2016-08-30 38 views
1

我只是一個初學者,試圖按照教程。我確實搜索了最接近的問題,嘗試了其中的幾個。他們都不適合我。我只需要一些可能出錯的線索。謝謝。C#web表單提交的數據無法插入sql數據庫

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.SqlClient; 

namespace Company 
{ 
    public partial class ContactUs : System.Web.UI.Page 
    { 
     protected void Page_Load(object sender, EventArgs e) 
     { 

     } 

     public void update(string FirstName, string LastName, string EmailAddress, string MobileNumber, string category, string message) 
     { 
      SqlConnection conn = new SqlConnection(GetConnectionString()); 
      string sql = "INSERT INTO ContactInformation (GUID,FirstName, LastName, EmailAddress, Number, Category, Message) VALUES " 
         + " (@GUID,@FirstName,@LastName,@EmailAddress,@Number,@category,@message)"; 

       Guid guid = Guid.NewGuid(); 
       string guidString = guid.ToString(); 

      try 
      { 
       conn.Open(); 
       SqlCommand cmd = new SqlCommand(sql, conn); 
       SqlParameter[] p = new SqlParameter[7]; 
       p[0] = new SqlParameter("@GUID", SqlDbType.UniqueIdentifier,50); 
       p[1] = new SqlParameter("@FirstName", SqlDbType.VarChar,50); 
       p[2] = new SqlParameter("@LastName", SqlDbType.VarChar,50); 
       p[3] = new SqlParameter("@EmailAddress", SqlDbType.VarChar,50); 
       p[4] = new SqlParameter("@Number", SqlDbType.VarChar,50); 
       p[5] = new SqlParameter("@Category", SqlDbType.VarChar,50); 
       p[6] = new SqlParameter("@Message", SqlDbType.VarChar,50); 
       p[0].Value = guid; 
       p[1].Value = FirstName; 
       p[2].Value = LastName; 
       p[3].Value = EmailAddress; 
       p[4].Value = MobileNumber; 
       p[5].Value = category; 
       p[6].Value = message; 

       for (int i = 0; i < p.Length; i++) 
       { 
        cmd.Parameters.Add(p[i]); 
       } 
       cmd.CommandType = CommandType.Text; 
       cmd.ExecuteNonQuery(); 
      } 
      catch (System.Data.SqlClient.SqlException ex) 
      { 
       string msg = "Insert Error:"; 
       msg += ex.Message; 
       throw new Exception(msg); 
      } 
      finally 
      { 
       conn.Close(); 
      } 
     } 

     public string GetConnectionString() 
     { 
      return System.Configuration.ConfigurationManager.ConnectionStrings["ContactInformation"].ConnectionString; 
     } 
     protected void btnSubmit_Click(object sender, EventArgs e) 
     { 
      if (!IsPostBack) 
      { 
       Validate(); 
       update(txtFirstName.Text, 
            txtLastName.Text, 
            txtEmail.Text, 
            txtNumber.Text, 
            DropDownList1.SelectedItem.Text, 
            txtMessage.Text); 
       Response.Write("Record was successfully added!"); 
       //ClearForm(Page); 
      } 
     } 
     public static void ClearForm(Control Parent) 
     { 
      if (Parent is TextBox) 
      { (Parent as TextBox).Text = string.Empty; } 
      else 
      { 
       foreach (Control c in Parent.Controls) 
        ClearForm(c); 
      } 
     } 
    } 
} 

連接字符串位於web.config文件中。

<connectionStrings> 
    <add name="ContactInformation" connectionString="datasource=database;initial catalog=ContactInformation;Integrated Security=SSPI;userid=xxx;password=xxx;"/> 
</connectionStrings> 

我部署了我的本地計算機上的網站,並嘗試測試它在互聯網上Explorer.And在另一邊,我打開數據庫管理工具,該表仍是空的。

+0

嗨@MethodMan對不起,我不熟悉張貼。我剛編輯我的代碼。我沒有收到代碼錯誤。我只是無法將任何數據插入數據庫。 – GoResistanceLi

+0

您是否嘗試過使用調試器逐行調試代碼行?它是否遵循了您期望的代碼路徑? – mason

+0

我認爲你應該看看如何正確創建參數,當你逐步完成應該能夠評估的代碼和所有參數值時。也在這一行上放置一個斷點,看看它是否到達這一行'string msg =「Insert Error:」; 1'你還應該將你的Sql對象包裝在一個'using(){}'的周圍,而不是處理所有的創建對象 – MethodMan

回答

0

您正在測試回發按鈕點擊。這可能是TRUE而不是FALSE。改變你的方法。如果您看到它寫出「回發」字符串,您需要將您的支票更改爲IsPostBack而不是!IsPostBack 此外,不確定什麼是Validate(),我沒有看到它的定義。可能會導致異常。如果您只是爲了測試插頁而出現問題,請將其註釋掉。

protected void btnSubmit_Click(object sender, EventArgs e) 
     { 
      if (!IsPostBack) 
      { 
       Validate(); 
       update(txtFirstName.Text, 
            txtLastName.Text, 
            txtEmail.Text, 
            txtNumber.Text, 
            DropDownList1.SelectedItem.Text, 
            txtMessage.Text); 
       Response.Write("Record was successfully added!"); 
       //ClearForm(Page); 
      } 
      else Response.Write("Postback"); 
     }