mysql union (all) 后order by的排序失效问题解决

上sql

select * FROM ( 
SELECT 
SUM(c.overtime_num) AS delay_num, 
ROUND((SUM(c.total_num) - SUM(c.overtime_num))*100/SUM(c.total_num),2) rate , 
'全网' as reaCodeFROM calc_vmap_repair_timely_rate_mon_stat c 
WHERE c.`type` = 22 
and c.MONTH BETWEEN '2019-01' AND '2019-01'
) t1 

UNION ALL 

SELECT t2.* FROM ( 
select tmp.* FROM ( 
SELECT SUM(c.overtime_num) AS delay_num, 
ROUND((SUM(c.total_num) - SUM(c.overtime_num))*100/SUM(c.total_num),2) rate , 
c.motorcade_area_code as reaCodeFROM calc_vmap_repair_timely_rate_mon_stat c 
WHERE c.`type` = 22 
and c.MONTH BETWEEN '2019-01' AND '2019-01' 
group by c.motorcade_area_code 
) tmp order by tmp.rate asc 
)  t2 

单独执行union all下面的结果如下：

单独执行union all上面的结果如下：

 

 

为了保证排序不乱，按照网上解决方案：

可是结果竟然还是：

 没能解决问题。

真正解决方案：

SELECT * FROM(

SELECT 
SUM(c.overtime_num) AS delay_num, 
ROUND((SUM(c.total_num) - SUM(c.overtime_num))*100/SUM(c.total_num),2) rate , 
'全网' AS reaCode,
0 AS od
FROM calc_vmap_repair_timely_rate_mon_stat c 
WHERE c.`type` = 22 
and c.MONTH BETWEEN '2019-01' AND '2019-01'

UNION ALL 

SELECT SUM(c.overtime_num) AS delay_num, 
ROUND((SUM(c.total_num) - SUM(c.overtime_num))*100/SUM(c.total_num),2) rate , 
c.motorcade_area_code AS reaCode,
1 AS od
FROM calc_vmap_repair_timely_rate_mon_stat c 
WHERE c.`type` = 22 
and c.MONTH BETWEEN '2019-01' AND '2019-01' 
group by c.motorcade_area_code 

) con ORDER BY od, rate;

先查询后排序

union 前的排序与union 后的顺序，采用加一个字段od来保证，然后再按rate排序则解决了以上的问题。