Find the count of words in string - sql

Question

SQL: How to find the count of words in following example?
declare #s varchar(55) = 'How to find the count of words in this string ?'
Subquestions:
How to count spaces?
How to count double/triple/... spaces as one? answer by Gordon Linoff here
How to avoid counting of special characters? Example: 'Please , don't count this comma'
Is it possible without string_split function (because it's available only since SQL SERVER 2016)?
Summary with the best solutions HERE

Answer 1

Thanks to Gordon Linoff's answer here
SELECT len(replace(replace(replace(replace(#s,' ','<>'),'><',''),'<>',' '),' ',','))
OutPut
-------
How,to,find,the,count,of,words,in,this,string?
SELECT replace(replace(replace(replace(replace(#s,' ','<>'),'><',''),'<>',' '),' ',','),',','')
OutPut
------
Howtofindthecountofwordsinthisstring?
Now you can find the difference between the length of both the output and add 1 for the last word like below.
declare #s varchar(55) = 'How to find the count of words in this string?'
SELECT len(replace(replace(replace(replace(#s,' ','<>'),'><',''),'<>',' '),' ',','))
-len(replace(replace(replace(replace(replace(#s,' ','<>'),'><',''),'<>',' '),' ',','),',',''))
+ 1 AS WORD_COUNT
WORD_COUNT
----------
10
http://sqlfiddle.com/#!18/06c1d/5

Answer 2

One method uses a recursive CTE:
declare #s varchar(55) = 'How to find the count of words in this string ?';
with cte as (
select convert(varchar(max), '') as word,
convert(varchar(max), ltrim(#s)) as rest
union all
select left(rest, patindex('%[ ]%', rest + ' ') - 1),
ltrim(stuff(rest, 1, patindex('%[ ]%', rest + ' '), ''))
from cte
where rest <> ''
)
select count(*)
from cte
where word not in ('', '?', ',')
--OPTION (MAXRECURSION 1000); -- use if number of words >99
;
Here is a db<>fiddle.

Answer 3

First thing is you need to remove the double/tripple.. or more count into one.
declare #str varchar(500) = 'dvdv sdd dfxdfd dfd'
select Replace(Replace(Replace( #str,' ',']['), '[]', ''), '][', ' ')
this will remove all the unnecessary space in between the word and you'll get your final word.
After that you may use string_split (for SQL SERVER 2016 and above). To count the number of word in your text from which minus 1 is your total count of spaces.
select count(value) - 1 from string_split( #str, ' ')
Final query looks like
declare #str varchar(500) = 'dvdv sdd dfxdfd dfd'
select count(value) - 1 from string_split( Replace(Replace(Replace( #str,' ',']['), '[]', ''), '][', ' '), ' ')

Answer 4

For only word count and if your MSSQL Version support STRING_SPLIT, you can use this simple script below-
DECLARE #s VARCHAR(55) = 'How to find the count of words in this string ?'
SELECT
COUNT(
IIF(
LTRIM(value)='',
NULL,
1
)
)
FROM STRING_SPLIT(#s, ' ')
WHERE value LIKE '%[0-9,A-z]%'

Answer 5

Using string_split (available only since SQL SERVER 2016):
declare #string varchar(55) = 'How to find the count of words in this string ?';
select count(*) WordCount from string_split(#string,' ') where value like '%[0-9A-Za-z]%'
The same idea is used in following answers:
https://stackoverflow.com/a/57783421/6165594
https://stackoverflow.com/a/57783743/6165594
Without using string_split:
declare #string varchar(55) = 'How to find the count of words in this string ?';
;with space as
( -- returns space positions in a string
select cast( 0 as int) idx union all
select cast(charindex(' ', #string, idx+1) as int) from space
where charindex(' ', #string, idx+1)>0
)
select count(*) WordCount from space
where substring(#string,idx+1,charindex(' ',#string+' ',idx+1)-idx-1) like '%[0-9A-Za-z]%'
OPTION (MAXRECURSION 0);
The same idea is used in following answers:
https://stackoverflow.com/a/57787850/6165594
As Inline Function:
ALTER FUNCTION dbo.WordCount
(
#string NVARCHAR(MAX)
, #WordPattern NVARCHAR(MAX) = '%[0-9A-Za-z]%'
)
/*
Call Example:
1) Word count for single string:
select * from WordCount(N'How to find the count of words in this string ? ', default)
2) Word count for set of strings:
select *
from (
select 'How to find the count of words in this string ? ' as string union all
select 'How many words in 2nd example?'
) x
cross apply WordCount(x.string, default)
Limitations:
If string contains >100 spaces function fails with error:
Msg 530, Level 16, State 1, Line 45
The statement terminated. The maximum recursion 100 has been exhausted before statement completion.
NB! OPTION (MAXRECURSION 0); -- don't work within inline function
*/
RETURNS TABLE AS RETURN
(
with space as
( -- returns space positions in a string
select cast( 0 as int) idx union all
select cast(charindex(' ', #string, idx+1) as int) from space
where charindex(' ', #string, idx+1)>0
)
select count(*) WordCount from space
where substring(#string,idx+1,charindex(' ',#string+' ',idx+1)-idx-1) like #WordPattern
-- OPTION (MAXRECURSION 0); -- don't work within inline function
);
go