将多个图像上传到服务器并将它们的文件名插入到数据库中

Question

我有以下代码将多个图像上传到服务器并在数据库中插入其名称。

内容页

<asp:FileUpload runat="server" ID="UploadImages" AllowMultiple="true" /> 
    <asp:Button runat="server" ID="uploadedFile" Text="Upload" OnClick="uploadFile_Click" /> 
    <asp:Label ID="listofuploadedfiles" runat="server" /> 

后面的代码

protected void uploadFile_Click(object sender, EventArgs e) 
    { 

     if (UploadImages.PostedFile != null) 

     { 

     string strConnString = System.Configuration.ConfigurationManager.ConnectionStrings["DefaultConnection"].ConnectionString; 
     SqlConnection con = new SqlConnection(strConnString); 
     SqlCommand cmd = new SqlCommand(); 

     try 
     { 
      con.Open();   

      foreach (HttpPostedFile uploadedFile in this.UploadImages.PostedFiles) 
      { 
       string newname = System.DateTime.Now.ToString("yyMMdd-hhmmss-") + uploadedFile.FileName; 
       uploadedFile.SaveAs(System.IO.Path.Combine(Server.MapPath("/Images/Editors/BG/"), newname)); 
       listofuploadedfiles.Text += string.Format("<br /><img width='100px' src='/Images/Editors/BG/{0}'/>{0}<br clear='all'/>", newname); 


       cmd.Connection = con; 
       cmd.CommandText = "INSERT INTO BackgroundImages([BG_fileName], [IDuser]) VALUES(" + newname + "," + HttpContext.Current.User.Identity.GetUserId() + ")"; 
       cmd.ExecuteNonQuery(); 
      } 
     } 
     catch (Exception ex) 
     { 
      Response.Write("Error while inserting record on table..." + ex.Message + "Insert Records"); 
     } 
     finally 
     { 
      con.Close(); 
      con.Dispose(); 

     } 

     } 

    } 

我能够将图像上传到服务器，但没有被添加到数据库中。我没有收到任何错误。怎么了？

其实我把上面的代码从VB翻译成C＃。我确定我错过了一些东西，因为我还不熟悉C＃。以下VB代码运行良好：

Protected Sub uploadFile_Click(sender As Object, e As EventArgs) 

    If UploadImages.HasFiles Then 

     Dim con As New SqlConnection(ConfigurationManager.ConnectionStrings("DefaultConnection").ConnectionString) 
     Dim cmd As New SqlCommand 
     Try 
      con.Open() 

      Dim newname As String 

      For Each uploadedFile As HttpPostedFile In UploadImages.PostedFiles 

       newname = System.DateTime.Now.ToString("yyMMdd-hhmmss-") + uploadedFile.FileName 
       uploadedFile.SaveAs(System.IO.Path.Combine(Server.MapPath("~/Images/"), newname)) 
       listofuploadedfiles.Text += [String].Format("<br /><img width='100px' src='Images/{0}'/>{0}<br clear='all'/>", newname) 

       cmd.Connection = con 
       cmd.CommandText = "INSERT INTO Images([filename], [userid]) VALUES('" & newname & "','" & userid & "')" 
       cmd.ExecuteNonQuery() 
      Next 

     Catch ex As Exception 
      'MessageBox.Show("Error while inserting record on table..." & ex.Message, "Insert Records") 
     Finally 
      con.Close() 
     End Try 

    End If 
End Sub 

Answer 1

下面是您需要将参数与参数绑定的解决方案。

protected void uploadFile_Click(object sender, EventArgs e) 
    { 

     if (UploadImages.PostedFile != null) 

     { 

      string strConnString = System.Configuration.ConfigurationManager.ConnectionStrings["DefaultConnection"].ConnectionString; 
     SqlConnection con = new SqlConnection(strConnString); 
     con.Open(); 
     try 
     {    

      foreach (HttpPostedFile uploadedFile in this.UploadImages.PostedFiles) 
      { 
       string newname = System.DateTime.Now.ToString("yyMMdd-hhmmss-") + uploadedFile.FileName; 
       uploadedFile.SaveAs(System.IO.Path.Combine(Server.MapPath("/Images/Editors/BG/"), newname)); 
       listofuploadedfiles.Text += string.Format("<br /><img width='100px' src='/Images/Editors/BG/{0}'/>{0}<br clear='all'/>", newname); 


       SqlCommand cmd = new SqlCommand(); 
       cmd.Connection = con; 
       cmd.CommandType = CommandType.Text; 
       cmd.CommandText = @"INSERT INTO BackgroundImages(BG_fileName, IDuser) 
       VALUES(@param1,@param2)"; 

       cmd.Parameters.AddWithValue("@param1", newname); 
       cmd.Parameters.AddWithValue("@param2", HttpContext.Current.User.Identity.GetUserId()); 
       cmd.ExecuteNonQuery(); 
      } 
     } 
     catch (Exception ex) 
     { 
      Response.Write("Error while inserting record on table..." + ex.Message + "Insert Records"); 
     } 
     finally 
     { 
      con.Close(); 
      con.Dispose(); 

     } 

     } 

    }