Run-time error "13" Type mismatch (1 Viewer)

P

Pavczech

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• Nov 13, 2012

• #1

Hi to all

I have trouble with this line in my code.

rs.FindFirst "[Surname] = " & Var(Nz(Me![Text185], 0))


I think it is because it is not number but text.

Could somebody help

Thanks

 

gemma-the-husky

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• Nov 13, 2012

• #2

not sure about "var" but you need the search string wrapped in "" chars.
i normally use chr(34) for this so you get


rs.FindFirst "[Surname] = " & chr(34) & nz([Text185], "") & chr(34)

note also that as text185 is text, the nz result must also be text, rather than zero as you had.

in fact maybe you ought to test for a valid string

Code:Copy to clipboard

if nz(text185,"")="" then
    msgbox("No search string")
    exit sub
else
[B]    rs.FindFirst "[Surname] = " & chr(34) & [Text185] & chr(34)[/B]
[B]    etc[/B]
end if

 

P

Pavczech

Registered User.

Local time

Today, 09:45

Joined

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Messages

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• Nov 13, 2012

• #3

Hi ya

i have just putted it in quotatins and took some out it works

rs.FindFirst "[Surname] = '" & Me![Text185]&"'"


Thanks for your reply