quote:

---

Originally posted by karthik0805

You can also find count of word in the sentence.
For example:

create proc charcount
@a varchar(8000),@b varchar(8000)
as
declare @i int = 1, @count int =0, @j int=0
select @j=LEN(@b)
while @i<=LEN(@a)
begin
if (select substring(@a,@i,@j))=@b
begin
set @count=@count+1
end
set @i=@i+1
end
select @count
go

EXEC charcount 'SQL DEVELOPERS ARE ONE OF THE BEST DEVELOPERS IN THE WORLD', 'DEVELOPERS'

The output is : 2

---

declare @a varchar(max), @b varchar(max)
set @a = 'SQL DEVELOPERS ARE ONE OF THE BEST DEVELOPERS IN THE WORLD'
set @b = 'DEVELOPERS'

select (len(@a) - len(replace(@a,@b,''))) / len(@b)