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Command string for opening a folder

Question asked by Dave LeCount on May 9, 2016
Latest reply on Aug 17, 2016 by Levi David

So I have a folder in my Vault titled ECN (full path is C:\Vault\ECN).  In this folder will eventually be Excel files for new Engineering Change Notices.  Also in this folder is a sub-folder called "ECN Reference Documents" (full path is C:\Vault\ECN\ECN Reference Documents).  My template is set up so that when a new ECN is created (1234.XLSX), a folder named "1234" will be created under "ECN Reference Documents" to contain any reference files (this folder would appear as C:\Vault\ECN\ECN Reference Documents\1234).


I'd like to create a button on the data card for the Excel file so that when it is clicked on, it opens up the associated reference folder.  So Excel file C:\Vault\ECN\1234.XLSX would have a button to open location C:\Vault\ECN\ECN Reference Documents\1234.  I've created a button set up to run this command:


That will get me as far as opening folder C:\Vault\ECN\ECN Reference Documents.  But I cannot figure out a way to get one more level down to C:\Vault\ECN\ECN Reference Documents\1234.  I do have a variable called ECN Number that is the ECN Number of 1234, is there any way to call this out at the tail end of my string to give me the full path that I want?