2012-06-29 50 views
0

在這裏,我寫的上傳圖片的編碼control.but獲得在SqlConnection的地方一些運行時error.error母豬 首先我
1.圖像名稱框 - 文本框
2.Image上傳控制 - ASP imageupload控件
3.上傳按鈕
的SqlConnection運行時錯誤

錯誤:從未同步的代碼塊調用對象同步方法。

代碼下面

public partial class ProfileDetails : System.Web.UI.Page 
{ 
    string connStr = ConfigurationManager.ConnectionStrings["ConnectionString"].ConnectionString; 
//SqlConnection con = new SqlConnection("Data Source=CHATHU-LAPTOP;Initial Catalog=ProfilemgtDB;User ID=sa;Password=sa123"); 
protected void Page_Load(object sender, EventArgs e) 
{ 

} 
protected void Button1_Click(object sender, EventArgs e) 
{ 

} 

protected void Upload_Click(object sender, EventArgs e) 
{ 
    string path = Server.MapPath("images/"); 
    if (FileUpload1.HasFile) 
    { 
     string ext = Path.GetExtension(FileUpload1.FileName); 
     if (ext == ".jpg" || ext == ".png") 
     { 
      FileUpload1.SaveAs(path + FileUpload1.FileName); 
      string name = "~/images/" + FileUpload1.FileName; 
      string s = "Insert into Profile values('" + TextBox12.Text.Trim() + " '.'" + name + "')"; 

      SqlConnection con = new SqlConnection(connStr); 
      SqlCommand cmd = new SqlCommand(s, con); 
      con.Open(); 
      cmd.ExecuteNonQuery(); 
      con.Close(); 
      Response.Write("File Uploaded"); 
     } 
     else 
     { 
      Response.Write("You can upload only JPG & PNG"); 
     } 

    } 
    else { 

     Response.Write("Please Select File"); 
    } 
    } 
} 

ERROR:對象同步方法是從代碼不同步塊調用。

+0

哪一行是拋出該錯誤? –

+0

SqlConnection con = new SqlConnection(connStr); –

+2

可能與你的錯誤沒有任何關係,但是這個代碼是一個等待發生的災難。切勿使用字符串連接來創建sql值以傳遞INSERT,UPDATE,DELETE,SELECT等操作 – Steve

回答