Differences between revisions 3 and 4
⇤ ← Revision 3 as of 2008-11-01 18:44:49 →
Size: 1329
Editor: anonymous
Comment:
← Revision 4 as of 2008-11-01 18:45:55 → ⇥
Size: 0
Editor: anonymous
Comment:
Deletions are marked like this. Additions are marked like this.
Line 1: Line 1:
#format wiki
#language en
= ZIPping with Windows Shell =
... in Dyalog APL

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}}}<<BR>>
 {{{(22↑80 75 5 6)⎕NAPPEND tn 83}}}<<BR>>
 {{{⎕NUNTIE tn}}}<<BR>>

Once the zip file is present, create an instance of the Shell COM...
 {{{'SHAPP'⎕WC'OLEClient' 'Shell.Application'}}}<<BR>>

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

 {{{ADD←SHAPP.NameSpace⊂add_folder_name}}}<<BR>>
 {{{FILES←ADD.Items}}}<<BR>>
 {{{ZIP←SHAPP.NameSpace⊂zip_file_name}}}<<BR>>

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

 {{{ZIP.CopyHere FILES(4+16)}}}<<BR>>

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.

ZippingWithWindowsShell (last edited 2017-02-16 19:22:55 by KaiJaeger)