Question

我必须在UploadFile.cshtml表单

的单个提交中上传文件和其他数据

我有一个基类，它是家庭控制器中mvc动作的输入。

我的基类，mvc动作方法和带有Razor脚本的cshtml部分在下面给出

我必须在提交UploadFile.cshtml表单时在mvc操作中将文件作为字节数组

我的基类

public class FileUpload
{
  public string Name {get;set;}
  public int Age {get;set;}
  public byte[] Photo {get;set;}
}

MyMVCAction

[HttpPost]
public void UploadFile(FileUploadobj)
{
  FileUploaddata=obj;
}

Mycshtmlform

@modelMVC.Models.FileUpload
@using(Html.BeginForm("UploadFile","Home",FormMethod.Post))
{

<fieldset>
<legend>File Upload</legend>

<div class="editor-label">
@Html.LabelFor(model=>model.Name)
</div>
<div class="editor-field">
@Html.EditorFor(model=>model.Name)
</div>

<div class="editor-label">
@Html.LabelFor(model=>model.Age)
</div>
<div class="editor-field">
@Html.EditorFor(model=>model.Age)
</div>
<div class="editor-label">
@Html.Label("UploadPhoto");
</div>
<div class="editor-field">
<input type="file" id="photo" name="photo"/>
</div>
<p>
<input type="submit" value="Create"/>
</p>

</fieldset>

}

Answer 1

我认为更好地实现它：

[HttpPost]
public ActionResult UploadFile(FileUpload obj) 
{
    using (var binaryReader = new BinaryReader(Request.Files[0].InputStream))
    {
       obj.Photo = binaryReader.ReadBytes(Request.Files[0].ContentLength);
    }  

    //return some action result e.g. return new HttpStatusCodeResult(HttpStatusCode.OK);
}

我希望它有所帮助。

mvc c＃file upload as byte array

1 个答案: