• ZippingWithWindowsShell

ZIPping with Windows Shell

... here in Dyalog APL syntax

Suddenly I came in need of zipping files. And what more obvious solution than to use the zip-facilities already present in Windows? In essence it can be accomplished in a 3 (+1 if you don't already have a zip file) step process as follows.

If need be first create an empty zip file, basically a file with these 22 byte values: 80 75 5 6 0 ... 0:

• tn←zip_file_name ⎕NCREATE 0
  (22↑80 75 5 6)⎕NAPPEND tn 83
  ⎕NUNTIE tn
  

Once the zip file is present, create an instance of the Shell COM...

• 'SHAPP'⎕WC'OLEClient' 'Shell.Application'
  

... get a handle to the files, here a folder and the zip archive, calling the NameSpace method...

• ADD←SHAPP.NameSpace⊂add_folder_name
  FILES←ADD.Items
  ZIP←SHAPP.NameSpace⊂zip_file_name
  

... and finally have all files in that folder copied to the ZIP archive,

• ZIP.CopyHere FILES(4+16)
  

where 4 means "don't display a progress dialog box", and 16 "respond with 'Yes to All' for any dialog box that is displayed".

Note that FILES can be a FolderItems object (as here), a FolderItem object (result from the Item method), or a string that represents a file name.