Comments on: SQL Puzzle: Write a Function to count the total number of word

By: VISHNUAUTO

VISHNUAUTO — Fri, 22 Sep 2017 06:14:18 +0000

ALTER FUNCTION [dbo].[fn_GetWordCount] (@InputStringData VARCHAR(100))
RETURNS INT
AS
BEGIN

DECLARE @i INT
DECLARE @Char CHAR(1)
DECLARE @PrevChar CHAR(1)
DECLARE @Counter INT

declare @input_xml xml

select @input_xml = ” + replace(@InputStringData,’ ‘,” ) + ”

SELECT @Counter=count(x.value(‘.’,’nvarchar’))
FROM @input_xml.nodes(‘//*:root/*:x’) y(x)
where y.x.value(‘.’,’nvarchar(100)’) ”

RETURN @Counter
END
GO

By: harini

harini — Mon, 18 Sep 2017 07:31:07 +0000

declare @q varchar(max) = ‘I am Database Engineer and I am the owner of dbrnd.com’;
select len(@q) – len(replace(@q,’ ‘,”)) + 1

By: Anvesh Patel

Anvesh Patel — Tue, 12 Sep 2017 07:46:30 +0000

In reply to Dinesh. Thank you Dinesh!

By: Dinesh

Dinesh — Tue, 12 Sep 2017 07:33:23 +0000

–Method-1

Declare @XML As XML;
Set @XML=”+Replace(‘I am Database Engineer and I am the owner of dbrnd.com’,’ ‘,”)+”
Select Count(r.value(‘.’,’NVarChar(Max)’)) As TotalWordCount
From @XML.nodes(‘t’) As Records(r)

–Method-2

Declare @SQL NvarChar(Max)=’I am Database Engineer and I am the owner of dbrnd.com’
Declare @Len As Int
Declare @Cnt Int
Set @Cnt=0
While Len(@SQL)>0
Begin

Set @Len=(Case CharIndex(‘ ‘,@SQL) When 0 Then Len(@SQL) Else CharIndex(‘ ‘,@SQL)-1 End)

Set @Cnt=@Cnt+1
Set @SQL=(Case Len(@SQL)-@Len When 0 Then NULL Else Right(@SQL,Len(@SQL)-@Len-1) End)

End
Select @Cnt As TotalWordCount
GO