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 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 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 reaCode 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 ) 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。