使用 jquery ajax 和 asp.net 处理程序上传文件

Question

我正在努力让它工作，但我在上传文件时遇到错误。

ASPX

<asp:FileUpload ID="FileUpload1" runat="server" CssClass="file-upload-dialog" />
<asp:Button runat="server" ID="btnUpload" CssClass="btn upload" Text="Upload" />

处理程序

public void ProcessRequest(HttpContext context)
      {
            context.Response.ContentType = "multipart/form-data";
            context.Response.Expires = -1;
            try
            {
                  HttpPostedFile postedFile = context.Request.Files["file"];
                  string savepath = HttpContext.Current.Server.MapPath("~/assets/common/CompetitionEntryImages/");
                  var extension = Path.GetExtension(postedFile.FileName);

                  if (!Directory.Exists(savepath))
                        Directory.CreateDirectory(savepath);

                  var id = Guid.NewGuid() + extension;
                  if (extension != null)
                  {
                        var fileLocation = string.Format("{0}/{1}",
                                                         savepath,
                                                         id);

                        postedFile.SaveAs(fileLocation);
                        context.Response.Write(fileLocation);
                        context.Response.StatusCode = 200;
                  }
            }
            catch (Exception ex)
            {
                  context.Response.Write("Error: " + ex.Message);
            }
      }

jQuery

$(document).ready(function () {
            email = $("input[id$='emailHV']").val();
            alert(email);
            $('#aspnetForm').attr("enctype", "multipart/form-data");
      });



$('#<%= btnUpload.ClientID %>').on('click', function (e) {
            e.preventDefault();
            var fileInput = $('#ctl00_PageContent_Signup_ctl06_MWFileUpload_FileUpload1');
            var fd = new window.FormData();
            fd.append('file', fileInput.files[0]);

            $.ajax({
                  url: '/charity-challenge/MWFileUploadHandler.ashx',
                  data: fd,
                  processData: false,
                  contentType: false,
                  type: 'POST',
                  success: function (data) {
                        alert(data);
                  }
            });
      });

错误

HTML

<input type="file" name="ctl00$PageContent$Signup$ctl06$MWFileUpload$FileUpload1" id="ctl00_PageContent_Signup_ctl06_MWFileUpload_FileUpload1" class="file-upload-dialog">

 <input type="submit" name="ctl00$PageContent$Signup$ctl06$MWFileUpload$btnUpload" 
value="Upload" onclick="javascript:WebForm_DoPostBackWithOptions(new 
WebForm_PostBackOptions(&quot;ctl00$PageContent$Signup$ctl06$MWFileUpload$btnUpload&quot;, 
&quot;&quot;, true, &quot;&quot;, &quot;&quot;, false, false))" 
id="ctl00_PageContent_Signup_ctl06_MWFileUpload_btnUpload" class="button">

编辑

最后，我通过做这些事情来形成数据来让它工作

var fileData = fileInput.prop("files")[0];   // Getting the properties of file from file field
        var formData = new window.FormData();                  // Creating object of FormData class
        formData.append("file", fileData); // Appending parameter named file with properties of file_field to form_data
        formData.append("user_email", email);

完整的工作代码

$('#<%= btnUpload.ClientID %>').on('click', function (e) {
            e.preventDefault();
            var fileInput = $('#<%= FileUpload1.ClientID %>');
            var fileData = fileInput.prop("files")[0];   // Getting the properties of file from file field
            var formData = new window.FormData();                  // Creating object of FormData class
            formData.append("file", fileData); // Appending parameter named file with properties of file_field to form_data
            formData.append("user_email", email);
            $.ajax({
                  url: '/charity-challenge/MWFileUploadHandler.ashx',
                  data: formData,
                  processData: false,
                  contentType: false,
                  type: 'POST',
                  success: function (data) {
                        var obj = $.parseJSON(data);
                        if (obj.StatusCode == "OK") {
                              $('#<%= imagePath.ClientID %>').val(obj.ImageUploadPath);
                              $('.result-message').html(obj.Message).show();
                        } else if (obj.StatusCode == "ERROR") {
                              $('.result-message').html(obj.Message).show();
                        }
                  },
                  error: function (errorData) {
                        $('.result-message').html("there was a problem uploading the file.").show();
                  }
            });
      });

Answer 1

你可以用这个：

$("#<% = FileUpload1.clientID %>")

Answer 2

在您的网络配置文件中使用它

<system.webServer>
 <validation validateIntegratedModeConfiguration="false" />
<handlers>
    <add name="AjaxFileUploadHandler" verb="*" 
      path="AjaxFileUploadHandler.axd"
      type="AjaxControlToolkit.AjaxFileUploadHandler, 
      AjaxControlToolkit"/>
</handlers>

Answer 3

在敲了一个下午之后，当我意识到您指定了“处理程序”而不是“服务”时，我回到了这个问题/解决方案。 :) 另外，对于可以在后面运行此 jquery 的工作处理程序，我去了https://github.com/superquinho/jQuery-File-Upload-ASPnet 并删除了我不需要的内容。这是我正在使用的处理程序代码（VS2012）。如您所见，我现在仅将其用于 csv 上传。

Imports System
Imports System.Web
Imports System.Data

Public Class FileUpload : Implements IHttpHandler, IReadOnlySessionState
    Public Sub ProcessRequest(ByVal context As HttpContext) Implements IHttpHandler.ProcessRequest
        Try
            Select Case context.Request.HttpMethod
                Case "POST"
                    Uploadfile(context)
                    Return

            Case Else
                context.Response.ClearHeaders()
                context.Response.StatusCode = 405
                Return
        End Select

    Catch ex As Exception
        Throw
    End Try

End Sub

Private Sub Uploadfile(ByVal context As HttpContext)

    Dim i As Integer
    Dim files As String()
    Dim savedFileName As String = String.Empty
    Dim js As New Script.Serialization.JavaScriptSerializer

    Try

        If context.Request.Files.Count >= 1 Then

            Dim maximumFileSize As Integer = ConfigurationManager.AppSettings("UploadFilesMaximumFileSize")

            context.Response.ContentType = "text/plain"
            For i = 0 To context.Request.Files.Count - 1
                Dim hpf As HttpPostedFile
                Dim FileName As String
                hpf = context.Request.Files.Item(i)

                If HttpContext.Current.Request.Browser.Browser.ToUpper = "IE" Then
                    files = hpf.FileName.Split(CChar("\\"))
                    FileName = files(files.Length - 1)
                Else
                    FileName = hpf.FileName
                End If


                If hpf.ContentLength >= 0 And (hpf.ContentLength <= maximumFileSize * 1000 Or maximumFileSize = 0) Then
                    Dim d As Date = Now
                    savedFileName = HttpRuntime.AppDomainAppPath & "CSVLoad\" + d.Year.ToString + d.DayOfYear.ToString + d.Hour.ToString + d.Minute.ToString + d.Second.ToString + d.Millisecond.ToString + ".csv"

                    hpf.SaveAs(savedFileName)

                Else

                End If
            Next

        End If

    Catch ex As Exception
        Throw
    End Try

End Sub

Public ReadOnly Property IsReusable() As Boolean Implements IHttpHandler.IsReusable
    Get
        Return False
    End Get
End Property

End Class

Answer 4

$("#file-upload") 

应该是

$("#ctl00_PageContent_Signup_ctl06_MWFileUpload_file-Upload")

查看使用ClientIdMode 属性将服务器代码上的文件上传控件更改为具有静态服务器端ID。像这样：

<asp:FileUpload ID="FileUpload1" runat="server" CssClass="file-upload-dialog" ClientIdMode="Static" />

那么就可以确定客户端代码中控件的ID是FileUpload1