| re: findRecord method
Try
Docmd.GotoControl "Info"
DoCmd.FindRecord FindBox, acStart
where FindBox is where you type your letter (This will only find the first
instance)
Buch better to use a ComboBox and jump to the right record on AfterUpdate
of the Combo
HTH
Phil
"Stu Mac" <wolffcastle@hotmail.com> wrote in message
news:e56b2423.0412140913.4e83e4a@posting.google.co m...[color=blue]
> Hi
>
> I have a table with two fields
>
> Key Info
> 1 a
> 2 b
> 3 c
>
>
> I've created a simple form to display the info and I want to have
> unbound text box and a find action button
>
> I want to enter a value into the text box (key or info) and then click
> the find button to jump the record matching that criteria. I'm trying
> to use the findRecord method as below
>
> DoCmd.FindRecord "b", acAnywhere, False, acSearchAll, False, acAll,
> True
>
> unfortunately I get a "Runtime error 2162 A macro set to one of the
> current fields properties failed because of an error in a FindRecord
> action agument
>
> N.B I'm not using a macro per say i'm using wrting my own action event
> for "on_click"
>
> Any help would be much appreciated. I don't really want to use Access
> existing Find and Replace tool
>
> thanks very much[/color]
| 
