使用 System.IO.Packaging 在 C# 中创建 zip

Question

我正在尝试在服务器上动态创建 zip 文件，其中包含来自在线服务的照片。但是当我尝试打开该文件时，WinRar 说该文件格式未知或已损坏。

这是代码：

MemoryStream stream = new MemoryStream();
Package package = ZipPackage.Open(stream, FileMode.Create, FileAccess.ReadWrite);

foreach (Order o in orders)
{
    o.GetImages();

    Parallel.ForEach(o.Images, (Dictionary<string, object> image) =>
    {
        string imageId = (string)image["ID"];
        int orderId = (int)image["OrderId"];
        string fileName = (string)image["FileName"];
        string url = (string)image["URL"];

        if (string.IsNullOrEmpty(fileName))
        {
            System.Net.WebClient wc = new System.Net.WebClient();
            byte[] data = wc.DownloadData(url);
            MemoryStream ms = new MemoryStream(data);
            ms.Write(data, 0, data.Length);

            string ext;

            switch(wc.ResponseHeaders["Content-Type"])
            {
                case "image/jpeg":
                case "image/jpg":
                ext = "jpg";
                    break;

                case "image/png":
                    ext = "png";
                    break;

                case "image/gif":
                    ext = "gif";
                    break;

                default :
                    ext = "un";
                    break;
            }

            Uri uri = new Uri("/" + orderId.ToString() + "/" + imageId + "." + ext, UriKind.Relative);

            var part = package.CreatePart(uri, wc.ResponseHeaders["Content-Type"], CompressionOption.NotCompressed);

            var pstream = part.GetStream(FileMode.Open, FileAccess.ReadWrite);

            pstream.Write(data, 0, data.Length);

            var rel = package.CreateRelationship(part.Uri, TargetMode.Internal, "http://example.com/AlbumImage");


        }
    });
}

package.Flush();

byte[] fileBytes = new byte[stream.Length];
stream.Read(fileBytes, 0, Convert.ToInt32(stream.Length));

Response.Clear();
Response.AddHeader("Content-Disposition", "attachment; filename=Orders.zip");
Response.AddHeader("Content-Length", stream.Length.ToString());
Response.ContentType = "application/octet-stream";
Response.BinaryWrite(fileBytes);
Response.End();

Answer 1

这都是关于帕克·乌里斯的。我们也被他们打过。这是在 zip 中存储两个文件的工作示例。代码刚刚从我们的项目复制，并删除了一些私人数据。

           var dataFilePath = Path.GetFileName(dataFileName);
           var dataFileUri = PackUriHelper.CreatePartUri(new Uri(dataFilePath, UriKind.Relative));
            // Create the Package
            using (var package = Package.Open(filePath, FileMode.Create))
            {
                // Add the diagram view part to the Package
                var pkgPart = package.CreatePart(dataFileUri, MediaTypeNames.Application.Octet);
                var pkgStream = pkgPart.GetStream();

                // Copy the data to the model view part
                // diagramFileName Encoding.Default.GetBytes(text)
                using (var modelStream = new FileStream(dataFileName, FileMode.Open, FileAccess.Read))
                {
                    const int bufSize = 0x1000;
                    var buf = new byte[bufSize];
                    int bytesRead;
                    while ((bytesRead = modelStream.Read(buf, 0, bufSize)) > 0)
                    {
                        pkgStream.Write(buf, 0, bytesRead);
                    }
                }

                // Add a context Part to the Package
                var pkgPartContext = package.CreatePart(ctxUri, MediaTypeNames.Application.Octet);
                var ctxPkgStream = pkgPartContext.GetStream();

                // Copy the data to the context part
                using (var ctxStream = new FileStream(ctxFileName, FileMode.Open, FileAccess.Read))
                {
                    const int bufSize = 0x1000;
                    var buf = new byte[bufSize];
                    int bytesRead;
                    while ((bytesRead = ctxStream.Read(buf, 0, bufSize)) > 0)
                    {
                        ctxPkgStream.Write(buf, 0, bytesRead);
                    }
                }
            }

            // remove tmp files
            File.Delete(ctxFileName);
            File.Delete(dataFileName);

Answer 2

我知道这个问题有点老了，但我也遇到过类似的问题。我用过

stream.Seek(0, SeekOrigin.Begin);

并且有效