Re: SQL Timeout Error


Good morning Robert,

I read the messages from the eggheadcafe link that you posted below, and I
have all the settings (Execution time-out = 0 and Set Lock Timeout = -1
millseconds) that should have been the solutions. Unfortunately, I am still
getting error when I run the View, but I do not get an error when I open that
View in Access as linked table. I am able to open it. That is what puzzles
me.

I am getting error on opening View because of the # of records in the source
table? The source table has over 10 million records. I think it is due to
the # of records because I just ran a View on a source table with much fewer
records than the first one, and I got the results with no error.

Thanks.

"Robert Lakinski" wrote:

Hi Accesshelp,

Now I understand what you are trying to achieve.

I'm not MS Access expert, but you can access data in Sql server via MS
Access client application in two ways:
- By linking tables. This approach is not limited to linking Sql Server
views as a linked tables in MS Access. You can also link Sql Server tables
as a linked tables in MS Access.
- By using pass-through queries ( using T-Sql ) . This approach is
usually more efficient.

You can also check these guides:
http://www.mssqltips.com/tip.asp?tip=1480
http://www.mssqltips.com/tip.asp?tip=1482


Regarding the problem related to your initial error while opening the view
in Sql Server Management Studio Express,
I have no idea what the problem could be - Tools/ Options / .....
Execution Timeout = 0 means 'wait forever'.
But it seems that someone else has also experienced similar problem.

Check this thread:

http://www.eggheadcafe.com/conversation.aspx?messageid=33740922&threadid=33740922


Robert

"Accesshelp" <Accesshelp@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:554730B6-B934-41D1-8960-E81C256ECC60@xxxxxxxxxxxxxxxx
Robert,

First of all, I want to thank you for continuing to help me.

Secondly, I want to use View as a Select query to access data in SQL from
Access. From what I learned, I can not link between SQL and Access with
regular SQL Select query. On the other hand, I can link between SQL and
Access via View to access data.

I just linked in Access to the View that I created in SQL via ODBC (as a
linked table in Access), and I was able to open that linked table in
Access
and retrieve data. Somehow, if I try to open the View in SQL, I got that
error message.

I don't know why I got the error message.

Thanks.

"Robert Lakinski" wrote:

AccessHelp,

If you are NOT trying to access the data in the view via your own
developed
..Net application (using VS or other tool), ignore my posts.

Robert

"Robert Lakinski" <rlakinski@xxxxxxxxx> wrote in message
news:uyHMRzB9JHA.4560@xxxxxxxxxxxxxxxxxxxxxxx

Thanks for your response. How can I check whether SqlConnection
object
is
estiablished

By debugging your .net client application. Can you post some code?
Which version of Visual Studio are you using ?
What kind of data access objects are you using: SqlDataSource,
SqlDataAdapter or SqlCommand, linq ...?

Robert Lakinski


, and how do I set the CommandTimeout property? I am sorry I am
still new to SQL Server.

Thanks.

"Robert Lakinski" wrote:

Hi, AccessHelp

If you are using .net client app to select data from the view, check
the
following:

1. Whether SqlConnection object establishes connection to the sql
server
successfully - check ConnectionString, ConnectionTimeout property
(default
= 15 sec)
2. If the connection was established, you may need to set the
CommandTimeout
property of the SqlCommand object to a value > 30 sec (default = 30
sec)

Robert Lakinski

"Accesshelp" <Accesshelp@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:8C19B329-0148-4EDE-A1EC-A2E3A3674C47@xxxxxxxxxxxxxxxx
Good morning all,

I just created a View, and when I tried to run it, I received the
following
error message:

error message: timeout expired. the timeout elapsed prior to
completion
of
the operation or the server is not responding.

error source: .Net SqlClient Data Provider

When I researched the above error message via Google, the solution
is
to
run
the CommandTimeout, and I don't know how to run it.

First of, if the CommandTimeout is the solution to the error, can
you
tell
me how to run it?

If the CommandTimeout is not the solution, can you help me with the
error?

Thanks.







.

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