问题
In my c# code I am trying to create a zip folder for the user to download in the browser. So the idea here is that the user clicks on the download button and he gets a zip folder.
For testing purpose I am using a single file and zipping it but when it works I will have multiple files.
Here is my code
var outPutDirectory = AppDomain.CurrentDomain.BaseDirectory;
string logoimage = Path.Combine(outPutDirectory, "images\\error.png"); // I get the file to be zipped
HttpContext.Current.Response.Clear();
HttpContext.Current.Response.BufferOutput = false;
HttpContext.Current.Response.ContentType = "application/zip";
HttpContext.Current.Response.AddHeader("content-disposition", "attachment; filename=pauls_chapel_audio.zip");
using (MemoryStream ms = new MemoryStream())
{
// create new ZIP archive within prepared MemoryStream
using (ZipArchive zip = new ZipArchive(ms))
{
zip.CreateEntry(logoimage);
// add some files to ZIP archive
ms.WriteTo(HttpContext.Current.Response.OutputStream);
}
}
When I try this thing it gives me this error
Central Directory corrupt.
[System.IO.IOException] = {"An attempt was made to move the position before the beginning of the stream."}
Exception occurs at
using (ZipArchive zip = new ZipArchive(ms))
Any thoughts?
回答1:
You're creating the ZipArchive
without specifying a mode, which means it's trying to read from it first, but there's nothing to read. You can solve that by specifying ZipArchiveMode.Create
in the constructor call.
Another problem is that you're writing the MemoryStream
to the output before closing the ZipArchive
... which means that the ZipArchive
code hasn't had a chance to do any house-keeping. You need to move the writing part to after the nested using
statement - but note that you need to change how you're creating the ZipArchive
to leave the stream open:
using (MemoryStream ms = new MemoryStream())
{
// Create new ZIP archive within prepared MemoryStream
using (ZipArchive zip = new ZipArchive(ms, ZipArchiveMode.Create, true))
{
zip.CreateEntry(logoimage);
// ...
}
ms.WriteTo(HttpContext.Current.Response.OutputStream);
}
来源:https://stackoverflow.com/questions/33687425/central-directory-corrupt-error-in-ziparchive