Format Result Of Sp_helpdb

Jul 23, 2005

Hello.

How to format field "create" (Jan 12 2005) in sp_helpdb
procedure to sth like yyyy-mm-dd (2005-01-12) in SQL?

bye...

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Sep 5, 2005

Hugh writes "Hi,
I'm using this in my query:

SELECT CAST(lotw AS varchar(5)) + ' x ' + CAST(lotl AS varchar(5)) + ' / ' + CAST(lotsq AS varchar(6)) AS lotsqft FROM testTable

lotw, lotl, lotsq are all of type int. Is it possible for sql server to format them in a number format:

LIke This
------------------------
lotsqft
------------------------
400 x 50 / 20,000
50 x 50 / 2,500
2,000 x 1,850 /3,700,000
========================


NOT LIke This
------------------------
lotsqft
------------------------
400 x 50 / 20000
50 x 50 / 2500
2000 x 1850 /3700000
========================

Thanx for your help, I do hope I did not stump you, lol.

-hugH"

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Sp_helpdb

Jul 9, 2002

Has anyone every seen this error message? If so, what caused it?

Server: Msg 515, Level 16, State 2, Procedure sp_helpdb, Line 53
Cannot insert the value NULL into column '', table ''; column does not allow nulls. INSERT fails.
The statement has been terminated.

Thanks

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Now when the user searchers for a  book with a keyword, how can I display the results which should show:
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Feb 16, 2005

i use sp_helpdb on one of my databases everyday to estimate growth.
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Feb 18, 2005

Well i have been using sp_helpdb 'mydatabasename' to get the info for 3 consecutive days now and it returns the same result every day in KB for data files and in MB overall database size. Wich is kinda impossible since this is a warehouse and there are about 30000 records insert every night what is going on. plz help

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Dec 6, 2005

Hi,
I have MOM 2005 installed to monitor all our sql servers. In one of the sql server on a specific db, I get error alerts when running the sp_helpdb error.

The error output is

"Server: Msg 515, Level 16, State 2, Procedure sp_helpdb, Line 53
Cannot insert the value NULL into column '', table ''; column does not allow nulls. INSERT fails.
The statement has been terminated."

I checked the sp_helpdb stored procedure and here is where it fails:

insert into #spdbdesc (dbname, owner, created, dbid, cmptlevel)
select name, suser_sname(sid), convert(nvarchar(11), crdate), dbid, cmptlevel from master.dbo.sysdatabases
where (@dbname is null or name = @dbname)

I understand this is a temporary table created by the stored procedure to insert all the db info.

I ran a query as follows:

select name, suser_sname(sid), crdate, dbid, cmptlevel
from master.dbo.sysdatabases

The output is

DB Name User Name Cr date db id cmpt
master sa2000-08-06 01:29:12.250180
model sa2000-08-06 01:40:52.437380
msdb sa2000-08-06 01:40:56.810480
RMSummaryNULL2002-06-13 16:40:32.203880
tempdb sa2005-11-29 19:10:48.450280

The problem is in RMSummary database and the "dbo" login does not
have a username.

How do I add a User Name "sa" to the DB Owner "dbo" for the RMSummary database?

I did try the following things with no success:

1. Try to delete the 'dbo' user in 'RMSummary' database so that
I can add a new login name as 'dbo' and username is 'sa'.
Error message is 'dbo' owns some objects and does not get deleted.
2. I tried sp_changedbowner but it does not work either.

Any help on this quickly is much appreciated.

Thanks
Murali

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Hey,

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How can I query the result of this stored procedure?

Thx

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When I run sp_helpdb against the master (or any other DB for that matter) I get the following error:

Server: Msg 515, Level 16, State 2, Procedure sp_helpdb, Line 53
Cannot insert the value NULL into column 'owner', table 'tempdb.dbo.#spdbdesc_ 0010001A6EF'; column does not allow nulls. INSERT fails. The statement has been terminated.

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UDF_table.sql:

USE AdventureWorks;

GO

IF OBJECT_ID(N'dbo.ufnGetContactInformation', N'TF') IS NOT NULL

DROP FUNCTION dbo.ufnGetContactInformation;

GO

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(

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ContactID int PRIMARY KEY NOT NULL,

FirstName nvarchar(50) NULL,

LastName nvarchar(50) NULL,

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)

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@FirstName nvarchar(50),

@LastName nvarchar(50),

@JobTitle nvarchar(50),

@ContactType nvarchar(50);

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SELECT

@ContactID = ContactID,

@FirstName = FirstName,

@LastName = LastName

FROM Person.Contact

WHERE ContactID = @ContactID;

SELECT @JobTitle =

CASE

-- Check for employee

WHEN EXISTS(SELECT * FROM HumanResources.Employee e

WHERE e.ContactID = @ContactID)

THEN (SELECT Title

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INNER JOIN Person.ContactType ct

ON vc.ContactTypeID = ct.ContactTypeID

WHERE vc.ContactID = @ContactID)

-- Check for store

WHEN EXISTS(SELECT * FROM Sales.StoreContact sc

INNER JOIN Person.ContactType ct

ON sc.ContactTypeID = ct.ContactTypeID

WHERE sc.ContactID = @ContactID)

THEN (SELECT ct.Name

FROM Sales.StoreContact sc

INNER JOIN Person.ContactType ct

ON sc.ContactTypeID = ct.ContactTypeID

WHERE ContactID = @ContactID)

ELSE NULL

END;

SET @ContactType =

CASE

-- Check for employee

WHEN EXISTS(SELECT * FROM HumanResources.Employee e

WHERE e.ContactID = @ContactID)

THEN 'Employee'

-- Check for vendor

WHEN EXISTS(SELECT * FROM Purchasing.VendorContact vc

INNER JOIN Person.ContactType ct

ON vc.ContactTypeID = ct.ContactTypeID

WHERE vc.ContactID = @ContactID)

THEN 'Vendor Contact'

-- Check for store

WHEN EXISTS(SELECT * FROM Sales.StoreContact sc

INNER JOIN Person.ContactType ct

ON sc.ContactTypeID = ct.ContactTypeID

WHERE sc.ContactID = @ContactID)

THEN 'Store Contact'

-- Check for individual consumer

WHEN EXISTS(SELECT * FROM Sales.Individual i

WHERE i.ContactID = @ContactID)

THEN 'Consumer'

END;

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BEGIN

INSERT @retContactInformation

SELECT @ContactID, @FirstName, @LastName, @JobTitle, @ContactType;

END;

RETURN;

END;

GO

----------------------------------------------------------------------
I executed it in my SQL Server Management Studio Express and I got: Commands completed successfully. I do not know where the result is and how to get the result viewed. Please help and advise.

Thanks in advance,
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Code Snippet
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