Question

Given a specific word pattern (say, “balloon”), I would like to find the number of n words before and after, group by them, with a count that exist in the title of my table

For, example if the data set was:

• red balloon sky
• yellow balloon sky road
• blue balloon chair

I’d like the results to be something like:

- red balloon | 1
- yellow balloon | 1
- blue balloon | 1
- balloon sky | 2
- balloon chair | 1

I figured the best way to accomplish this would be with regex in my sproc. So, I added the great regex functions listed here, and the FindWordsInContext function.

To start with:

WITH Words_CTE (Title)
AS
-- Define the CTE query.
(
    SELECT Title
    FROM ItemData
    WHERE Title LIKE '%balloon%'
)
-- Define the outer query referencing the CTE name.
SELECT Title
FROM Words_CTE

So I figured I would start with that and work the FindWordsInContext function into the mix, then do a grouping on the words/before a given word.

— UPDATE —

Thanks to Adrian Iftode below… but the code doesn’t exactly do what I’m looking for.

declare @table table(Sentence varchar(250))

insert into @table(sentence)
    values ('I have another red balloon in the car.'),
            ('Here is a new balloon for you.'),
            ('A red balloon is in the other room.'),
            ('Is there another balloon for me?')


select TOP(5) SentencePart, NumberOfWords
from @table
cross apply dbo.fnGetPartsFromSentence(Sentence, 'balloon') f
order by
  NumberOfWords DESC,
  case when f.Side = 'R' then 0
  else 1 end

Outputs:

balloon is in the other room.       5
I have another red balloon          4
Here is a new balloon               4
Is there another balloon            3
balloon in the car.                 3

I would like to be able to set the range on either side of “balloon”. In this case, let’s say one word, the output should be:

red balloon      2
new balloon      1
another balloon  1
balloon in       1
balloon for      2
balloon is       1

Answer 1

Is a bit a lot of code, I’ll try to explain

First I used a split function, is going to split a varchar by a given varchar

CREATE FUNCTION [dbo].[fnSplitString](@str NVARCHAR(MAX),@sep NVARCHAR(MAX))
RETURNS TABLE
AS
RETURN
    WITH a AS(
        SELECT CAST(0 AS BIGINT) AS idx1,
               CHARINDEX(@sep,@str) idx2, 
               1 as [Level]
        UNION ALL
        SELECT idx2 + coalesce(nullif(LEN(@sep),0),1),
               CHARINDEX(@sep,@str, idx2 + 1), 
               [Level] + 1 as [Level]
        FROM a
        WHERE idx2 > 0
    )
    SELECT SUBSTRING(@str,idx1,COALESCE(NULLIF(idx2,0),LEN(@str)+1)-idx1) AS Value, 
           [Level], 
           case when idx1 = 0 then 'R' when idx2 != 0 then 'LR' else 'L' end as Side
    FROM a  

Given the varchar ‘red balloon sky’ and when the split is the space character it will output :

select *
from dbo.fnSplitString('red balloon sky', ' ')

Value   Level   Side
red      1       R
balloon  2       LR
sky      3       L

The Side part means: if R then the space is on the right side of the word, if L then the space is on the left side of the word and if LR then the word is surrounded by spaces.

When the split is ‘balloon’

select *
from dbo.fnSplitString('red balloon sky', 'balloon')

red     1   R
 sky    2   L

So the balloon appears on the right side of red and appears on the left side of sky

Having this helpful function I created another function which will output the required format for a single sentence (varchar)

create FUNCTION [dbo].[fnGetPartsFromSentence](@sentence NVARCHAR(MAX),@word NVARCHAR(MAX))
RETURNS TABLE
AS
RETURN


with RawData as
(select rtrim(ltrim(f.Value)) as LR, 
       (select COUNT (*) from dbo.fnSplitString(rtrim(ltrim(f.Value)), ' ')) as NumberOfWords,
       f.Side,
       0 as SideLevel
from dbo.fnSplitString(@sentence, @word) as f
where f.Side = 'R' or f.Side = 'L'
union all
(
    select rtrim(ltrim(f.Value)) as LR, 
       (select COUNT (*) from dbo.fnSplitString(rtrim(ltrim(f.Value)), ' ')) as NumberOfWords,
       f.Side,
       sl.no as SideLevel
    from dbo.fnSplitString(@sentence, @word) as f
    join (select 1 as no union all select 2) sl on 1 = 1
    where f.Side = 'LR'
)
)
select (case when Side = 'R' then LR + ' ' + @word 
             when Side = 'L' then @word + ' ' + LR
             when Side = 'LR' then  
                    (
                        case when SideLevel  = 1 then @word + ' ' + LR
                        when SideLevel  = 2 then LR + ' ' + @word 
                        end
                    )
            end) as SentencePart,
        (case when Side = 'R' or Side = 'L' then Side
              else           
                   (    case when SideLevel  = 1 then 'L'
                        when SideLevel  = 2 then 'R'
                        end
                    )
            end) as Side,
        NumberOfWords           
from RawData

This function uses the previous one. First it splits the sentence by word and counts the words in the splits by doing another split by space. When a word appears on both sides of the split, it repeats the split (that join with 1, 2 values).

This function will also output split concatenated with the word, depending on which side it is: left, right or both. It will also output the Side, this time is Left or Right.

select *
from [dbo].[fnGetPartsFromSentence]('yellow balloon sky road','balloon')

SentencePart        Side    NumberOfWords

yellow balloon      R           1
balloon sky road    L           2

Now using this function I can cross apply it with a table

declare @table table(Sentence varchar(250))

insert into @table(sentence)
    values ('red balloon sky'),
            ('yellow balloon sky road'),
            ('blue balloon chair')


select SentencePart, NumberOfWords
from @table
cross apply dbo.fnGetPartsFromSentence(Sentence, 'balloon') f
order by
  case when f.Side = 'R' then 0
  else 1 end

The output is

red balloon           1
yellow balloon        1
blue balloon          1
balloon chair         1
balloon sky road      2
balloon sky           1

Works and when there are multiple occurrences