ExecuteNonQuery：连接属性尚未初始化？ [英] ExecuteNonQuery: Connection property has not been initialized?
问题描述
在我的代码中出现错误：
ExecuteNonQuery：连接属性尚未初始化。
这可能是由于此代码中的行：
OdbcCommand cmd = new OdbcCommand（INSERT INTO Pictures（UserID，picturepath）VALUES（'+ theUserId +'，'+ fileuploadpath +'））;
完整代码：
{
string theUserId = Session [UserID]。ToString（）;
{
OdbcConnection cn = new OdbcConnection（Driver = {MySQL ODBC 3.51 Driver}; Server = localhost; Database = gymwebsite2; User = root; Password = commando;）;
cn.Open（）;
if（FileUploadControl.HasFile）
{
try
{
string filename = Path.GetFileName（FileUploadControl.FileName）;
//FileUploadControl.SaveAs(Server.MapPath(\"~/userdata/+ theUserId +/ uploadedimage /）+ filename）;
string fileuploadpath = Server.MapPath（〜/ userdata /+ theUserId +/ uploadedimage /）;
FileUploadControl.SaveAs（Path.Combine（fileuploadpath，filename））;
StatusLabel.Text =上传状态：文件已上传！;
$ b OdbcCommand cmd = new OdbcCommand（INSERT INTO Pictures（UserID，picturepath）VALUES（''+ theUserId +''，'+ fileuploadpath +'） ）;
cmd.ExecuteNonQuery（）;
}
catch（Exception ex）
{
StatusLabel.Text =上传状态：文件无法上传，出现以下错误：+ ex 。信息;
$ / code> 还有一个问题，我不认为它是我想要的插入，因为这将在我的数据库中创建一个重复条目，是否将它从INSERT INTO更改为更新？
在上传图片时是否有覆盖的方法？ Atm它只是将图像保存到我已有的文件夹中？第一个图像或任何图像显然不会有相同的文件名，所以我怎么会用一个im上传覆盖文件夹中的任何图像？
编辑：
新错误（fileupload作为它存储在正确的区域，但传递fileupload到插入语句是abit wonky）
我得到错误
无法上传文件。出现以下错误：错误[42000] [MySQL] [ODBC 3.51驱动程序] [mysqld-5.5.9]您的SQL语法错误;检查与您的MySQL服务器版本相对应的手册，以找到在C：\ Users \Garrith\Documents\Visual Studio 2010\WebSites\WebSite1\userdata\1\uplo附近使用的正确语法'在第1行
这有点奇怪吗？
所有我试图做的是保存文件路径+文件名在mydb我尝试传递给插入显然失败。
protected void UploadButton_Click（object sender，EventArgs e）
{
if（FileUploadControl.HasFile）
{
try
{
string theUserId = Session [UserID]。ToString（）;
OdbcConnection cn = new OdbcConnection（Driver = {MySQL ODBC 3.51 Driver}; Server = localhost; Database = gymwebsite2; User = root; Password = commando;）;
cn.Open（）;
string filename = Path.GetFileName（FileUploadControl.FileName）;
//FileUploadControl.SaveAs(Server.MapPath(\"~/userdata/+ theUserId +/ uploadedimage /）+ filename）;
string fileuploadpath = Server.MapPath（〜/ userdata /+ theUserId +/ uploadedimage /）;
FileUploadControl.SaveAs（Path.Combine（fileuploadpath，filename））;
StatusLabel.Text =上传状态：文件已上传！;
//某种功能可以将路径输入到我的插入语法中？
OdbcCommand cmd = new OdbcCommand（INSERT INTO Pictures（UserID，picturepath）VALUES（''+ + UserUd +''，'+ fileuploadpath +'），cn）;
cmd.ExecuteNonQuery（）;
}
catch（Exception ex）
{
StatusLabel.Text =上传状态：文件无法上传，出现以下错误：+ ex.Message;
code
你可以在这行看到：
VALUES（''+ theUserId +''，'+ fileuploadpath +'） ，cn）;
我错过了文件名我试过了：
VALUES（'+ theUserId +'，'+ fileuploadpath，filename +''），cn）;
便宜的拍摄大声笑但值得一试我猜，它总是这样哭泣！
解决方案使用 CreateCommand 创建 OdbcCommand code>在 OdbcConnection 上。发布的代码不会将 cmd 与 cn 绑定。此外，您应该使用CommandParameters而不是内联值（以防止SQL注入攻击）。
OdbcCommand cmd = cn.CreateCommand （）;
cmd.CommandText =INSERT INTO图片（UserID，图片路径）VALUES（？，？）;
cmd.Parameters.Add（新的OdbcParameter（@ UserID，OdbcType.Int，theUserID））;
cmd.Parameters.Add（new OdbcParameter（@ picturepath，OdbcType.VarChar，fileuploadpath））;
cmd.ExecuteNonQuery（）;
I get an error in my code:
ExecuteNonQuery: Connection property has not been initialized.
This could be due to the line in this code:
OdbcCommand cmd = new OdbcCommand("INSERT INTO Pictures (UserID, picturepath) VALUES ('" + theUserId + "' , '" + fileuploadpath + "')");
Full code:
{
string theUserId = Session["UserID"].ToString();
{
OdbcConnection cn = new OdbcConnection("Driver={MySQL ODBC 3.51 Driver}; Server=localhost; Database=gymwebsite2; User=root; Password=commando;");
cn.Open();
}
if (FileUploadControl.HasFile)
{
try
{
string filename = Path.GetFileName(FileUploadControl.FileName);
//FileUploadControl.SaveAs(Server.MapPath("~/userdata/" + theUserId + "/uploadedimage/") + filename);
string fileuploadpath = Server.MapPath("~/userdata/" + theUserId + "/uploadedimage/");
FileUploadControl.SaveAs(Path.Combine(fileuploadpath, filename));
StatusLabel.Text = "Upload status: File uploaded!";
OdbcCommand cmd = new OdbcCommand("INSERT INTO Pictures (UserID, picturepath) VALUES ('" + theUserId + "' , '" + fileuploadpath + "')");
cmd.ExecuteNonQuery();
}
catch (Exception ex)
{
StatusLabel.Text = "Upload status: The file could not be uploaded. The following error occured: " + ex.Message;
}
}
}
}
There is another problem, I dont think its an insert I want as this is going to make a duplicate entry within my database, would it just be a case of changing it from INSERT INTO to UPDATE?
Also is there a way to overwrite upon uploading the image? Atm it just saving the image into the same folder as the one I already have? The first image or any image obviously isnt going to have the same file name so how would I go about overwriting any image in the folder with the one im uploading?
EDIT:
New error (fileupload works as its stored in the correct area but passing the fileupload to the insert statement is abit wonky)
I get the error
The file could not be uploaded. The following error occured: ERROR [42000] [MySQL][ODBC 3.51 Driver][mysqld-5.5.9]You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''C:\Users\Garrith\Documents\Visual Studio 2010\WebSites\WebSite1\userdata\1\uplo' at line 1
Which is kind of strange?
All im trying to do is save the filepath+filename in mydb my attempt at passing to the insert has obviously failed.
protected void UploadButton_Click(object sender, EventArgs e)
{
if (FileUploadControl.HasFile)
{
try
{
string theUserId = Session["UserID"].ToString();
OdbcConnection cn = new OdbcConnection("Driver={MySQL ODBC 3.51 Driver}; Server=localhost; Database=gymwebsite2; User=root; Password=commando;");
cn.Open();
string filename = Path.GetFileName(FileUploadControl.FileName);
//FileUploadControl.SaveAs(Server.MapPath("~/userdata/" + theUserId + "/uploadedimage/") + filename);
string fileuploadpath = Server.MapPath("~/userdata/" + theUserId + "/uploadedimage/");
FileUploadControl.SaveAs(Path.Combine(fileuploadpath, filename));
StatusLabel.Text = "Upload status: File uploaded!";
//some kind of function to take the path then enter it into my insert syntax?
OdbcCommand cmd = new OdbcCommand("INSERT INTO Pictures (UserID, picturepath) VALUES ('" + theUserId + "' , '" + fileuploadpath + "')", cn);
cmd.ExecuteNonQuery();
}
catch (Exception ex)
{
StatusLabel.Text = "Upload status: The file could not be uploaded. The following error occured: " + ex.Message;
}
}
}
}
As you can see on this line:
VALUES ('" + theUserId + "' , '" + fileuploadpath + "')", cn);
im missing the "filename" i tryed this:
VALUES ('" + theUserId + "' , '" + fileuploadpath, filename + "')", cn);
Cheap shot lol but worth a go I guess and it cryed as it always does!
解决方案 You want to create the OdbcCommand using CreateCommand on OdbcConnection. The code as posted does not tie cmd to cn. Also, you should use CommandParameters instead of inlining the values (to guard against SQL injection attacks).
OdbcCommand cmd = cn.CreateCommand();
cmd.CommandText = "INSERT INTO Pictures (UserID, picturepath) VALUES (?, ?)";
cmd.Parameters.Add(new OdbcParameter("@UserID", OdbcType.Int, theUserID));
cmd.Parameters.Add(new OdbcParameter("@picturepath", OdbcType.VarChar, fileuploadpath));
cmd.ExecuteNonQuery();
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