First, I want to create three files without actually creating a file like memorystream. I want to compress those three files and put them in a memoeystream. It is possible to put a zip file in memoeystream, but like memorystream, can it actually contain three files without creating a file?
Here is my code,
using (MemoryStream ms = new MemoryStream()) {
using (ZipArchive fileContainer = new ZipArchive(ms, ZipArchiveMode.Create, true)) {
using(MemoryStream fileMS = new MemoryStream()){
//I want to create file to like memorystream, Not local
//file txt1.txt contain 123456789
//file txt2.txt contain 12345
//file txt1.txt contain 6789
}
}
return File(ms.ToArray(), System.Net.Mime.MediaTypeNames.Application.Octet, "Result.zip");
}
Worked fine for me
using (MemoryStream ms = new MemoryStream())
{
using (ZipArchive archive = new ZipArchive(ms, ZipArchiveMode.Create, true))
{
for (int i = 0; i < 3; i++)
{
ZipArchiveEntry readmeEntry = archive.CreateEntry($"text{i}.txt");
using (StreamWriter writer = new StreamWriter(readmeEntry.Open()))
{
writer.WriteLine("text");
}
}
}
return File(ms.ToArray(), System.Net.Mime.MediaTypeNames.Application.Octet, "Result.zip");
}