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。 加上limit问题也是可以解决的。

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。