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Kristen
Test

United Kingdom
22415 Posts

Posted - 09/23/2006 :  03:15:38  Show Profile  Reply with Quote
Going back to your original query this would work, because it is a boolean test (but it is probably solving a different problem to the one you wanted):

declare @table table(ad_num varchar(20), ad_str1 varchar(20))

insert @table
select '20', '20 apple avenue' union all
select '20', 'apple avenue'

select * from @table

if (EXISTS(select * from @table where substring(ad_str1, 1, 1) like '[0-9]'))

print 'At least one address starts with a number'
else
print 'None of the addresses starts with a number'

Kristen
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