如果有人想用Ajax上传多个文件，下面是我的文章在Asp.NetMVC中使用Ajax进行多文件上传

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。

我的方法和上面差不多，我将向您展示我的代码以及如何使用MYSSQL数据库。。。

数据库中的文档表-

int Id（PK），字符串Url，字符串描述，创建者，租户ID上传日期

上面的代码ID是主键，URL是文件的名称（末尾有文件类型），要在文档视图中输出的文件描述，CreatedBy是上载文件的人，tenncyId，dateUploaded

在视图中，必须定义enctype，否则它将无法正常工作。

@using (Html.BeginForm("Upload", "Document", FormMethod.Post, new { enctype = "multipart/form-data" }))
{
<div class="input-group">
    <label for="file">Upload a document:</label>
    <input type="file" name="file" id="file" />
</div>
}

上面的代码将给你一个浏览按钮，然后在我的项目中，我有一个基本上名为IsValidImage的类，它只是检查文件大小是否在指定的最大大小之下，检查它是否是IMG文件，这都在类bool函数中。所以，如果true返回true。

public static bool IsValidImage(HttpPostedFileBase file, double maxFileSize, ModelState ms )
{
    // make sur the file isnt null.
    if( file == null )
        return false;

// the param I normally set maxFileSize is 10MB  10 * 1024 * 1024 = 10485760 bytes converted is 10mb
var max = maxFileSize * 1024 * 1024;

// check if the filesize is above our defined MAX size.
if( file.ContentLength > max )
    return false;

try
{
    // define our allowed image formats
    var allowedFormats = new[] { ImageFormat.Jpeg, ImageFormat.Png, ImageFormat.Gif, ImageFormat.Bmp };

    // Creates an Image from the specified data stream.      
    using (var img = Image.FromStream(file.InputStream))
    {
        // Return true if the image format is allowed
        return allowedFormats.Contains(img.RawFormat);
    }
}
catch( Exception ex )
{
    ms.AddModelError( "", ex.Message );                 
}
return false;   
}

因此，在控制器中：

if (!Code.Picture.IsValidUpload(model.File, 10, true))
{                
    return View(model);
}

// Set the file name up... Being random guid, and then todays time in ticks. Then add the file extension
// to the end of the file name
var dbPath = Guid.NewGuid().ToString() + DateTime.UtcNow.Ticks + Path.GetExtension(model.File.FileName);

// Combine the two paths together being the location on the server to store it
// then the actual file name and extension.
var path = Path.Combine(Server.MapPath("~/Uploads/Documents/"), dbPath);

// set variable as Parent directory I do this to make sure the path exists if not
// I will create the directory.
var directoryInfo = new FileInfo(path).Directory;

if (directoryInfo != null)
    directoryInfo.Create();

// save the document in the combined path.
model.File.SaveAs(path);

// then add the data to the database
_db.Documents.Add(new Document
{
    TenancyId = model.SelectedTenancy,
    FileUrl = dbPath,
    FileDescription = model.Description,
    CreatedBy = loggedInAs,
    CreatedDate = DateTime.UtcNow,
    UpdatedDate = null,
    CanTenantView = true
});

_db.SaveChanges();
model.Successfull = true;

如果有人想用Ajax上传多个文件，下面是我的文章在Asp.NetMVC中使用Ajax进行多文件上传

public ActionResult FileUpload(upload mRegister) {
    //Check server side validation using data annotation
    if (ModelState.IsValid) {
        //TO:DO
        var fileName = Path.GetFileName(mRegister.file.FileName);
        var path = Path.Combine(Server.MapPath("~/Content/Upload"), fileName);
        mRegister.file.SaveAs(path);

        ViewBag.Message = "File has been uploaded successfully";
        ModelState.Clear();
    }
    return View();
}