C# Zip 将字节 [] 保存为 zip，转换为流，将邮件作为邮件附件发送 - 附件文档的正确方法

Question

所以下面是我正在使用的代码。基本上它只是遍历一个数组，将文件添加到 zip 中，然后将 zip 保存到内存流中，然后通过电子邮件发送附件。

当我在调试中查看项目时，我可以看到 zip 文件有大约 20 兆字节的数据。当我收到附件时，它只有大约 230 位数据并且没有内容。有什么想法吗？

byteCount = byteCount + docs[holder].FileSize;
            if (byteCount > byteLimit)
            {
                //create a new stream and save the stream to the zip file
                System.IO.MemoryStream attachmentstream = new System.IO.MemoryStream();
                zip.Save(attachmentstream);

                //create the attachment and send that attachment to the mail
                Attachment data = new Attachment(attachmentstream, "documentrequest.zip");
                theMailMessage.Attachments.Add(data);

                //send Mail
                SmtpClient theClient = new SmtpClient("mymail");
                theClient.UseDefaultCredentials = false;
                System.Net.NetworkCredential theCredential = new System.Net.NetworkCredential("bytebte", "2323232");
                theClient.Credentials = theCredential;
                theClient.Send(theMailMessage);
                zip = new ZipFile();

                //iterate Document Holder
                holder++;
            }
            else
            {
                //create the stream and add it to the zip file
                //System.IO.MemoryStream stream = new System.IO.MemoryStream(docs[holder].FileData);
                zip.AddEntry("DocId_"+docs[holder].DocumentId+"_"+docs[holder].FileName, docs[holder].FileData);
                holder++;

            }

问题出在Attachment data = new Attachment(attachmentstream, "documentrequest.zip");，一旦我查看附件，它的大小为-1 那么附加此项目的正确方法是什么？

Answer 1

我怀疑对zip.Save 的调用会在写入后关闭流。您可能最终不得不将字节复制到一个数组中，然后创建一个新的MemoryStream 以供读取。例如：

//create a new stream and save the stream to the zip file
byte[] streamBytes;
using (var ms = new MemoryStream())
{
    zip.Save(ms);
    // copy the bytes
    streamBytes = ms.ToArray();
}

// create a stream for the attachment
using (var attachmentStream = new MemoryStream(streamBytes))
{
    //create the attachment and send that attachment to the mail
    Attachment data = new Attachment(attachmentstream, "documentrequest.zip");
    theMailMessage.Attachments.Add(data);

    // rest of your code here
}