How to Use the Seek Method on Linked Tables 2

[easycode]

Hi, All
I am trying to use the seek method on a linked tables, but is not working and gives me this weird message:
"Current provider does not support the necessary interface for index functionality"
The database worked fine before, i use the database splitter utility.
Here is the code


Function validatelogin(localuserid As String, floginpas) As Boolean
Dim rst As ADODB.Recordset

Set rst = New ADODB.Recordset
rst.CursorLocation = adUseServer

rst.Open "users", CurrentProject.Connection, adOpenKeyset, adLockPessimistic, adCmdTableDirect
rst.Index = "PrimaryKey"
rst.Seek localuserid, adSeekFirstEQ

Troubleshooting the procedure i got the error message in the last line of the instructions above.

Any help would be appreciated.
thanks

 

[jebry]

Hi!

I don't think you can use the seek method with linked tables because Access will open them as dynasets and not tables. You need to use the FindFirst method.

hth


Jeff Bridgham
Purdue University
Graduate School
Data Analyst

 

[RoyVidar]

It doesn't work on linked tables...

Either rewrite to use the .Find method, or you could open a connection on the other db, where it should work OLE DB Provider for Microsoft Jet

Roy-Vidar

 

[easycode]

Thanks for replying,

I just finish coding a few days ago, it took me more than 6 months to build the software and changing all the seek instructions to a find it's gonna take me a while, do you guys thing is there any other way to solve this problem?.

Otherwise i thing i might have to avoid doing the split and just use one database as front-back-end.

Thanks again

 

[RoyVidar]

Reread my answer...

Roy-Vidar

 

[RoyVidar]

It's either
- .Find
- "unsplit"
- or...

[tt] Dim rst As ADODB.Recordset
dim cn as adodb.connection
set cn = new adodb.connection
cn.open <string from the link>

Set rst = New ADODB.Recordset
rst.CursorLocation = adUseServer
rst.Open "users", cn, adOpenKeyset, adLockPessimistic, adCmdTableDirect

rst.Index = "PrimaryKey"
rst.Seek localuserid, adSeekFirstEQ[/tt]

as far as I know...

Roy-Vidar