Comments on: SQL SERVER – Interesting Interview Questions – Part 2 – Puzzle

By: Rakesh Shrivastava

Rakesh Shrivastava — Tue, 25 Aug 2009 07:25:41 +0000

select p.name from Personcolors p
left outer join SelectedColors s on p.colorcode=s.colorcode
group by p.name
having count(s.colorcode)=(select count(d.colorcode) from SelectedColors d)

By: Atin Srivastava

Atin Srivastava — Mon, 10 Aug 2009 11:02:32 +0000

How about this–
What I noticed some blogs above given by bloggers doenst have dynamic queries means their queries will fail if we insert or update more data in either tables… Queries should be general that goes write no matter what data table contains

select name from personcolors a, selectedcolors b where a.colorcode=b.colorcode
group by [name] having count(*)= (select count(*) from selectedcolors)

By: JOBY CHERIYAN

JOBY CHERIYAN — Tue, 09 Jun 2009 12:59:00 +0000

Guys,

My Advice for you all / Correct me if I’m wrong
———————————————————
1. Join Query will have Perfomance issue when data is huge.
2. If you use COUNT(*) this also Affect the Query perfomace.
3. name is already a SQL key word so use Tbl.[name] to improve perfomance insted just use name in a query.

Query can be like this
—————————-

SELECT P.[name] FROM PersonColors P, SelectedColors C
WHERE P.colorcode = C.colorcode
GROUP BY P.[name] HAVING count(P.[name]) >= (SELECT count(c2.colorcode) FROM SelectedColors c2)

By: JOBY CHERIYAN

JOBY CHERIYAN — Tue, 09 Jun 2009 12:44:55 +0000

SELECT P.[name] FROM PersonColors P, SelectedColors C
WHERE P.colorcode = C.colorcode
GROUP BY P.[name] HAVING count(P.[name]) >= (SELECT count(c2.colorcode) FROM SelectedColors c2)

By: Mike

Mike — Mon, 09 Mar 2009 15:09:27 +0000

Using the INNER JOIN this one is only processing PersonColor records that meet the SelectedColors criteria. You don’t necessarily need the DISTINCT keyword, but depending on the source of your data, you never know if you’re going to get duplicates in your table (no constraints on the table DDL).

SELECT Name FROM PersonColors A
INNER JOIN SelectedColors B ON A.ColorCode = B.ColorCode
GROUP BY Name
HAVING COUNT(DISTINCT A.ColorCode) = (SELECT COUNT(DISTINCT ColorCode) FROM SelectedColors)

By: Uday Chowdary

Uday Chowdary — Sat, 14 Feb 2009 06:31:16 +0000

Hi Pinal

Here is my kind of solution:

SELECT NAME
FROM
PERSONCOLORS PC,SELECTEDCOLORS SC
WHERE PC.COLORCODE=SC.COLORCODE GROUP BY PC.NAME HAVING COUNT(DISTINCT(PC.COLORCODE))>=(SELECT COUNT(*) FROM SELECTEDCOLORS)

By: A.A

A.A — Sat, 31 Jan 2009 21:56:22 +0000

select distinct name from personcolors p where not exists ( select colorcode from selectedcolors where colorcode not in
(select colorcode from personcolors p2 where p.Name=p2.name)
)

Another long solution

By: Hema

Hema — Fri, 09 Jan 2009 17:03:41 +0000

Or even this one

declare @ln_count int
set @ln_count = (select count(*) from #tabb)
select name, count(*)
from #taba
group by name
having count (*) >= @ln_count

By: Hema

Hema — Fri, 09 Jan 2009 15:14:16 +0000

Hi Pinal,

How about

select name, count(*)
from #personcolors
group by name having count(*) >= 3

since I know there is a maximum of 3 colors in the table SelectedColors

By: Jerry

Jerry — Mon, 05 Jan 2009 09:24:10 +0000

Hi Pinal,

I am a beginner in SQL Server 2005.
Kindly suggest me the way of improving the same as I am very much interested in SQL server.

Regards,
Jerry

By: pallavi

pallavi — Sat, 27 Dec 2008 07:31:45 +0000

select distinct(name) from
PersonColors a
where (select count(colorcode) from PersonColors b where a.name=b.name)>=(select count(colorcode) from SelectedColors)

By: Scott K

Scott K — Thu, 11 Dec 2008 20:07:29 +0000

I came up with the following:

SELECT Name
FROM dbo.PersonColors P
LEFT JOIN dbo.SelectedColors S
ON P.ColorCode = S.ColorCode
GROUP BY Name
HAVING COUNT(*) =
(
SELECT DISTINCT COUNT(*)
FROM dbo.SelectedColors
)

Then I read the below comments. I was happy to note that Tejas Shah was close to your answer.

I will say this took me more than 3 minutes to answer, but I am happy that I completed the exercise.

By: santosh

santosh — Thu, 11 Dec 2008 10:54:44 +0000

Good post pinal
thanks once again

By: Gajanan

Gajanan — Thu, 11 Dec 2008 05:40:29 +0000

For duplicate entries we just need to change the having clause in tejas’ query

HAVING COUNT(DISTINCT pc.ColorCode) >= (SELECT COUNT(ColorCode) FROM SelectedColors)

By: mumu

mumu — Thu, 11 Dec 2008 05:01:59 +0000

FYI – Tejas’s SQLSvr 2005 script left out a ‘Tom’, ‘Blue’.

By: Tejas Shah

Tejas Shah — Thu, 11 Dec 2008 04:46:23 +0000

Hi CB,

Thanks for this great solution.

I missed this situation, thanks for this solution. your solution is very correct.

Tejas

By: Ajay Sinha

Ajay Sinha — Thu, 11 Dec 2008 04:24:48 +0000

Hi Pinal
I solved this problem within 5 mins.answer given below
select [Name] ,count(color_code) from
(select [name],color_main1.color_code from color_main inner join color_main1
on color_main.color_code=color_main1.color_code) cat1
group by [Name]
having count(color_code)=(select count(color_code) from color_main1)

By: CB

CB — Thu, 11 Dec 2008 01:39:15 +0000

The only drawback to Tejas’ solution is that you must ensure that duplicate entries in the PersonColor table are prevented.

See what happens if you add another row to the table with the values ‘Mike’, ‘Red’. Mike will now appear in the result query because his count is 3, even though he only has 2 colors.

To address that issue, this seemed to work:

SELECT [Name]
FROM
(select distinct [name],[colorcode] from PersonColors) pc
INNER JOIN SelectedColors sc ON pc.ColorCode = sc.ColorCode
GROUP BY [Name]
HAVING COUNT(*) = (SELECT COUNT(*) FROM SelectedColors)

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Thu, 11 Dec 2008 01:31:06 +0000

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By: Gabriel Rodriguez

Gabriel Rodriguez — Thu, 11 Dec 2008 00:20:28 +0000

The ones that use

COUNT(ColorCode) >= (SELECT COUNT(*) FROM SelectedColors)

are probably wrong. THe reason is because there is no validation to make sure that 3 of the COlorCodes that the person has are all in the SelectedColors table.

I mean, the results are coming up correct, but if you update the ColorCodes for the persons that have all three colors to have ColorCodes that are not in the SelectedColors table, the results will still show and one condition that Pinal specified was that they had to have all the colors in the other table.

again this is a good excercise.