我通过查询得到这样的输出:
ID customer_name_Now customer_name_Before MOVEMENT
123451 Rustle Bock ltd N £2,121
123451 N Rustle Bock ltd -£25,666,899
123452 Little Garage Ltd N £6,987
123453 N The Big Shop £15,850
故事是这样的,我有 2 个月的数据。在这两个月中,客户可能有也可能没有移动,这取决于上个月是我们的客户还是现在的客户。在许多情况下,这两个月都是客户,因此我得到了 2 行,如上所示。
理想的输出应该是:
ID customer_name_Now customer_name_Before MOVEMENT
123451 Rustle Bock ltd Rustle Bock ltd -£25,664,778
123452 Little Garage Ltd N £6,987
123453 N The Big Shop £15,850
因此,移动应该求和以给出月份的实际移动,并且鉴于客户在两个月份都有关系,因此客户的姓名应该在两列中。
@DMK 我用来获取初始输出的查询是:
select /*+ NO_REWRITE */
customer_id,
customer_name_now,
customer_name_before,
movement
from
(select /*+ NO_REWRITE */
main.customer_id,
main.customer_name_now,
main.customer_name_before,
main.limits_before,
main.limits_now,
sum(main.limits_now-main.limits_before) as movement
from
(select /*+ NO_REWRITE */
customer_id,
(customer_name_before) as customer_name_before,
(customer_name_now) as customer_name_now,
sum(limits_current) as limits_now,
sum(limits_previous) as limits_before
from
(select /*+ NO_REWRITE */
sub.customer_id,
sub.customer_name_now,
sub.customer_name_before,
sub.limits_current,
sub.limits_previous
from
(select /*+ NO_REWRITE */
T2.customer_ID,
(T2.customer_name) customer_name_now,
'N' customer_name_before,
sum(T26.AGREED_LIMIT) limits_current,
0 limits_previous
from
DWH_customer_HISTORY T2,
DWH_TIME_DIM T25,
DWH_FACILITY_MONTHLY T2
where
---some internal filters are applied here, i habe ot shown coz of security reasons----
and
T25.MONTH_END = '2012-11-30' and
group by
T2.customer_ID,
T2.customer_name,
) sub
union all
select /*+ NO_REWRITE */
sub.customer_id,
sub.customer_name_now,
sub.customer_name_before,
sub.limits_current,
sub.limits_previous
from
(select /*+ NO_REWRITE */
T2.customer_ID,
'N' as customer_name_now,
(T2.customer_name)customer_name_before,
0 limits_current,
sum(T2.AGREED_LIMIT) limits_previous,
from
DWH_customer_HISTORY T2,
DWH_TIME_DIM T25,
DWH_FACILITY_MONTHLY T2
where
---some internal filters are applied here, i habe ot shown coz of security reasons----
and
T25.MONTH_END = '2012-10-31'
group by
T2.customer_ID,
T2.customer_name,) sub
) un
group by
customer_id,
customer_name_now,
customer_name_before,) main
group by
main.customer_id,
main.customer_name_now,
main.customer_name_before,
main.limits_before,
main.limits_now)
最佳答案
我假设您使用 SqlServer,但下面的查询也适用于 MySql。
Select c1.ID, c1.customer_name_Now, c2.customer_name_Before, Total
from Customers c1
left Join Customers c2
on c2.ID = c1.ID
left join
(select ID as ID2, sum(MOVEMENT) as Total, count(*) as Cnt
from Customers
group by ID) t1
on ID2 = c1.ID
where (c1.customer_name_Now <> 'N' and c2.customer_name_Before <> 'N')
or CNT = 1
如果您不确定,请查看以下演示SqlFiddle
查看您刚刚添加的查询后,上面的内容应该仍然有效。你要么需要
Customers
Customers
我会去第二个。保存重新运行相同的查询。
关于sql - 合并 2 行或更多行有字符和其他有货币,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/13970275/