T-SQL ：计算列中每个唯一子字符串的出现次数

Question

我想计算列中每个唯一子字符串的出现次数。

SELECT DISTINCT LEFT(code, 3) 
FROM table-with-codes

输出：

code
------------------
VJCrandomthings
PASrandomthings
CAArandomthings
PASrandomthings2
PASrandomthings3

预期输出：

caa 1
pas 3
vjc 1

我试过了

SELECT COUNT(DISTINCT LEFT(code, 3)) 
FROM table-with-codes

但结果是 returns 3。

Answer 1

您需要 count 和 group by

select Left(code,3), Count(*)
from [table-with-codes]
group by Left(code,3)

Answer 2

您使用 left 是正确的，但您应该 group by

而不是 distinct

declare @table_with_codes table (code varchar(50))
insert into @table_with_codes values ('VJCrandomthings'), ('PASrandomthings'), ('CAArandomthings'), ('PASrandomthings2'), ('PASrandomthings3')

select left(code, 3) as code_3, 
       count(*) as cnt
from   @table_with_codes 
group by left(code, 3)

这是如何工作的
通过使用 group by 你可以计算所有被分组的行（因此只在结果中返回一次），在这种情况下，所有在 lef(col, 3)

中具有相同值的行

结果会是

code_3  cnt
------  ---
CAA      1
PAS      3
VJC      1

Answer 3

你应该保留select语句和Group by中提取的3个字母。

查询

Select left(code, 3) as code3, count(left(code, 3)) as cnt
From tablename
Group by left(code, 3)
Order by 1;

Answer 4

我不喜欢 LEFT(code, 3) 在查询中出现不止一次。为了提高可维护性和可读性，我将使用 CROSS APPLY 重构它，如下所示：

SELECT subcode, COUNT(*)
FROM table-with-codes
CROSS APPLY (SELECT LEFT(code, 3)) AS T(subcode)
GROUP BY subcode

Answer 5

您可以使用 Common Transaction Expression 来避免两次提及 left 函数：

create table [table-with-codes] (code nvarchar(100))
go
insert into [table-with-codes] values('VJCrandomthings'),('PASrandomthings'),('CAArandomthings'),('PASrandomthings2'),('PASrandomthings3');


with cte as (select LEFT(code,3) mycode from [table-with-codes])
select mycode,count(1) N from cte group by mycode

T-SQL ：计算列中每个唯一子字符串的出现次数

T-SQL : count occurrences per unique substring in a column

sql

tsql

sql-server