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How open form with certain circumstances

Posted on 2014-01-09
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Last Modified: 2014-01-10
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?
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Question by:SteveL13
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Rey Obrero (Capricorn1) earned 2000 total points
ID: 39769728
if dcount("*", "NameOfTable","PetId=" & me.txtPetID)>0 then

DoCmd.OpenForm "frmRequestForPickup", acNormal, ,"[PetID]=" & me.txtPetID , , acWindowNormal


else

DoCmd.OpenForm "frmRequestForPickup", acNormal, , , acFormAdd, acWindowNormal

End if



if PetID is Text, use this


DoCmd.OpenForm "frmRequestForPickup", acNormal, ,"[PetID]='" & me.txtPetID & "'" , , acWindowNormal
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Expert Comment

by:PatHartman
ID: 39769976
You don't actually need to do the dLookup().  Just open the form normally with a where argument.   If the record is found, the form will open to the found record.  If it wasn't found, the form will open to a new record.  So,
DoCmd.OpenForm "frmRequestForPickup", acNormal, ,"[PetID]=" & me.txtPetID , , acWindowNormal

Should do it.
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