50分！sql查询解决思路

50分！！sql查询
select  
count(1) as '北京'
 , sum(TFT_TRNAMT) as 金额
 ,left( TFT_DTE ,6) as month 
 from ABST_TrnFlw group by left( TFT_DTE ,6)
order by month  

select sum(TFT_TRNAMT) as 分发金额
,left( TFT_DTE ,6) as month 
from ABST_TrnFlw where TFT_TRNFIG='1' 
group by left( TFT_DTE ,6)

现在把2个表格利用 month 左连接，
谢谢 帮忙拼起来。

------解决方案--------------------------------------------------------
select * from 
(select
count(1) as '北京' 
, sum(TFT_TRNAMT) as 金额 
,left( TFT_DTE ,6) as month 
from ABST_TrnFlw group by left( TFT_DTE ,6) 
order by month ) a1
left join

(select sum(TFT_TRNAMT) as 分发金额 
,left( TFT_DTE ,6) as month 
from ABST_TrnFlw where TFT_TRNFIG='1' 
group by left( TFT_DTE ,6)) a2 
on a1.month=a2.month
------解决方案--------------------------------------------------------

SQL code

select *from (select count(1) as '北京', sum(TFT_TRNAMT) as 金额 ,left( TFT_DTE ,6) as month from ABST_TrnFlw group by left( TFT_DTE ,6)) a left join (select sum(TFT_TRNAMT) as 分发金额,left( TFT_DTE ,6) as month from ABST_TrnFlw where TFT_TRNFIG='1' group by left( TFT_DTE ,6)) b on a.month=b.month