SQL Server 2000 SP4.

Running the script below prints 'Unexpected':

-----------------------------
DECLARE @String AS varchar(1)

SELECT @String = 'z'

IF @String LIKE '[' + CHAR(32) + '-' + CHAR(255) + ']'
PRINT 'Expected'
ELSE
PRINT 'Unexpected'
-----------------------------

If the @String variable is set to 'y' (or in fact any ANSI character other
than 'z'), the result is 'Expected'. The comparison also evaluates as
expected if CHAR(255) is replaced with CHAR(254). The server collation, if
that matters, is SQL_Latin1_General_CP1_CI_AS.

It would be helpful to find the explanatin of this behavior. Thanks.

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Dimitri Furman wrote:

CHAR(255) = 'y' with an umlaut
http://englishplus.com/xascii.htm

Dimitri Furman (dfurman (AT) cloud99 (DOT) net) writes:
I suspect that this is a case of wrong expectations. When you have a
range in a pattern, it is not defined from ASCII codes, but from the
collation. Only if the collation is a binary collation, character codes
apply. As Ed said char(255) is ÿ, so z is definitely outside the range.

This prints "Expected":

DECLARE @String AS varchar(1)

SELECT @String = 'z'

IF @String COLLATE Latin1_General_BIN
LIKE '[' + CHAR(32) + '-' + CHAR(255) + ']'
PRINT 'Expected'
ELSE
PRINT 'Unexpected'


--
Erland Sommarskog, SQL Server MVP, esquel (AT) sommarskog (DOT) se

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