AnsweredAssumed Answered

Function IF and part type

Question asked by Lorenzo Ponti on Aug 19, 2019
Latest reply on Aug 19, 2019 by Deepak Gupta

Goodmorning,

 

I'm having a problem while trying to create a macro. I need it to add properties to a file depending on the type of the file itself  (If Part/Assembly add "Data Dis", if Drawing add "Data Disegnato") but when I try to use it it gives me back a 424 error. What I'm doing wrong? Thanks in advice for the help

 

Dim retval As String

 

Set swApp = Application.SldWorks

 

MsgBox ("Firmato!")

 

Set swModel = swApp.ActiveDoc

 

retval = swModel.DeleteCustomInfo("Disegnato")
retval = swModel.DeleteCustomInfo("Data Disegnato")
retval = swModel.DeleteCustomInfo("Data Dis")
retval = swModel.DeleteCustomInfo("Dwg Type")

 

retval = swModel.AddCustomInfo3("", "Disegnato", swCustomInfoText, "SIB/LP")
retval = swModel.AddCustomInfo3("", "Dwg Type", swCustomInfoText, "Costruttivo")

 

Set Part.Type = Part.GetType

 

If Part.Type = swDocASSEMBLY Or Part.Type = swDocPART Then

retval = swModel.AddCustomInfo3("", "Data Dis", swCustomInfoText, Date)

End If

 

If Part.Type = swDocDRAWING Then

retval = swModel.AddCustomInfo3("", "Data Disegnato", swCustomInfoText, Date)

End If

 

End Sub

 

EDIT: I would also like to add a line to enter and exit "Modify sheet format" under the Drawing if. How can I do this?

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