Hi,
I want to ZIP one file contained in a PC's local directory.
Can anyone advise how to convert this into cobol?
//Creates a new, blank zip file to work with - the file will be
//finalized when the using statement completes
using (ZipArchive newFile = ZipFile.Open(zipName, ZipArchiveMode.Create))
{
//Here are two hard-coded files that we will be adding to the zip
//file. If you don't have these files in your system, this will
//fail. Either create them or change the file names.
newFile.CreateEntryFromFile(@"C:\\Temp\\File1.txt", "File1.txt");
newFile.CreateEntryFromFile(@"C:\\Temp\\File2.txt", "File2.txt", CompressionLevel.Fastest);
}
Taken from the website:-
www.codeproject.com/.../Creating-Zip-Files-Easily-in-NETHi Neil, should be something like:
declare zipName = "File.dat"
perform using newFile as type ZipArchive = type ZipFile::Open(zipName, type ZipArchiveMode::Create)
*>Here are two hard-coded files that we will be adding to the zip
*>file. If you don't have these files in your system, this will
*>fail. Either create them or change the file names.
invoke newFile::CreateEntryFromFile("C:\\Temp\\File1.txt", "File1.txt");
invoke newFile::CreateEntryFromFile("C:\\Temp\\File2.txt", "File2.txt", type CompressionLevel::Fastest);
end-perform
Hi,
I want to ZIP one file contained in a PC's local directory.
Can anyone advise how to convert this into cobol?
//Creates a new, blank zip file to work with - the file will be
//finalized when the using statement completes
using (ZipArchive newFile = ZipFile.Open(zipName, ZipArchiveMode.Create))
{
//Here are two hard-coded files that we will be adding to the zip
//file. If you don't have these files in your system, this will
//fail. Either create them or change the file names.
newFile.CreateEntryFromFile(@"C:\\Temp\\File1.txt", "File1.txt");
newFile.CreateEntryFromFile(@"C:\\Temp\\File2.txt", "File2.txt", CompressionLevel.Fastest);
}
Taken from the website:-
www.codeproject.com/.../Creating-Zip-Files-Easily-in-NETthanks for that works great.
