特定字符在字符串中出现的次数

try that :

declare @t nvarchar(max)
set @t='aaaa'


select len(@t)-len(replace(@t,'a',''))

There's no direct function for this, but you can do it with a replace:

declare @myvar varchar(20)
set @myvar = 'Hello World'


select len(@myvar) - len(replace(@myvar,'o',''))

Basically this tells you how many chars were removed, and therefore how many instances of it there were.

Extra:

The above can be extended to count the occurences of a multi-char string by dividing by the length of the string being searched for. For example:

declare @myvar varchar(max), @tocount varchar(20)
set @myvar = 'Hello World, Hello World'
set @tocount = 'lo'


select (len(@myvar) - len(replace(@myvar,@tocount,''))) / LEN(@tocount)

You can do that using replace and len.

Count number of x characters in str:

len(str) - len(replace(str, 'x', ''))

Look at the length of the string after replacing the sequence

declare @s varchar(10) = 'aabaacaa'
select len(@s) - len(replace(@s, 'a', ''))
>>6

function for sql server:

CREATE function NTSGetCinC(@Cadena nvarchar(4000), @UnChar nvarchar(100))
Returns int


as


begin


declare @t1 int


declare @t2 int


declare @t3 int


set @t1 = len(@Cadena)


set @t2 = len(replace(@Cadena,@UnChar,''))


set @t3 = len(@UnChar)




return (@t1 - @t2)  / @t3


end

Code for visual basic and others:

Public Function NTSCuentaChars(Texto As String, CharAContar As String) As Long


NTSCuentaChars = (Len(Texto) - Len(Replace(Texto, CharAContar, ""))) / Len(CharAContar)


End Function

Use this function begining from SQL SERVER 2016

Select Count(value) From STRING_SPLIT('AAA AAA AAA',' ');


-- Output : 3

When This function used with count function it gives you how many character exists in string

Use this code, it is working perfectly. I have create a sql function that accept two parameters, the first param is the long string that we want to search into it,and it can accept string length up to 1500 character(of course you can extend it or even change it to text datatype). And the second parameter is the substring that we want to calculate the number of its occurance(its length is up to 200 character, of course you can change it to what your need). and the output is an integer, represent the number of frequency.....enjoy it.

CREATE FUNCTION [dbo].[GetSubstringCount]
(
@InputString nvarchar(1500),
@SubString NVARCHAR(200)
)
RETURNS int
AS
BEGIN
declare @K int , @StrLen int , @Count int , @SubStrLen int
set @SubStrLen = (select len(@SubString))
set @Count = 0
Set @k = 1
set @StrLen =(select len(@InputString))
While @K <= @StrLen
Begin
if ((select substring(@InputString, @K, @SubStrLen)) = @SubString)
begin
if ((select CHARINDEX(@SubString ,@InputString)) > 0)
begin
set @Count = @Count +1
end
end
Set @K=@k+1
end
return @Count
end

You can do it inline, but you have to be careful with spaces in the column data. Better to use datalength()

SELECT
ColName,
DATALENGTH(ColName) -
DATALENGTH(REPLACE(Col, 'A', '')) AS NumberOfLetterA
FROM ColName;

-OR- Do the replace with 2 characters

SELECT
ColName,
-LEN(ColName)
+LEN(REPLACE(Col, 'A', '><')) AS NumberOfLetterA
FROM ColName;

It will count occurrences of a specific character

DECLARE @char NVARCHAR(50);
DECLARE @counter INT = 0;
DECLARE @i INT = 1;
DECLARE @search NVARCHAR(10) = 'o'
SET @char = N'Hello World';
WHILE @i <= LEN(@char)
BEGIN
IF SUBSTRING(@char, @i, 1) = @search
SET @counter += 1;


SET @i += 1;
END;


SELECT @counter;