本文介绍了我总是犯这个错误。的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.UI;
using System.Web.UI.WebControls;
using System.Data;
using System.Data.OleDb;
using System.Web.Security;
public partial class Admin_uyeekle : System.Web.UI.Page
{
string baglantistringi = "Provider=Microsoft.ACE.OLEDB.12.0;Data Source=" +
HttpContext.Current.Server.MapPath("~/App_Data/VT.accdb") +
";Persist Security info=false";
OleDbConnection Baglanti;
OleDbDataAdapter Adaptor;
DataTable qwe;
OleDbCommand ab;
public static int IDKullanici;
protected void cusval1_ServerValidate(object source, ServerValidateEventArgs args)
{
string cevap =" 16";
string kulcevabı = args.Value;
if(cevap!=kulcevabı)
{
args.IsValid = false;
}
}
protected void btn_kaydet_Click(object sender, EventArgs e)
{
string isim = txt_ad.Text;
string soyisim = txt_soyad.Text;
string Email = txt_email.Text;
string Sifre = txt_sifre.Text;
string SifreTekrar = txt_sifretekrar.Text;
string kayitekle;
if (KullaniciEmailKontrolEt() != 0)
{
lbl_uyari.Visible = true;
lbl_uyari.Text = "ayni email adresi ile kayıt yapılmıştır.";
return;
}
kayitekle = "INSERT into Kullanicilar(isim,soyisim,mail,sifre,kullanicitipi) Values(@isim,@soyisim,@mail,@sifre,@kullanicitipi)";
ab = new OleDbCommand(kayitekle, Baglanti);
ab.Parameters.Add("@isim", OleDbType.Char, 100).Value = isim;
ab.Parameters.Add("@soyisim", OleDbType.Char, 100).Value = isim;
ab.Parameters.Add("@mail", OleDbType.Char, 100).Value = Email;
ab.Parameters.Add("@sifre", OleDbType.Char, 50).Value = Sifre;
ab.Parameters.Add("@kullanicitipi", OleDbType.Char, 50).Value = "UYE";
try
{
Baglanti.Open();
ab.ExecuteNonQuery();
Baglanti.Close();
lbl_uyari.Text = "Kayıt Başarılı";
}
catch (Exception hata)
{
lbl_uyari.Text = "Kayıt yapılamadı";
}
}
public int KullaniciEmailKontrolEt()
{
DataTable dataEmailKontrol = new DataTable();
string abc = "Select mail From uye Where mail=@mail";
OleDbCommand ab = new OleDbCommand(abc, Baglanti);
ab.Parameters.Add("@mail", OleDbType.Char, 50).Value = txt_email.Text;
Adaptor = new OleDbDataAdapter(ab);
if (Baglanti.State == ConnectionState.Closed)
{
Baglanti.Open();
}
Adaptor.Fill(dataEmailKontrol);
if (Baglanti.State == ConnectionState.Open)
{
Baglanti.Close();
}
return dataEmailKontrol.Rows.Count;
}
}
system.NullReferenceException
system.NullReferenceException
推荐答案
OleDbConnection Baglanti;
ab = new OleDbCommand(kayitekle, Baglanti);
Baglanti.Open();
Baglanti.Close();
OleDbCommand ab = new OleDbCommand(abc, Baglanti);
if (Baglanti.State == ConnectionState.Closed)
Baglanti.Open();
if (Baglanti.State == ConnectionState.Open)
Baglanti.Close();
你实际上从来没有给它赋值......在某个地方,你需要添加
You never actually assign a value to it...somewhere, you need to add
Baglanti = new OleDbConnection(baglantistringi);
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