.net - 压缩电子邮件附件的 MemoryStream

Question

我正在尝试压缩 XML 树并将其用作电子邮件附件。带附件的电子邮件发送成功，但创建的 zip 文件始终损坏 - 它不是有效的 zip 文件，但包含二进制数据。

问题重现如下，具体参见BuildAttachment():

static void Main(string[] args)
{
    try
    {
        var report = new XElement("Report",
            new XElement("Product",
                new XElement("ID", "10000001"),
                new XElement("Name", "abcdefghijklm"),
                new XElement("Group", "nopqrstuvwxyz")
            )
        );
        var mailMessage = BuildMessage(report);
        EmailMessage(mailMessage);
        Thread.Sleep(10000);
    }
    catch (Exception e) { Console.WriteLine(e.Message); }
}
static MailMessage BuildMessage(XElement report)
{
    string from = "email1@address.com";
    string to = "email2@address.com";
    var message = new MailMessage(from, to, "Subject text", "Body text");

    var attachment = BuildAttachment(report);
    message.Attachments.Add(attachment);

    return message;
}
static Attachment BuildAttachment(XElement report)
{
    var inStream = new MemoryStream();
    report.Save(inStream);
    inStream.Position = 0;

    var outStream = new MemoryStream();
    var compress = new DeflateStream(outStream, CompressionMode.Compress);
    inStream.CopyTo(compress);

    outStream.Position = 0;
    return new Attachment(outStream, "report.zip", "application/zip");
}
static void EmailMessage(MailMessage message)
{
    var smtpClient = new SmtpClient("127.0.0.1");
    smtpClient.SendCompleted += SendCompletedCallback;
    smtpClient.SendAsync(message, null);
}
static void SendCompletedCallback(object sender, AsyncCompletedEventArgs e)
{
    if (e.Error != null)
        Console.WriteLine(e.Error.ToString());
}

将问题放在上下文中:它是 Windows 服务应用程序的一部分，所以我不想在磁盘上创建文件，并且电子邮件还包含 xml 树的 xslt 转换的备用 View ，所以我不想一个完全不同的解决方案。

对于 zip 文件损坏的原因有什么建议吗？

Answer 1

我知道这是一个老问题，但当我搜索相同的问题时它出现了。

这是我使用 System.IO.Compression.ZipArchive (requires .NET 4.5 or higher) 添加压缩 (zip) 附件的解决方案[基于acraig5075的回答]:

byte[] report = GetSomeReportAsByteArray();
string fileName = "file.txt";

using (MemoryStream memoryStream = new MemoryStream())
{
    using (ZipArchive zipArchive = new ZipArchive(memoryStream, ZipArchiveMode.Update))
    {
        ZipArchiveEntry zipArchiveEntry = zipArchive.CreateEntry(fileName);
        using (StreamWriter streamWriter = new StreamWriter(zipArchiveEntry.Open()))
        {
            streamWriter.Write(Encoding.Default.GetString(report));
        }
    }

    MemoryStream attachmentStream = new MemoryStream(memoryStream.ToArray());

    Attachment attachment = new Attachment(attachmentStream, fileName + ".zip", MediaTypeNames.Application.Zip);
    mail.Attachments.Add(attachment);
}