asked on How open form with certain circumstances
I have a 2nd form which opens from a command button on a 1st form:
DoCmd.OpenForm "frmRequestForPickup", acNormal, , , acFormAdd, acWindowNormal
if the form has never been filled out before.
But if the form has been filled out before I want it to open and show that record which I suppose means it can't open with the acFormAdd in the code.
The way to determine if the form has been filled out before and there is a record entered already is via a field named "PetID". If the table has a record with PetID matching the field txtPetID in form 1 then display the already existing record.
How?
DoCmd.OpenForm "frmRequestForPickup", acNormal, , , acFormAdd, acWindowNormal
if the form has never been filled out before.
But if the form has been filled out before I want it to open and show that record which I suppose means it can't open with the acFormAdd in the code.
The way to determine if the form has been filled out before and there is a record entered already is via a field named "PetID". If the table has a record with PetID matching the field txtPetID in form 1 then display the already existing record.
How?
Microsoft Access
Last Comment
PatHartman
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DoCmd.OpenForm "frmRequestForPickup", acNormal, ,"[PetID]=" & me.txtPetID , , acWindowNormal
Should do it.