MySQL对不同/唯一值进行排序

Question

决定将我的查询分解成小块，以帮助传达我想要实现的目标。

我有一些关于客户和他们使用的优惠券代码的信息，格式如下

Customer	Coupon Code
1	FreeDel
1	FreeDel
1	FreeDel
1	1562733
1	8842939
1	847hr64
1	83jd63j
1	FreeDel
1	8eh33jr
1	AA-2637
1	AA-9837
1	Save200
1	Save200

我想对其进行排序：如果已知优惠券代码前缀可用，则使用如下语法：

CASE WHEN Coupon Code LIKE 'AA-%' THEN 'AA-' 优惠券代码唯一的情况，然后是“唯一” 优惠券代码不是唯一的情况，然后是“非唯一”

然后输出

Count	Customer	Coupon Code Type
6	1	Non-Unique
2	1	AA-
5	1	Unique

在这种理想情况下，首先添加已知前缀作为案例，然后将未知的、未重复的优惠券代码标记为唯一，然后将重复的优惠券代码标记为非唯一。

任何帮助将不胜感激！ :)

表格编辑

Answer 1

您可以分别计算它们：

select sum(cnt), customer, 'Non-Unique'
from (
  select customer, coupon_code, count(*) as cnt
  from ccodes
  where coupon_code not like 'AA-%'
  group by customer, coupon_code
  having cnt>1
) as q
group by customer
union
select count(*), customer, 'AA-'
from ccodes
where coupon_code like 'AA-%'
group by customer
union
select sum(cnt), customer, 'Unique'
from (
  select customer, coupon_code, count(*) as cnt
  from ccodes
  where coupon_code not like 'AA-%'
  group by customer, coupon_code
  having cnt=1
) as q
group by customer

见db-fiddle

Answer 2

您可以使用聚合和条件逻辑：

(case when sum( code like 'AA-%' ) > 0 then 'AA-'
      when min(code) = max(code) then 'UNIQUE'
      else 'NON-UNIQUE'
 end)

您可以将其合并到聚合查询中：

select grp, count(*)
from (select c.customer,
             (case when sum( code like 'AA-%' ) > 0 then 'AA-'
                   when min(code) = max(code) then 'UNIQUE'
                   else 'NON-UNIQUE'
              end) as grp,
             count(*)
      from c
      group by customer
     ) x
group by grp;