存储文件上传到MemoryStream（C＃）的最佳做法是什么？ [英] What is the best practice for storing a file upload to a MemoryStream (C#)?

问题描述

我需要临时将文件上传到MemoryStream。

使用asp.net（3.5）做这个最好的方法是什么？

这是我到目前为止。它工作（本地测试），但它看起来不正确的我。

 保护无效lnkUploadFile_Click（对象发件人，EventArgs e） 
 {
 MemoryStream memStream = new MemoryStream（）; 
 
 BinaryWriter sWriter = new BinaryWriter（memStream）; 
 
 foreach（在flUpload.FileBytes中的字节b）
 {
 sWriter.Write（b）; 
} 
 
 sWriter.Flush（）; 
 
 //写入文件来验证正确转换的文件流
 FileStream fstream = new FileStream（@C：/ test /+ flUpload.FileName，FileMode.Create）; 
 memStream.WriteTo（fstream）; 
 fstream.Dispose（）; 
 
 memStream.Dispose（）; 
 
  

解决方案

> flUpload.FileBytes 是一个字节数组，您可以使用接受包含的数据作为参数的MemoryStream构造函数：

  MemoryStream memStream = new MemoryStream（flUpload.FileBytes）;如果不是（如果它只是实现了IEnumerable），那么可以使用Linq将它转换为一个字节数组：
 
 
 
 
 
 
  
 
 
  MemoryStream memStream = new MemoryStream（flUpload.FileBytes.ToArray（））; 
  
 
I need to temporary store a file upload to a MemoryStream. 

What would be the best method to do this using asp.net (3.5)?

Here's what I have so far. It works (locally tested), but it does not look right to me.
protected void lnkUploadFile_Click(object sender, EventArgs e)
    {
        MemoryStream memStream = new MemoryStream();

        BinaryWriter sWriter = new BinaryWriter(memStream);

        foreach (byte b in flUpload.FileBytes)
        {
            sWriter.Write(b);
        }

        sWriter.Flush();

        // writing to file to verify file stream converted correctly
        FileStream fstream = new FileStream(@"C:/test/" + flUpload.FileName, FileMode.Create);          
        memStream.WriteTo(fstream);         
        fstream.Dispose();

        memStream.Dispose();
    }

解决方案
If flUpload.FileBytes is a byte array, you can use the MemoryStream constructor that accepts the contained data as a parameter:
MemoryStream memStream = new MemoryStream(flUpload.FileBytes);
If not (if it just implements IEnumerable), you can convert it to a byte array using Linq:
MemoryStream memStream = new MemoryStream(flUpload.FileBytes.ToArray());


                        
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