Sorry, in relation to this....
"Also, keep in mind that if your results page (2) is
initially being fed by
the results of a FORM and you then, on the link back from
page 3, are trying
to feed it with a QUERYSTRING, you will get an error with the
recordset. If
that's happening, I think you just need to change the METHOD
of the original
form to GET rather than POST."
I should have also mentioned that you'd need to change your
recordset on the
results list page (page 2) from Request.Form("searchfield")
to
Request.QueryString("searchfield")
Doops!
Nath.
"Tag2007" <webforumsuser@macromedia.com> wrote in
message
news:eauant$50e$1@forums.macromedia.com...
> Hi I Tried it and get an error..
>
> I search on page 1
>
> Results list on page 2
>
> Details on page 3
>
> All work, But my new link on page 3 back to page 2 works
But when back on
> page
> 2 if you click a result again to view the details it
errors out saying
>
> Microsoft OLE DB Provider for ODBC Drivers error
'80040e14'
>
> [Microsoft][ODBC Microsoft Access Driver] Syntax error
(comma) in query
> expression 'field_id = 14417, 14417'.
>
>
> Here are my 2 links I tried as suggested.
>
> Link On Results page :page2
> To Detail Page
>
> <A HREF="/sbt_results2.asp?<%=
Server.HTMLEncode(MM_keepURL) &
> MM_joinChar(MM_keepURL) & "searchID=" &
> RS_sbt_results.Fields.Item("field_id").Value %>"
& "searchID2=" &
> Request.Form("txt_sbt")"
class="style4">View</A>
>
>
------------------------------------------------------------------
>
> Back To results
>
>
> <A HREF="/sbt_results1.asp?<%=
Server.HTMLEncode(MM_keepURL) &
> MM_joinChar(MM_keepURL) & "searchID2=" &
Request.Form("txt_sbt")%>">View
> Search
> Results</A>
>
> Thanks again for any help
> Tag
>