^ : T-SQL

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^ : T-SQL

VM-723206

SSCrazy

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November 19, 2009 at 10:07 pm

#228041

Comments posted to this topic are about the item ^ : T-SQL

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Bhavesh_Patel SSCrazy Points: 2259 More actions · Answer 1

This was easy one. You can find more about wildcard use with LIKE search condition at below mentioned link:

http://bhaveshgpatel.wordpress.com/2009/10/03/sql-server-select-statement-with-where-like-clause/

[font="System"]Bhavesh Patel[/font]

http://bhaveshgpatel.wordpress.com/

rajaneeshk Old Hand Points: 349 More actions · Answer 2

There are no start and end for that regular expression na... so the answer should be three..???

Jamie-2229 SSCrazy Eights Points: 8156 More actions · Answer 3

I took the approach that the answer was easy. Not a wise approach. Good subject.

Jamie

ronmoses@gmail.com SSCarpal Tunnel Points: 4480 More actions · Answer 4

I got tripped up on the fact that the string didn't end with %. So I answered the question for 'Jo[^n]%' instead of 'Jo[^n]'. Oops.

-----
a haiku...

NULL is not zero
NULL is not an empty string
NULL is the unknown

daniel.gardiner Old Hand Points: 305 More actions · Answer 5

I answered correctly but didn't get the point, the answer page tells me that the correct selection is '1' because 'Joe' is the only match (which is the fourth selection when I look at the list)

ronmoses@gmail.com SSCarpal Tunnel Points: 4480 More actions · Answer 6

daniel.gardiner (11/20/2009)
I answered correctly but didn't get the point, the answer page tells me that the correct selection is '1' because 'Joe' is the only match (which is the fourth selection when I look at the list)

The query is pulling a count of matches, not the id value.

-----
a haiku...

NULL is not zero
NULL is not an empty string
NULL is the unknown

daniel.gardiner Old Hand Points: 305 More actions · Answer 7

ronmoses (11/20/2009)
daniel.gardiner (11/20/2009)
I answered correctly but didn't get the point, the answer page tells me that the correct selection is '1' because 'Joe' is the only match (which is the fourth selection when I look at the list)
The query is pulling a count of matches, not the id value.

D'oh! Focused on the regex and not enough on the sql.

jcrawf02 SSC-Insane Points: 24198 More actions · Answer 8

ronmoses (11/20/2009)
I got tripped up on the fact that the string didn't end with %. So I answered the question for 'Jo[^n]%' instead of 'Jo[^n]'. Oops.

Thank you ron, I did the same thing, but got confused for a second and thought that I had been using this incorrectly the entire time, and that the brackets were somehow referring to any character, not just the third. Your response clarified it for me.

Good simple question, keeping us in the details. 😀

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"stewsterl 80804 (10/16/2009)I guess when you stop and try to understand the solution provided you not only learn, but save yourself some headaches when you need to make any slight changes."

paul.knibbs SSCoach Points: 15320 More actions · Answer 9

ronmoses (11/20/2009)
I got tripped up on the fact that the string didn't end with %. So I answered the question for 'Jo[^n]%' instead of 'Jo[^n]'. Oops.

Same mistake here...

Elly Odell SSC Enthusiast Points: 172 More actions · Answer 10

does this change from SQL2000 to SQL2005/8? when i try this in my SQL2000 if i leave the % out of the select- i get zero rows. needless to say when I try the test question in my database only the ones that have an 'n' in the third position are dropped so the answer appears to be three. and according the MSDN site it says:

The following example uses the [^] operator to find all the people in the Contact table who have first names that start with Al and have a third letter that is not the letter a.

Copy Code USE AdventureWorks;

GO

SELECT FirstName, LastName

FROM Person.Contact

WHERE FirstName LIKE 'Al[^a]%'

ORDER BY FirstName;

not trying to be difficult- trying to understand what i am missing.. thanks

rajaneeshk Old Hand Points: 349 More actions · Answer 11

Now is that confirmed....

May i know what the answer is...???

If it is 3.... then atleast say thanks na.... 🙂

Alan. T. Hall of Fame Points: 3221 More actions · Answer 12

Good question; I had forgotten about all of the pattern matching built into LIKE. I'll have to try to use it more.

JF1081 SSC Eights! Points: 982 More actions · Answer 13

I almost slected 4 (Joe's ID) then I noticed the * and realized there was only one column in the answer. I had to go back and read the SQL again to catch the count() function. I have learned to think twice about my answers to these questions.

Good question.

kramaswamy SSCoach Points: 18181 More actions · Answer 14

rajaneeshk (11/20/2009)
Now is that confirmed....
May i know what the answer is...???
If it is 3.... then atleast say thanks na.... 🙂

The answer is 1 - there is only one record which matches the LIKE clause.