Question

我有一个非常复杂的查询，我已经加入了很多表格来查看

是的报告

我终于有了一个视图report_final

当我这样查询时

SELECT * FROM `reports_final` LIMIT 0,30

我得到的结果，但是当我这样给出时

SELECT count(*) FROM `reports_final`

或

SELECT count(1) FROM `reports_final`

或

SELECT SQL_CALC_FOUND_ROWS * FROM `reports_final` LIMIT 0,30

为了使用

SELECT FOUND_ROWS()

查询挂起！

db optimization and restructuring我知道你会说
但是在这个时间表中这是不可能的

修改

表格结构

NULLDATA    PAIDDATA    nousers     paidstatus  payment_userid  payment_date    endtime     name

这些是这里的领域

实际上这些是被操纵的列说通过做各种if循环和switch case 比如啄掉所需的柱子。所以，我在问是否有任何进一步的相关方法可以获得计数。

好的我会发布查询

CREATE  VIEW `report_final` AS 
select sql_big_result 
`a`.`NULLDATA` AS `NULLDATA`,`a`.`PAIDDATA` AS `PAIDDATA`,
`a`.`nousers` AS `nousers`,

`a`.`paidstatus` AS `paidstatus`,`a`.`payment_userid`,`a`.`payment_date`,
date(`b`.`endtime`) AS `endtime`,
concat_ws(' ',`um`.`user_fname`,`um`.`user_lname`) AS `name`,
`um`.`user_email` AS `user_email`,
 (case `um`.`user_gender` when 0 then 'MALE' else 'FEMALE' end) AS `gender`,
`um`.`user_dob` AS `user_dob`,`um`.`user_country` AS ,`um`.`user_language` ,
`um`.`user_doj` AS `user_doj`,
if(
    (`c`.`payment_amount` is not null),
    if(
        (`c`.`payment_amount` < 9.91),
        if(
            (`c`.`payment_amount` not in (7.9,9.9)),'DISCOUNT','NORMAL'
        ),'GIFTCODE'
    ),
    if(isnull(`b`.`endtime`),'REGISTER','FREE WORK')
) AS `type`,
`c`.`payment_amount`,`bm`.`brand_name` AS `brand_name`,`bu`.`brand_master_id`,
`disc`.`discount_code` AS `discount_code`,`sp`.`programs_id` AS `programs_id` 
from 
`report1` `a` 
left join `report2` `b` on `a`.`payment_userid` = `b`.`user_id` 
left join `report3` `c` on `a`.`payment_userid` = `c`.`payment_userid`  
join `user_master` `um` on (
    `um`.`user_id` = `a`.`payment_userid` 
     and `um`.`user_id` NOT IN(0,1,'NULL','') 
     and `um`.`user_type` = 1
) 
left join `brand_user` `bu` on `bu`.`user_id` = `um`.`user_id`  
left join `brand_master` `b` on `bu`.`brand_master_id`=`b`.`brand_master_id`  
left join `disc_net` `disc` on `disc`.`user_id` = `um`.`user_id`  
left join `sub_prog` `sp` on `sp`.`user_id` = `um`.`user_id`

N.B

report1,report2,report3

是 mysql VIEWS

查询挂起COUNT（）

0 个答案: