[sql-server] How do you count the number of occurrences of a certain substring in a SQL varchar?

I have a column that has values formatted like a,b,c,d. Is there a way to count the number of commas in that value in T-SQL?

I finally write this function that should cover all the possible situations, adding a char prefix and suffix to the input. this char is evaluated to be different to any of the char conteined in the search parameter, so it can't affect the result.

CREATE FUNCTION [dbo].[CountOccurrency]
(
@Input nvarchar(max),
@Search nvarchar(max)
)
RETURNS int AS
BEGIN
    declare @SearhLength as int = len('-' + @Search + '-') -2;
    declare @conteinerIndex as int = 255;
    declare @conteiner as char(1) = char(@conteinerIndex);
    WHILE ((CHARINDEX(@conteiner, @Search)>0) and (@conteinerIndex>0))
    BEGIN
        set @conteinerIndex = @conteinerIndex-1;
        set @conteiner = char(@conteinerIndex);
    END;
    set @Input = @conteiner + @Input + @conteiner
    RETURN (len(@Input) - len(replace(@Input, @Search, ''))) / @SearhLength
END 

usage

select dbo.CountOccurrency('a,b,c,d ,', ',')

Perhaps you should not store data that way. It is a bad practice to ever store a comma delimited list in a field. IT is very inefficient for querying. This should be a related table.

Darrel Lee I think has a pretty good answer. Replace CHARINDEX() with PATINDEX(), and you can do some weak regex searching along a string, too...

Like, say you use this for @pattern:

set @pattern='%[-.|!,'+char(9)+']%'

Why would you maybe want to do something crazy like this?

Say you're loading delimited text strings into a staging table, where the field holding the data is something like a varchar(8000) or nvarchar(max)...

Sometimes it's easier/faster to do ELT (Extract-Load-Transform) with data rather than ETL (Extract-Transform-Load), and one way to do this is to load the delimited records as-is into a staging table, especially if you may want an simpler way to see the exceptional records rather than deal with them as part of an SSIS package...but that's a holy war for a different thread.

You can compare the length of the string with one where the commas are removed:

len(value) - len(replace(value,',',''))

In SQL 2017 or higher, you can use this:

declare @hits int = 0
set @hits = (select value from STRING_SPLIT('F609,4DFA,8499',','));
select count(@hits)

this T-SQL code finds and prints all occurrences of pattern @p in sentence @s. you can do any processing on the sentence afterward.

declare @old_hit int = 0
declare @hit int = 0
declare @i int = 0
declare @s varchar(max)='alibcalirezaalivisualization'
declare @p varchar(max)='ali'
 while @i<len(@s)
  begin
   set @hit=charindex(@p,@s,@i)
   if @hit>@old_hit 
    begin
    set @old_hit =@hit
    set @i=@hit+1
    print @hit
   end
  else
    break
 end

the result is: 1 6 13 20

If we know there is a limitation on LEN and space, why cant we replace the space first? Then we know there is no space to confuse LEN.

len(replace(@string, ' ', '-')) - len(replace(replace(@string, ' ', '-'), ',', ''))

You can use the following stored procedure to fetch , values.

IF  EXISTS (SELECT * FROM sys.objects 
WHERE object_id = OBJECT_ID(N'[dbo].[sp_parsedata]') AND type in (N'P', N'PC'))
    DROP PROCEDURE [dbo].[sp_parsedata]
GO
create procedure sp_parsedata
(@cid integer,@st varchar(1000))
as
  declare @coid integer
  declare @c integer
  declare @c1 integer
  select @c1=len(@st) - len(replace(@st, ',', ''))
  set @c=0
  delete from table1 where complainid=@cid;
  while (@c<=@c1)
    begin
      if (@c<@c1) 
        begin
          select @coid=cast(replace(left(@st,CHARINDEX(',',@st,1)),',','') as integer)
          select @st=SUBSTRING(@st,CHARINDEX(',',@st,1)+1,LEN(@st))
        end
      else
        begin
          select @coid=cast(@st as integer)
        end
      insert into table1(complainid,courtid) values(@cid,@coid)
      set @c=@c+1
    end

Quick extension of cmsjr's answer that works for strings with more than one character.

CREATE FUNCTION dbo.CountOccurrencesOfString
(
    @searchString nvarchar(max),
    @searchTerm nvarchar(max)
)
RETURNS INT
AS
BEGIN
    return (LEN(@searchString)-LEN(REPLACE(@searchString,@searchTerm,'')))/LEN(@searchTerm)
END

Usage:

SELECT * FROM MyTable
where dbo.CountOccurrencesOfString(MyColumn, 'MyString') = 1

Building on @Andrew's solution, you'll get much better performance using a non-procedural table-valued-function and CROSS APPLY:

SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
/*  Usage:
    SELECT t.[YourColumn], c.StringCount
    FROM YourDatabase.dbo.YourTable t
        CROSS APPLY dbo.CountOccurrencesOfString('your search string',     t.[YourColumn]) c
*/
CREATE FUNCTION [dbo].[CountOccurrencesOfString]
(
    @searchTerm nvarchar(max),
    @searchString nvarchar(max)

)
RETURNS TABLE
AS
    RETURN 
    SELECT (DATALENGTH(@searchString)-DATALENGTH(REPLACE(@searchString,@searchTerm,'')))/NULLIF(DATALENGTH(@searchTerm), 0) AS StringCount

for SQL Server 2017

declare @hits int = 0;
select @hits = count(*) from STRING_SPLIT('F609,4DFA,8499',',')
select @hits;

The answer by @csmjr has a problem in some instances.

His answer was to do this:

Declare @string varchar(1000)
Set @string = 'a,b,c,d'
select len(@string) - len(replace(@string, ',', ''))

This works in most scenarios, however, try running this:

DECLARE @string VARCHAR(1000)
SET @string = 'a,b,c,d ,'
SELECT LEN(@string) - LEN(REPLACE(@string, ',', ''))

For some reason, REPLACE gets rid of the final comma but ALSO the space just before it (not sure why). This results in a returned value of 5 when you'd expect 4. Here is another way to do this which will work even in this special scenario:

DECLARE @string VARCHAR(1000)
SET @string = 'a,b,c,d ,'
SELECT LEN(REPLACE(@string, ',', '**')) - LEN(@string)

Note that you don't need to use asterisks. Any two-character replacement will do. The idea is that you lengthen the string by one character for each instance of the character you're counting, then subtract the length of the original. It's basically the opposite method of the original answer which doesn't come with the strange trimming side-effect.

Accepted answer is correct , extending it to use 2 or more character in substring:

Declare @string varchar(1000)
Set @string = 'aa,bb,cc,dd'
Set @substring = 'aa'
select (len(@string) - len(replace(@string, @substring, '')))/len(@substring)

The following should do the trick for both single character and multiple character searches:

CREATE FUNCTION dbo.CountOccurrences
(
   @SearchString VARCHAR(1000),
   @SearchFor    VARCHAR(1000)
)
RETURNS TABLE
AS
   RETURN (
             SELECT COUNT(*) AS Occurrences
             FROM   (
                       SELECT ROW_NUMBER() OVER (ORDER BY O.object_id) AS n
                       FROM   sys.objects AS O
                    ) AS N
                    JOIN (
                            VALUES (@SearchString)
                         ) AS S (SearchString)
                         ON
                         SUBSTRING(S.SearchString, N.n, LEN(@SearchFor)) = @SearchFor
          );
GO

---------------------------------------------------------------------------------------
-- Test the function for single and multiple character searches
---------------------------------------------------------------------------------------
DECLARE @SearchForComma      VARCHAR(10) = ',',
        @SearchForCharacters VARCHAR(10) = 'de';

DECLARE @TestTable TABLE
(
   TestData VARCHAR(30) NOT NULL
);

INSERT INTO @TestTable
     (
        TestData
     )
VALUES
     ('a,b,c,de,de ,d e'),
     ('abc,de,hijk,,'),
     (',,a,b,cde,,');

SELECT TT.TestData,
       CO.Occurrences AS CommaOccurrences,
       CO2.Occurrences AS CharacterOccurrences
FROM   @TestTable AS TT
       OUTER APPLY dbo.CountOccurrences(TT.TestData, @SearchForComma) AS CO
       OUTER APPLY dbo.CountOccurrences(TT.TestData, @SearchForCharacters) AS CO2;

The function can be simplified a bit using a table of numbers (dbo.Nums):

   RETURN (
             SELECT COUNT(*) AS Occurrences
             FROM   dbo.Nums AS N
                    JOIN (
                            VALUES (@SearchString)
                         ) AS S (SearchString)
                         ON
                         SUBSTRING(S.SearchString, N.n, LEN(@SearchFor)) = @SearchFor
          );

Declare @MainStr nvarchar(200)
Declare @SubStr nvarchar(10)
Set @MainStr = 'nikhildfdfdfuzxsznikhilweszxnikhil'
Set @SubStr = 'nikhil'
Select (Len(@MainStr) - Len(REPLACE(@MainStr,@SubStr,'')))/Len(@SubStr)

Declare @string varchar(1000)

DECLARE @SearchString varchar(100)

Set @string = 'as as df df as as as'

SET @SearchString = 'as'

select ((len(@string) - len(replace(@string, @SearchString, ''))) -(len(@string) - 
        len(replace(@string, @SearchString, ''))) % 2)  / len(@SearchString)

DECLARE @records varchar(400)
SELECT @records = 'a,b,c,d'
select  LEN(@records) as 'Before removing Commas' , LEN(@records) - LEN(REPLACE(@records, ',', '')) 'After Removing Commans'

The Replace/Len test is cute, but probably very inefficient (especially in terms of memory). A simple function with a loop will do the job.

CREATE FUNCTION [dbo].[fn_Occurences] 
(
    @pattern varchar(255),
    @expression varchar(max)
)
RETURNS int
AS
BEGIN

    DECLARE @Result int = 0;

    DECLARE @index BigInt = 0
    DECLARE @patLen int = len(@pattern)

    SET @index = CHARINDEX(@pattern, @expression, @index)
    While @index > 0
    BEGIN
        SET @Result = @Result + 1;
        SET @index = CHARINDEX(@pattern, @expression, @index + @patLen)
    END

    RETURN @Result

END

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