MySQL排序不同/唯一值

决定将我的查询分解成若干部分，以帮助传达我正在努力实现的目标

我有一些关于客户和他们使用的优惠券代码的信息，采用这种格式

顾客 优惠码 1. 弗里德尔 1. 弗里德尔 1. 弗里德尔 1. 1562733 1. 8842939 1. 847hr64 1. 83jd63j 1. 弗里德尔 1. 8eh33jr 1. AA-2637 1. AA-9837 1. 节省200 1. 节省200

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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，如下所示：

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;

---

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，如下所示：

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;

---

您可以分别计算：

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

---

请参见

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

---

请参见

我不明白-这将如何计算唯一和非唯一实例的数量？@UsmaanBhatti。您可以将逻辑合并到聚合查询中。我编辑了答案。我不明白-这将如何计算唯一和非唯一实例的数量？@UsmaanBhatti。您可以将逻辑合并到聚合查询中。我编辑了答案。