另一种传输到字节[]的方法（用于保存到DB）。

@Arthur的方法工作得很好，但不能完美复制，因此MS Office文档在从数据库检索后可能无法打开。MemoryStream.GetBuffer（）可以在字节[]的末尾返回额外的空字节，但可以使用MemoryStream.ToArray（）来解决这个问题。然而，我发现这个替代方案可以完美地适用于所有文件类型：

using (var binaryReader = new BinaryReader(file.InputStream))
{
    byte[] array = binaryReader.ReadBytes(file.ContentLength);
}

这是我的完整代码：

文档类别：

public class Document
{
    public int? DocumentID { get; set; }
    public string FileName { get; set; }
    public byte[] Data { get; set; }
    public string ContentType { get; set; }
    public int? ContentLength { get; set; }

    public Document()
    {
        DocumentID = 0;
        FileName = "New File";
        Data = new byte[] { };
        ContentType = "";
        ContentLength = 0;
    }
}

文件下载：

[HttpGet]
public ActionResult GetDocument(int? documentID)
{
    // Get document from database
    var doc = dataLayer.GetDocument(documentID);

    // Convert to ContentDisposition
    var cd = new System.Net.Mime.ContentDisposition
    {
        FileName = doc.FileName, 

        // Prompt the user for downloading; set to true if you want 
        // the browser to try to show the file 'inline' (display in-browser
        // without prompting to download file).  Set to false if you 
        // want to always prompt them to download the file.
        Inline = true, 
    };
    Response.AppendHeader("Content-Disposition", cd.ToString());

    // View document
    return File(doc.Data, doc.ContentType);
}

文件上载：

[HttpPost]
public ActionResult GetDocument(HttpPostedFileBase file)
{
    // Verify that the user selected a file
    if (file != null && file.ContentLength > 0)
    {
        // Get file info
        var fileName = Path.GetFileName(file.FileName);
        var contentLength = file.ContentLength;
        var contentType = file.ContentType;

        // Get file data
        byte[] data = new byte[] { };
        using (var binaryReader = new BinaryReader(file.InputStream))
        {
            data = binaryReader.ReadBytes(file.ContentLength);
        }

        // Save to database
        Document doc = new Document()
        {
            FileName = fileName,
            Data = data,
            ContentType = contentType,
            ContentLength = contentLength,
        };
        dataLayer.SaveDocument(doc);

        // Show success ...
        return RedirectToAction("Index");
    }
    else
    {
        // Show error ...
        return View("Foo");
    }
}

视图（代码段）：

@using (Html.BeginForm("GetDocument", "Home", FormMethod.Post, new { enctype = "multipart/form-data" }))
{
    <input type="file" name="file" />
    <input type="submit" value="Upload File" />
}

MemoryStream.GetBuffer() can return extra empty bytes at the end of the byte[], but you can fix that by using MemoryStream.ToArray() instead. However, I found this alternative to work perfectly for all file types:

using (var binaryReader = new BinaryReader(file.InputStream))
{
    byte[] array = binaryReader.ReadBytes(file.ContentLength);
}
Here's my full code:

Document Class:

public class Document
{
    public int? DocumentID { get; set; }
    public string FileName { get; set; }
    public byte[] Data { get; set; }
    public string ContentType { get; set; }
    public int? ContentLength { get; set; }

    public Document()
    {
        DocumentID = 0;
        FileName = "New File";
        Data = new byte[] { };
        ContentType = "";
        ContentLength = 0;
    }
}
File Download:

[HttpGet]
public ActionResult GetDocument(int? documentID)
{
    // Get document from database
    var doc = dataLayer.GetDocument(documentID);

    // Convert to ContentDisposition
    var cd = new System.Net.Mime.ContentDisposition
    {
        FileName = doc.FileName, 

        // Prompt the user for downloading; set to true if you want 
        // the browser to try to show the file 'inline' (display in-browser
        // without prompting to download file).  Set to false if you 
        // want to always prompt them to download the file.
        Inline = true, 
    };
    Response.AppendHeader("Content-Disposition", cd.ToString());

    // View document
    return File(doc.Data, doc.ContentType);
}
File Upload:

[HttpPost]
public ActionResult GetDocument(HttpPostedFileBase file)
{
    // Verify that the user selected a file
    if (file != null && file.ContentLength > 0)
    {
        // Get file info
        var fileName = Path.GetFileName(file.FileName);
        var contentLength = file.ContentLength;
        var contentType = file.ContentType;

        // Get file data
        byte[] data = new byte[] { };
        using (var binaryReader = new BinaryReader(file.InputStream))
        {
            data = binaryReader.ReadBytes(file.ContentLength);
        }

        // Save to database
        Document doc = new Document()
        {
            FileName = fileName,
            Data = data,
            ContentType = contentType,
            ContentLength = contentLength,
        };
        dataLayer.SaveDocument(doc);

        // Show success ...
        return RedirectToAction("Index");
    }
    else
    {
        // Show error ...
        return View("Foo");
    }
}
View (snippet):

@using (Html.BeginForm("GetDocument", "Home", FormMethod.Post, new { enctype = "multipart/form-data" }))
{
    <input type="file" name="file" />
    <input type="submit" value="Upload File" />
}

我在做文件上传概念时也遇到过同样的错误。我知道开发人员为这个问题提供了很多答案。

尽管我回答这个问题的原因是，由于下面提到的疏忽错误，我犯了这个错误。

<input type="file" name="uploadedFile" />

在指定name属性时，请确保控制器参数也具有相同的名称值“uploadedFile”。这样地：

   [HttpPost]
            public ActionResult FileUpload(HttpPostedFileBase uploadedFile)
            {

            }

否则它不会被映射。

Html：

@using (Html.BeginForm("StoreMyCompany", "MyCompany", FormMethod.Post, new { id = "formMyCompany", enctype = "multipart/form-data" }))
{
   <div class="form-group">
      @Html.LabelFor(model => model.modelMyCompany.Logo, htmlAttributes: new { @class = "control-label col-md-3" })
      <div class="col-md-6">
        <input type="file" name="Logo" id="fileUpload" accept=".png,.jpg,.jpeg,.gif,.tif" />
      </div>
    </div>

    <br />
    <div class="form-group">
          <div class="col-md-offset-3 col-md-6">
              <input type="submit" value="Save" class="btn btn-success" />
          </div>
     </div>
}  

背后代码：

public ActionResult StoreMyCompany([Bind(Exclude = "Logo")]MyCompanyVM model)
{
    try
    {        
        byte[] imageData = null;
        if (Request.Files.Count > 0)
        {
            HttpPostedFileBase objFiles = Request.Files["Logo"];

            using (var binaryReader = new BinaryReader(objFiles.InputStream))
            {
                imageData = binaryReader.ReadBytes(objFiles.ContentLength);
            }
        }

        if (imageData != null && imageData.Length > 0)
        {
           //Your code
        }

        dbo.SaveChanges();

        return RedirectToAction("MyCompany", "Home");

    }
    catch (Exception ex)
    {
        Utility.LogError(ex);
    }

    return View();
}

在控制器中

 if (MyModal.ImageFile != null)
                    {
                        MyModal.ImageURL = string.Format("{0}.{1}", Guid.NewGuid().ToString(), MyModal.ImageFile.FileName.Split('.').LastOrDefault());
                        if (MyModal.ImageFile != null)
                        {
                            var path = Path.Combine(Server.MapPath("~/Content/uploads/"), MyModal.ImageURL);
                            MyModal.ImageFile.SaveAs(path);
                        }
                    }

在视图中

<input type="hidden" value="" name="..."><input id="ImageFile" type="file" name="ImageFile" src="@Model.ImageURL">

在模态类中

 public HttpPostedFileBase ImageFile { get; set; }

在项目的Content文件夹中创建上载文件夹

如果你碰巧像我一样在这里跌跌撞撞，想知道尽管代码正确，为什么你的代码仍然不工作。然后，请在输入控件中查找name属性，您可能会意外错过或从未将其放在首位。

<input class="custom-file-input" name="UploadFile" id="UploadFile" type="file" onchange="ValidateFile(this);" accept=".xls, .xlsx">