Access Command to run access, open to specific form and specific record

Oct 3, 2018
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was posted over 10 years ago and it is almost getting me to where I need to go. Was wondering if anyone other there can assist and help me to get this working?

I am trying to generate an HTML email from Access that will include a link to open another Access database to a specific record.

this was my line of code for the outgoing email:

"<TD align=center width=""10%""><a href=""C:\Program Files (x86)\Microsoft Office\root\Office16\MSACCESS.EXE"" ""J:\SW Follow-up.accdb"" ;frmSW,SWID=" & rsBody.Fields![STWID] & """>" & rsBody.Fields![SWID].Value & "</a></TD>" & _

But that just opens MS Access and not the database.

Then I tried:

"<TD align=center width=""10%""><a href=""J:\SW Follow-up.accdb"" ;frmSW,SWID=" & rsBody.Fields![STWID] & """>" & rsBody.Fields![SWID].Value & "</a></TD>" & _

Even though that opens the database, it does not go to the desired record.

I created a module in the destination database - SW Follow-up following the example in the thread I noted above:

Dim varParms As Variant

varParms = Split(Command, ",")
If UBound(varParms) = 1 Then
DoCmd.OpenForm varParms(0), acNormal, , varParms(1), OpenArgs:=True
End If

and called it from an AutoExec Macro. The frmSW is set to be the Display form when the database opens.

What I don't understand is how the SWID gets passed from the email to the OpenArgs for the database so that the form opens on that record. Can anyone help me? Thank you.

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