Viewing 15 posts - 1,216 through 1,230 (of 3,221 total)
I do not know what response you will receive from users of SSC, but might I suggest you determine where there is a local chapter of SQL Server users, i.e.,...
May 17, 2011 at 7:09 pm
I do not know what response you will receive from users of SSC, but might I suggest you determine where there is a local chapter of SQL Server users, i.e.,...
May 17, 2011 at 7:06 pm
In the business world, I would first consult with qualified lawyers to determine if per chance with a single database, if confidential. or business confidential data was visible by one...
May 17, 2011 at 6:55 pm
For someone to assist you, they will need more information... post the create table statement, your T-SQL statement that is having the problem and the sample data that is giving...
May 16, 2011 at 1:55 pm
I have also used this module in VB5 and simply moved it into VB6 to keep up to date.
What I would suggest to you is to just execute...
May 16, 2011 at 7:09 am
Welsh Corgi (5/15/2011)
Ninja's_RGR'us (5/15/2011)
Welsh Corgi (5/15/2011)
I'm wondering about misuse of the forum.What misuse??
Just forget it
Lets see 677 points as of your post .. next name will occur when you reach...
May 15, 2011 at 7:25 pm
Last time I could find a copy:
NameThreshold Points
Newbie0
Grasshopper10
SSC Rookie25
Valued Member50
SSC Journeyman75
SSC Enthusiastic100
SSC Veteran200
Old Hand 300
SSC Addicted400
Mr. or Mrs....
May 15, 2011 at 5:42 pm
Is this what you are looking fo:
DECLARE @T VARCHAR(12)
SET @T = '9^1234567890'
SELECT CHARINDEX ( '^' ,@T,1) -- just so you understand what this performs
SELECT SUBSTRING(@T,CHARINDEX ( '^' ,@T,1)+1 ...
May 15, 2011 at 10:34 am
It would have been nice had you posted the error message returned when you executed your code, now it is my guess that your connection string is invalid.
I...
May 14, 2011 at 6:12 am
Without further information it is difficult to assist you. Please click on the first link in my signature block and learn how to post table definitions, some sample data,...
May 13, 2011 at 1:42 pm
Use the REPLACE fimction
REPLACE(find, replace, string)
find
Specifies the string that contains the substring to replace all instances of with another.
replace
Specifies the substring to locate.
string
...
May 12, 2011 at 7:29 pm
Try this code to determine if it is what you require.
SELECT
u.name COLLATE Latin1_General_CI_AS AS 'User',
schema_name(o.schema_id) As 'Schema',
o.name COLLATE Latin1_General_CI_AS...
May 12, 2011 at 9:37 am
Nice question - required carefully reading .. but that is how the QOD should .. so again nice question.
May 10, 2011 at 9:14 pm
Jeff Moden (5/10/2011)
Steve Jones - SSC Editor (5/10/2011)
May 10, 2011 at 5:00 pm
Viewing 15 posts - 1,216 through 1,230 (of 3,221 total)