Open Form problem

[emad981]

I have two forms
- frmVList with ID (from tblV - Table)
-frmVDetails (with ID as "tblV.ID) as tblV is the table name that contains the ID.
(That is because i made a query between some tables in the control source of [frmVDetails]

I want to click ID in [frmVList] to open [frmVDetails] where ID = tblV.ID

I wrote the code like the following but did not work:

Private Sub ID_Click()
DoCmd.OpenForm "frmVDetails", acNormal, , , acFormReadOnly, , "ID" & Me.ID
         

End Sub

what is the mistake

 

[sneuberg]

You are missing the equals sign. Try

DoCmd.OpenForm "frmVDetails", acNormal, , , acFormReadOnly, , "ID = " & Me.ID

 

[emad981]

I also tried adding = but still the same problem

 

[sneuberg]

I didn't notice that you had the condition in the openargs argument. If you want it as a where condition then try:

DoCmd.OpenForm "frmVDetails", acNormal, , "ID = " & Me.ID, acFormReadOnly

Otherwise the "frmVDetails" will have to have some code to do something with the passed openarg.

 

[emad981]

This also does not work

 

[sneuberg]

Could you upload your database?

 

[emad981]

Thanks for your help actually you did a lot to me. I decided finally to solve my problem by creating one main form which is linked directly to the table and then I added some some forms do [Id] field is unique

Sent from my G-TiDE E77 using Tapatalk