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我正在使用 system.io.packaging.package 创建一个包含一些文件的包。目的是创建导入/导出功能。

我这样创建包：

        // Create the compressed file.
        using (FileStream outFile = File.Create(this.packageName))
        {
            using (Package Compress = Package.Open(outFile, FileMode.Create))
            {
                foreach (string file in Directory.GetFiles(Path.GetDirectoryName(this.packageName)))
                {
                    FileInfo fi = new FileInfo(file);

                    // Prevent compressing hidden and already compressed files.
                    if ((File.GetAttributes(fi.FullName) & FileAttributes.Hidden) != FileAttributes.Hidden & fi.Extension != ".gz")
                    {
                        PackagePart packagePartDocument;

                        if (fi.Extension == ".xml")
                        {
                            packagePartDocument = Compress.CreatePart(PackUriHelper.CreatePartUri(new Uri(@"/" + Path.GetFileName(file), UriKind.Relative)), System.Net.Mime.MediaTypeNames.Text.Xml);
                        }
                        else
                        {
                            packagePartDocument = Compress.CreatePart(PackUriHelper.CreatePartUri(new Uri(@"/" + Path.GetFileName(file), UriKind.Relative)), System.Net.Mime.MediaTypeNames.Text.Plain);
                        }

                        using (FileStream inFile = fi.OpenRead())
                        {
                            inFile.CopyTo(packagePartDocument.GetStream());
                        }

                        Compress.CreateRelationship(packagePartDocument.Uri, TargetMode.Internal, fi.FullName);
                    }
                }
            }
        }

我是这样读的：

using (Package zip = Package.Open(PackageStream, FileMode.Open, FileAccess.Read, FileShare.None))
        {
            Directory.CreateDirectory(Path.Combine(App.Configuration.TempDirectory, this.packageName.Split('.')[0]));
            foreach (PackagePart file in zip.GetParts())
            {
                string fileName = Path.Combine(App.Configuration.TempDirectory, this.packageName.Split('.')[0], file.Uri.ToString().TrimStart('/'));
                using (FileStream fileStream = new FileStream(fileName, FileMode.Create))
                {
                    file.GetStream().CopyTo(fileStream);
                }
            }
        }

包尝试打开文件时出现我的问题

Package zip = Package.Open(PackageStream, FileMode.Open, FileAccess.Read, FileShare.None)

我有一个错误“文件包含损坏的数据”

试了好多遍还是没看懂。。。

编辑：相同的代码正在使用控制台应用程序，但不适用于 Web 应用程序。

4

1 回答 1

0

最后我发现了我的错误：

当我返回这个生成的包时，我会返回包，但也会返回整个网页。

    Response.AddHeader("content-disposition", "attachment; filename=" + export.PackageName);
    Response.ContentType = "application/octet-stream";
    Response.WriteFile(export.PackageFullName);

所以我纠正了它

    Response.Clear();
    Response.ClearHeaders();
    Response.ClearContent();
    Response.AddHeader("content-disposition", "attachment; filename=" + export.PackageName);
    Response.ContentType = "application/octet-stream";
    Response.WriteFile(export.PackageFullName);
    Response.Flush();
    Response.End();

有用 ：）

于 2013-07-25T11:58:56.470 回答