A Function Gotcha with the Use of an Asterisk
This article describes and demonstrates the problem of using asterisk to select all fields from a table within a table function. (from Apr 2008)
2009-09-11 (first published: 2008-04-30)
18,716 reads
This article describes and demonstrates the problem of using asterisk to select all fields from a table within a table function. (from Apr 2008)
2009-09-11 (first published: 2008-04-30)
18,716 reads
In an interesting exercise in data conversion, Stephen Lasham brings us a nice article on basic manipulations.
2008-04-25 (first published: 2007-05-17)
13,032 reads
Do you use identity values? Do missing values ever cause problems? Maybe you need to look for missing invoices? Stephen Lasham brings us a technique for finding those missing values in a sequence.
2006-04-03
14,744 reads
Selecting a menu structure from a series of tables is pretty easy. If you bound the number of levels in the menu. However if you have potentially variable numbers of levels. Stephen Lasham brings us a new piece on using recursion to select the data.
2005-10-11
8,827 reads
T-SQL can be a complex language to work within and code solutions with. Author Stephen Lasham brings us a twist on his last solution of extracting a string from within another string. Read about his method of working with strings in T-SQL.
2004-09-13
7,863 reads
String manipulation in SQL Server or T-SQL can be a challenage. There are multiple ways to accomplish every task and often learning a new one may spark some creativity in yourself or teach you a better way to accomplish an old task. New Author Stephen Lasham brings us his technique for extracting a string that is delimited from within a larger string. Something I'm sure we'll all have use for in the future.
2004-06-14
23,404 reads
By Steve Jones
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In SQL Server 2025, I have a table (dbo.UserPermission) that contains this data:
UserID UserPermissions 15 23 37What is returned when I run this code:
select bit_count(UserPermissions) as PermissionCount from dbo.UserPermission where UserID = 3;See possible answers