我想給你帶來這個頁面的完整代碼。
想法是:我有登記頁面。我通過存儲過程將數據傳遞給數據庫(DataInserter
)。
問題在於:在註冊期間,用戶從他希望回答的字母的下拉列表中選擇數字,最後,其他DB上的字母(IncomeLetters.tbl
)的狀態將從「未決「('1')改爲」已發佈「('2')。
我想,我可以清楚我點你,感謝您的支持!
protected void Button1_Click(object sender, EventArgs e)
{
SqlConnection conn = new SqlConnection(ConfigurationManager.ConnectionStrings["MaktubhoConnectionString2"].ConnectionString);
using (SqlCommand comm = new SqlCommand("DataInserter", conn))
{
comm.CommandType = CommandType.StoredProcedure;
comm.Connection = conn;
SqlParameter employeeparam = new SqlParameter("EmployeeSentIndex", int.Parse(ddlemployee.SelectedItem.Value));
SqlParameter doctypeparam = new SqlParameter("doctype_ID", int.Parse(ddldoctype.SelectedItem.Value));
SqlParameter doccharparam = new SqlParameter("docchar_ID", int.Parse(ddldocchar.SelectedItem.Value));
SqlParameter authorityparam = new SqlParameter("authority", txtauthority.Text);
SqlParameter subjectparam = new SqlParameter("subject", txtsubject.Text);
DateTime dt = DateTime.Now;
string todasdate = dt.ToString("d", CultureInfo.CreateSpecificCulture("de-DE"));
SqlParameter entrydateparam = new SqlParameter("entrydate", todasdate);
string Pathname = "UploadImages/" + Path.GetFileName(FileUpload1.PostedFile.FileName);
SqlParameter imagepathparam = new SqlParameter("image_path", Pathname);
SqlParameter loginparam = new SqlParameter("login", "jsomon");
comm.Parameters.Add(employeeparam);
comm.Parameters.Add(doctypeparam);
comm.Parameters.Add(doccharparam);
comm.Parameters.Add(authorityparam);
comm.Parameters.Add(subjectparam);
comm.Parameters.Add(entrydateparam);
comm.Parameters.Add(imagepathparam);
comm.Parameters.Add(loginparam);
comm.Parameters.Add("@forlabel", SqlDbType.VarChar, 100);
comm.Parameters["@forlabel"].Direction = ParameterDirection.Output;
FileUpload1.SaveAs(Server.MapPath("~/UploadImages/" + FileUpload1.FileName));
string ansTo = ddlAnswerTo.SelectedItem.Value;
SqlParameter answertoparam = new SqlParameter("answertoparam", ansTo);
string commandText = "update IncomeLetters set IncomeLetters.docState_ID = '2' where income_number = @answertoparam";
SqlCommand findincomelett = new SqlCommand(commandText, conn);
findincomelett.Parameters.Add(answertoparam);
conn.Open();
findincomelett.ExecuteNonQuery();
comm.ExecuteNonQuery();
lblresult.Visible = true;
Image1.Visible = true;
lblresult.Text = "Document number:";
lblnumber.Visible = true;
lblnumber.Text = (string)comm.Parameters["@forlabel"].Value; ;
conn.Close();
}
txtauthority.Text = "";
txtsubject.Text = "";
}
如果你需要修改你的問題,**編輯**你的問題 - 不要發佈*虛擬*回答你自己的問題..... –