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T-SQL : Variables

Posted on 2011-03-04
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Last Modified: 2012-05-11
Hi !

There's a part of the code from a report a run everyday that I can't understand.

	DECLARE @CRLD CHAR(2) = CHAR(13) + CHAR(10)
	DECLARE @LineBreak VARCHAR(10) = @CRLD + '<br />'

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Can someone explain me the prupose of this.

Thank You !

-M
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Question by:Rubicon2009
8 Comments
 
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Expert Comment

by:amenkes
ID: 35036499
It appears you are storing a variable @CRLD as 2 characters, a carriage return and a line feed.
The 2nd variable @LineBreak adds the first variable plus an HTML break for output to the web.
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Accepted Solution

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tlovie earned 125 total points
ID: 35036521
What it does is declare a @CRLD as a CHAR(2), and assigns the ascii values chr(13) and chr(10) to it (CR + LF)
then it does the same for @LineBreak, it defines it as a varchar(10) and assigns the value of @CRLD concatenated with '<br />'

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Author Comment

by:Rubicon2009
ID: 35036523
Butt how is it possible to store 23 characters in a 2 characters space ?
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Assisted Solution

by:derekkromm
derekkromm earned 125 total points
ID: 35036532
char(13) and char(10) are 1 character each

char(13) represents a carriage return
char(10) represents a line feed

for example, do "select char(70)" in query analyzer and you'll get "F", a single character. 70 is the ascii representation of "F"
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LVL 19

Assisted Solution

by:amenkes
amenkes earned 125 total points
ID: 35036534
It is not 23 characters. CHAR(2) is defining the storage type of 2 characters.

CHAR(13) is the ASCII code for a Carriage return, it is not 13 characters.
CHAR(10) is the ASCII code for Line Feed, not 10 characters.
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LVL 18

Assisted Solution

by:deighton
deighton earned 125 total points
ID: 35036611
CHAR(2) means 'two character variable', but then very confusingly CHAR(13) + CHAR(10) means concatenate ASCII character 13 to ASCII character 10

CHAR has different meaning on each side
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Author Comment

by:Rubicon2009
ID: 35036618
Thank you ! I'm very surprised, I was not expecting this kind of logic at all.
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Author Closing Comment

by:Rubicon2009
ID: 35036642
Thank a lot !
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