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 odFROM 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 odFROM 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。