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VB- ADO Returns a recordcount of -1

Posted on 2003-02-24
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Last Modified: 2010-05-01
This should be a really easy one. I have the following code:

SQLConnect = "Provider=SQLOLEDB;Data Source=CLUSTERNAME\CLUSTERINSTANCE;Initial Catalog=MyTable;Integrated Security=SSPI;"
Conn1.ConnectionString = SQLConnect
Conn1.Open
Cmd1.ActiveConnection = Conn1
Cmd1.CommandText = "Select MyColumn from MyTable where Need2Publish = 1"
Set Rs1 = Cmd1.Execute
Rs1.MoveFirst
i = 1
x = 1
intCount = Rs1.RecordCount
MsgBox intCount

This always produces a RecordCount of -1, even though there are 2 records? I know for a fact there are two records because if I do this:

intCount = 2
If intCount > 0 Then
    ReDim strVarName(intCount)
    Do Until Rs1.EOF
        strVarName(i) = Rs1.Fields(0).Value
        MsgBox (strVarName(i))
        Rs1.MoveNext
        i = i + 1
    Loop
End If

It returns my 2 values in the message box...

Does anyone have an explanation to this?

TIA,

Owen
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Question by:eojhan
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3 Comments
 
LVL 18

Expert Comment

by:Sethi
ID: 8012910
This is becuase you are not specifying the type of recordset that you want to open. ADO by default will open a Forward only recordset and it will always return -1. Open a dynamic or keyset recordset and it will give you a recordcount. Another way of getting the total no of records is by opening another recordset:
strSQL="Select Count(*) from tblTableName"
rsGetData.Open
0
 
LVL 70

Accepted Solution

by:
Éric Moreau earned 200 total points
ID: 8013096
Do not use a Command object for a simple query like this. Conisder this:

SQLConnect = "Provider=SQLOLEDB;Data Source=CLUSTERNAME\CLUSTERINSTANCE;Initial Catalog=MyTable;Integrated Security=SSPI;"
Conn1.ConnectionString = SQLConnect
Conn1.Open
set rs1 = new adodb.recordset
rs1.cursorlocation = aduseclient
rs1.open "Select MyColumn from MyTable where Need2Publish = 1", conn1
msgbox Rs1.RecordCount

0
 

Author Comment

by:eojhan
ID: 8017328
Why does aduseclient work and aduseserver not? Why is this the only way to get at the recordcount property?
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