2016-05-31 57 views
0

我一直试图整理这一整天与大量的谷歌搜索,但几乎没有影响,所以提前感谢,如果你阅读并能够提供帮助。试图用ashx来获取查询字符串变量

我有一个简单的页面,它允许用户选择文件并上传到亚马逊桶,第一页是这样的:

var checkError = false; 
$(window).load(
    function() { 
     $("#<%=FileImages.ClientID %>").fileUpload({ 
      'fileSizeLimit': '100MB', 
      'uploader': 'scripts/uploader.swf', 
      'cancelImg': 'images/cancel.png', 
      'buttonText': 'Upload Files', 
      'script': 'UploadVB.ashx?gid=56', 
      'folder': 'uploads', 
      'multi': true, 
      'auto': true, 
      'onAllComplete': function (uploads) { 
       if (!checkError) { 
        location.href = "finished"; 
       } 
      }, 
      'onComplete': function (event, ID, fileObj, response, data) { 
       if (200 != parseInt(response)) { 
        checkError = true; 
        $('#FileImages' + ID).addClass('uploadifyError'); 
        $('#FileImages' + ID + ' .percentage').text(' - ' + response); 

        return false; 
       } 
      } 
     }); 
    } 

);

注意gid=56。然后,将文件发送到UploadVB.ashx这看起来是这样的:

Public Sub ProcessRequest(ByVal context As HttpContext) Implements IHttpHandler.ProcessRequest 
    Dim awsRegion As Amazon.RegionEndpoint = Amazon.RegionEndpoint.EUWest1 
    Dim client As New AmazonS3Client(System.Configuration.ConfigurationManager.AppSettings("AWSAccessKey").ToString(), System.Configuration.ConfigurationManager.AppSettings("AWSSecretKey").ToString(), awsRegion) 
    Dim fileID As String = Guid.NewGuid().ToString("N") 
    Dim postedFile As HttpPostedFile = context.Request.Files("Filedata") 

    Dim galleryid = context.Request.QueryString("gid") 

    Try 
     Dim filename As String = Path.GetFileName(postedFile.FileName) 
'### I WANT TO SUBMIT THE FILENAME AND ID INTO A TABLE HERE ###### 
    AddPortfolioPhoto(filename, Convert.ToInt16(galleryid)) 
    '################################################################# 
     Dim S3_KEY As [String] = filename.Replace(" ", "") 
     Dim request As New PutObjectRequest() 
     With request 
      .BucketName = System.Configuration.ConfigurationManager.AppSettings("AWSBucketName").ToString() 
      .Key = S3_KEY 
      .InputStream = postedFile.InputStream 
     End With 

     client.PutObject(request) 
     context.Response.Write("200") 
     context.Response.StatusCode = 200 
    Catch ex As Exception 
     context.Response.Write("AWS ERROR :: " + ex.Message) 
    End Try 
End Sub 
Public ReadOnly Property IsReusable() As Boolean Implements IHttpHandler.IsReusable 
    Get 
     Return False 
    End Get 
End Property 

我每次运行这个上面它只是没有做任何事情,如果我删除此行:

AddPortfolioPhoto(filename, Convert.ToInt16(galleryid)) 

它的工作原理很好,但它是至关重要的,我把它插入到数据库中,我似乎无法抓住gid值。

任何帮助,非常感谢。

Public Shared Sub AddPortfolioPhoto(ByVal fn As String, gid As Integer) 

    Dim Writer As New Data.SqlClient.SqlCommand 
    Writer.CommandType = Data.CommandType.StoredProcedure 
    Writer.CommandText = "BCMS_AddGalleryPhoto" 
    Writer.Connection = New Data.SqlClient.SqlConnection(System.Configuration.ConfigurationManager.AppSettings("sqlstr")) 
    Writer.Parameters.AddWithValue("@imgstr", Data.SqlTypes.SqlString.Null).Value = fn 
    Writer.Parameters.AddWithValue("@galleryid", Data.SqlTypes.SqlInt32.Null).Value = gid 
    Writer.Parameters.AddWithValue("@so", Data.SqlTypes.SqlInt32.Null).Value = 0 
    Writer.Parameters.AddWithValue("@featured", Data.SqlTypes.SqlInt32.Null).Value = 0 
    Writer.Prepare() 
    Writer.Connection.Open() 
    Writer.ExecuteNonQuery() 
    Writer.Connection.Close() 

End Sub 
+0

context.Request.QueryString(“gid”)的值是什么? –

+0

它总是一个整数,在这种情况下:56 –

+0

错误来自AddPortfolio方法。你可以发布该代码吗? –

回答

0

看起来像你的帮助,我可能已经到这条底线,我期待值(56)不是我实际上得到在同一个变量,我其实是让“上传”所以发送一个字符串到这个方法是永远不会工作的,谢谢@ matt-dot-net提出建议并感谢Andrew Morton提供的处理技巧,我将使用它而不是关闭。