getting file storage path in the XML output

jmcmanus

was told to create a data field and name it storage path, and that there was a expression to use that would put the path of the file in the XML output.    can anyone point me to that expression, can't seem to find it.   thx. 

GrooperGuru

All documents imported from file shares will maintain a link back to the original filename and path of the imported file. If you right click on an imported document within Grooper and examine the Item Properties, you'll see the Content Link.



Now, in order to populate this into a field in Grooper, you'll need to add a Data Field to the Data Model and set the Field's Default Value Expression to: Link.Path
This will populate the field with the complete file path of the imported file.

tmartin

Our customer actually needs their Export directory path in their xml file. I can manually build this path in the expression mentioned above and I can name the file by the content name. An issue occurs however, when there are two "like" content types in the same batch. The "Name Uniquely" feature in Document Export takes care of the image naming problem but the xml file does not reflect this as this field is populated at extraction.

I say all that to give some background, but the customer has said that they don't care what the file name is as long as it is unique. I tried using the Guid but I can't seem to be able to find the correct item to place into the expression to return the guid. Is there a keyword or variable that remains unique(sequence number or DOCID) can be used by the expression and Document Export?

GrooperGuru

In a Default Value Expression, Folder.Id will return the guid of the document.