再去链接其它表

出现问题：

    最终主表并无按照预想进行顺序输出，可是分页的数据是正确的。

来自  stackflow 解答：

No, the JOIN by order is changed during optimization.

个人解决办法（经测试有效）

分页排序在主表中进行，这样就mysql在执行的过程当中分根据咱们的理想按字段排序且选出指定分页。

可是在Join时，mysql系统作了优化，因此最终出来的结果又是乱序，此时，对最终被mysql Join打乱的结果顺序再作一次排序，这样就能获得咱们想要的结果了。

SELECT a.id AS "id", a.code AS "code", a.name AS "name", a.type AS "type", a.importance_degree AS "importanceDegree", a.tech_state AS "techState", a.customerid AS "customerid", a.departmentid AS "departmentid", a.project_managerid AS "projectManagerid", a.plan_starttime AS "planStarttime", a.plan_endtime AS "planEndtime", a.state AS "state", a.act_starttime AS "actStarttime", a.act_endtime AS "actEndtime", a.sending_time AS "sendingTime", a.old_plan_starttime AS "oldPlanStarttime", a.old_plan_endtime AS "oldPlanEndtime", a.delivery_time AS "deliveryTime", a.complete_status AS "completeStatus", a.marketerid AS "marketerid", a.create_by AS "createBy.id", a.create_date AS "createDate.id", a.update_by AS "updateBy.id", a.update_date AS "updateDate.id", a.remarks AS "remarks", a.del_flag AS "delFlag", department.id AS "department.id", department.name AS "department.name", projectManager.id AS "projectManager.id", projectManager.name AS "projectManager.name", projectManager.mobile AS "projectManager.mobile", marketer.id AS "marketer.id", marketer.name AS "marketer.name", marketer.mobile AS "marketer.mobile", customer.id AS "customer.id", customer.code AS "customer.code", customer.name AS "customer.name", push.id AS "push.id", push.projectid AS "push.project.id", push.userid AS "push.user.id", push.is_readed AS "push.isReaded"
FROM (SELECT * FROM ps_project WHERE del_flag = 0 ORDER BY code DESC  limit 0,2 ) a
LEFT  JOIN ps_customer customer ON a.customerid = customer.id
LEFT  JOIN sys_office department ON a.departmentid = department.id
LEFT  JOIN sys_user projectManager ON a.project_managerid = projectManager.id
LEFT  JOIN sys_user marketer ON a.marketerid = marketer.id
LEFT  JOIN ps_project_push push ON a.id = push.projectid 
AND push.del_flag = 0 AND customer.del_flag= 0 AND department.del_flag= 0 AND projectManager.del_flag= 0 AND marketer.del_flag= 0
ORDER BY code DESC