Lynn Pettis (7/16/2014)

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Eirikur Eiriksson (7/16/2014)

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Lynn Pettis (7/16/2014)

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TomThomson (7/16/2014)

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Jack Corbett (7/16/2014)

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Steve Jones - SSC Editor (7/16/2014)

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GilaMonster (7/16/2014)

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Jeff Moden (7/15/2014)

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To be honest, I'm glad this person is taking such caution...

Lynn Pettis (7/16/2014)

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TomThomson (7/16/2014)

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Jack Corbett (7/16/2014)

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Steve Jones - SSC Editor (7/16/2014)

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GilaMonster (7/16/2014)

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Jeff Moden (7/15/2014)

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To be honest, I'm glad this person is taking such caution and has such FUD...

Did you try the code from this post?

😎

TomThomson (7/16/2014)

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Welsh Corgi (7/16/2014)

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Tom, thank you very much for your help.

If you were to execute the code that I posted you will find that it does not not store a...

There are some errors in your code, mainly table an column references (non existing table and/or column)

Here is an example using code I pointed to earlier

😎

USE tempdb;

GO

CREATE TABLE dbo.Name_and_SSN

(Full_Name VARCHAR(50),

encodedSSN...

On the line of George Orwell, "one-size-fits-all as not all sizes are equal"

😎

Sean Pearce (7/16/2014)

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I did some investigation and found the problem, XPStar.dll contains a bug.

File Version 2014.120.2000.8

Product Version 12.0.2000.8

Size 409 KB

As a workaround, and the method I used to prove my...

Valbuenito (7/16/2014)

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Hi Eirikur,

Thanks for your reply.

But, I do not think that answers my problem because I don't master on my parameter.

This request comes of Reporting, and the parameter is a...

Quick thought, use the NULLIF function

😎

USE tempdb;

GO

DECLARE @TEST TABLE (STR_VAL VARCHAR(50) NOT NULL);

INSERT INTO @TEST(STR_VAL) VALUES ('ABC'),('DEF'),('NULL'),('JKL'),('NULL');

SELECT

T.STR_VAL

,NULLIF(T.STR_VAL,'NULL') AS STR_VAL_NULL

FROM @TEST T

Results

STR_VAL STR_VAL_NULL

--------...

Hmm, 4,85 x 10^14, that would be one row for each mile of blood vessel in every human on earth:w00t:

😎

Here is an example of encryption - decryption of an XML node, the principle is the same for a column.

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Quick suggestion, add the .value() method to get the correct output

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select

mult1.VisitID

,mult1.AbstractID

,(select

Response + ','

from

[livedb_daily].[dbo].[AbsProjectsQueriesMultCs] mult2

where

mult2.VisitID=mult1.VisitID

and mult2.AbstractID=mult1.AbstractID

order by

mult2.VisitID

,mult2.AbstractID

for xml path (''), TYPE).value('.[1]','NVARCHAR(MAX)') as overall

from

[livedb_daily].[dbo].[AbsProjectsQueriesMultCs] mult1

where

mult1.VisitID='RA0-20120603025910735'

and mult1.[Query]='Additional Notes'

group by

mult1.VisitID

,mult1.AbstractID

This should get you started, find where lines drawn between every other point defining the circle crosses

😎

USE tempdb;

GO

DECLARE @CIRCLE GEOMETRY = 'CIRCULARSTRING (52.6107417597068 -1.19053984363516, 52.6107417597068 -1.19185949047026, 52.6114844449159 -1.19185949047026, 52.6114844449159 -1.19053984363516,...

panneermca35 (7/15/2014)

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Hi All,

Please read the following query.

DECLARE @MyXML XML

SET @MyXML = '<Item>

<Item accountnumber="900044010163" versionnumber="1" repaymentdate="2013-09-05" />

</Item>'

DECLARE @i INT

EXEC sp_xml_preparedocument @i OUTPUT, @MyXML

select * from OPENXML(@i, '/Item/Item') order...

Simplest method is to use STBuffer, the parameter passed is effectively the radius.

😎

DECLARE @CIRCLE GEOMETRY = geometry::Point(1.00,1.00,0).STBuffer(1);

SELECT @CIRCLE