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CouinCouin avatar CouinCouin commented on May 27, 2024 1

Hi wqweto,

Thanks for help :)

I though that using memory was nrequiered to push the files to zip into a subfolder of my choice.

Just tested with changes you posted, it looks ok now , all files seems to be in the archive file, and this last is not corrupted.

(Also, mp3 files was just for testing with "big" files instead of several smalls).

Rest to implent creating subfolder when unzipping (in my project), but I think it will be also fine onece done.

Keep you informed :)

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wqweto avatar wqweto commented on May 27, 2024

This works here. Try getting the latest version of the whole repository (all files).

If still not working the you can ZIP your non-working project and attach it here so I can take a peek.

Btw, there is no reason to use memory streams for what you are doing. Just using plain filenames is more efficient in your case.

Instead of

.AddFile pvInitMemStream("C:\TEMP\Ep.5 2017-06-25 (via E90).mp3"), "Ep.5 2017-06-25 (via E90).mp3"

just call

.AddFile "C:\TEMP\Ep.5 2017-06-25 (via E90).mp3", "Ep.5 2017-06-25 (via E90).mp3"

or even shorter

.AddFile "C:\TEMP\Ep.5 2017-06-25 (via E90).mp3"

With CompressArchive call its the same -- no need to output to a memory stream at all, just call it with a string filename like this

bResult = .CompressArchive("C:\TEMP\Podcasts.zip", Level:=lLevel)

Don't fiddle with ReadBinaryFile, WriteBinaryFile and cMemoryStream unless you have good reason to use these i.e. the source of the data to be compressed is not in files but for instance coming from the internet request or you need the output of the compression not in a file but sent to a serial port for instance.

The use-case for in-memory operations (which is what you are using here) is very specific and compressing MP3 files is not one of these.

from ziparchive.

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