Is it possible to Open EasyLabel template using MS Access command button / VBA code.
Using MS Access form ,I want to create a drop down list (i.e. Combo Box : “CmbTemplateName”) of template names that are stored on network path say “T:\Projects\EasylabelTemplates”. If the user selects a template from combo box and click command button (“OpenTemplate”) it should open the specific template using EasyLabel.
Private Sub OpenTemplate_Click()
Dim dbsCurrent As Database
Dim EasyLabelStartPath As String
Dim EasyLabelTemplatePath As String
Dim TemplateName As String
Dim objEasyLabel As EasyLabel.Application
EasyLabelStartPath = “C:\Program Files\Tharo\EASYLABEL Platinum\easy.exe”
EasyLabelTemplatePath = “T:\Projects\EasylabelTemplates”
TemplateName = Me.CmbTemplateName
Set objEasyLabel = CreateObject("EasyLabel.Application")
objEasyLabel.Visible = True
With objEasyLabel
.Activate
‘ Here I need to say the path for the template to open , but I don’t know how
.Template "CmbTemplateName.fmt"
.Refresh
End With
Exit Sub
End Sub