Possible If statement for loading of a form

[computer_man20037]

Hello everyone I have question,

I am trying to go about making a copying button for one of my forms in current database that I am building and I recently went about incorporating a "Copy" button which would take certain fields from one form (Item Form) and copy them over to a new record in a different form (Detail Form).



I got the copy button to open up a "middleman" form which would be like a pop up box to ask them if they are wanting to actually copy over the data or not. Upon doing so and choosing the extra option I require them to choose (type of item drop down combo box). I would then take all the data in there and copy it to the (Detail Form) where it would populate the required fields that I have already filled out on the (Item Form) itself.

Hopefully that is easy to understand a little bit it seems like a lot. my actual question is this up loading up (Detail Form) I want to go about doing a IF statement so that pressing the copy button it will run through the IF statement depending on if its a Yes/No and do accordingly.

My thought about this that if I didn't want to copy populated fields over and wanted to go straight to the (Detail Form) how could I set up my code so that it just open the form and doesn't error out cause the following variable I am using isn't populated and thus would be Null and access pops up a error.

So overall I want to find out if I can go about creating a way that depending on how you load up the form determines via the DoCmd code or just opening it normally from a switchboard or double clicking on the form itself.

Here is my form_load code I have:

 Private Sub Form_Load()
Dim StrYesorNo As String

'This is where I pre-determine the varible for my IF statement'
StrYesorNo = "No"

'This is where I am wanting to pull my variable from ***Only if I am going about using the copy button***'
StrYesorNo = Forms!CopyTestForm2.Form!YesOrNo

'Double check the value for the variable'
MsgBox StrYesorNo

'My IF statement' '<--- want this to only run if I go through the copy button on my (Item Form)'
If StrYesorNo = "Yes" Then
'Me.Item_Desc2.Text = Form!Item_Single_Add!Item_Desc
'Me.Orig_Price_1.Text = Form!Item_Single_Add!Price1
'Me.Orig_Price_2.Text = Form!Item_Single_Add!Price2
'Me.Orig_Price_3.Text = Form!Item_Single_Add!Price3
'Me.Orig_Price_4.Text = Form!Item_Single_Add!Price4
ElseIf StrYesorNo = "No" Then
End If

End Sub

[June7]

I use OpenArgs argument of DoCmd.OpenForm (or OpenReport) to pass a parameter to form/report then If code can refer to the passed value.

DoCmd.OpenForm "formname", , , , , "button" 'or whatever string value you want

Then in the opening form:

If Me.OpenArgs = "button" Then
'do this
End If

[computer_man20037]

Ok, so after a little looking into it and after reading your suggestion. I was able to figure out a way to load up my form correctly and have it load without any errors coming up.

Private Sub Form_Load()
'Way this form is setup to load is that depending on if they copy from Items Table
'or if you just load this form up normally
'Define the variables here:
Dim StrYesorNo As String

'When the form loads we see that it is looking for
'a value for the variable which by defualt is set
'to "No"
StrYesorNo = "No"
'IF statement that defines how the form should load up and if values need to
'be copied over if coming from previous form (Items Table)

If StrYesorNo = "No" Then
'"No" = that it will open the form to see if anything was copied over
'since there isn't it will return a "No" value

DoCmd.OpenForm "CopyTestForm2", acNormal, , , acFormAdd
StrYesorNo = Forms!CopyTestForm2.Form!YesOrNo
DoCmd.Close acForm, "CopyTestForm2", acSaveNo
End If
If StrYesorNo = "Yes" Then
'"Yes" = that you have chosen to go about copying over data
'from the Items form and have chosen the correct publications
'will return a "Yes"
End If
End Sub

Here is my code that I used to get it working and also will marked this forum as solved!