Question

我需要编写一个查询，告诉我有多少（COUNT）个唯一客户购买了$ 200以上的商品。我的表有一个customer_id，purchase_id和total。 purchase_id对于每一行都是唯一的。 Customer_id可以重复，也可以有空值。

“唯一客户被认为是唯一的customer_id或空的customer_id。因此在列表中：

1
2
1
null
null
null

将有5个唯一客户。

Answer 1

此SQL应该给您结果

select (select count(distinct customer_id)) + 
       (select count(*) from a where customer_id is null)
  from your table

Answer 2

这可以做到：

select
  count(distinct customer_id) + 
  count(case when customer_id is null then 1 end) as counter
from tablename 
where total > 200

第一个count（）将对所有不同的非null customer_id进行计数，第二个count（）将对空值进行计数。

Answer 3

您需要进行过滤器内部查询并使用union all。 Distinct仅在客户ID不为null时获取一次。另一方面，您将每个空值都计为单独的客户。下面的查询将为您提供帮助。

select count(*) from (
   select distinct customerid 
    from table_name 
   where customerid is not null 
         and total > 200 
    union all 
   select customerid 
    from table name 
    where customerid is null 
      and total>200)

Answer 4

SELECT (
(SELECT COUNT DISTINCT customer_id FROM TABLE where total > 200 and customer_id IS NOT NULL) +
(SELECT COUNT customer_id FROM TABLE where total > 200 and customer_id IS NULL)
)

Answer 5

您可以使用简单的SQL来做到这一点。

SELECT customer_id, count(*) AS count
FROM t1
WHERE total > 200
GROUP BY customer_id

或者获得简单的唯一客户数...

SELECT count(*) AS totalUniqueCustomers
FROM (
  SELECT customer_id
  FROM t1
  WHERE total > 200
  GROUP BY customer_id
) s1

SQL Fiddle

MS SQL Server 2017架构设置：

CREATE TABLE t1 (purchase_id int IDENTITY, customer_id int, total int);

INSERT INTO t1 (customer_id, total)
VALUES 
      (1, 202)
    , (2, 250)
    , (1, 10)
    , (null, 1000)
    , (null, 20)
    , (null, 500)
    , (3, 10)
;
/* NOTE: I added orders with both over and under the 200 mark to show excluding 
   the ones you don't want. */

查询1 ：

SELECT customer_id, count(*) AS count
FROM t1
WHERE total > 200
GROUP BY customer_id

Results ：

| customer_id | count |
|-------------|-------|
|      (null) |     2 |
|           1 |     1 |
|           2 |     1 |

查询2 ：

/* TO GET A SIMPLE COUNT OF UNIQUE CUSTOMERS */
SELECT count(*) AS totalUniqueCustomers
FROM (
  SELECT customer_id
  FROM t1
  WHERE total > 200
  GROUP BY customer_id
) s1

Results ：

| totalUniqueCustomers |
|----------------------|
|                    3 |

Answer 6

只需：

select count(distinct customer_id) + sum(case when customer_id is null then 1 else 0 end)
from t;

count(distinct)计算不同的非NULL值的数量。第二个增加了NULL的数量。

在某些数据库中，您可以简化第二个表达式。例如，在MySQL中：

select count(distinct customer_id) + sum( customer_id is null )
from t;

SQL将所有空值计数为不同

6 个答案: