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我有这个 ASPX 代码：

<form name="AddArticle" action="Register.aspx" method="post">
<b>
title :<br /></b><input id="Text1" type="text" name="ArticleTitle"/><p></p>
<b>

date: <br /></b><input id="Text2" type="text" name="ArticleDate"/><p></p>

<b>
author : <br /></b><input id="Text3" type="text" name="ArticleAuthor"/><p></p>
<b>
text: <br /></b> <textarea rows="10" cols="60"  name="ArticleBody"></textarea>
<br />
<input id="Reset1" type="reset" value="clean" />
<input id="Submit1" type="submit" value="send" /></form>

我有这个 ASPX.C# 代码：

using System;
using System.Collections;
using System.Configuration;
using System.Data;
using System.Linq;
using System.Web;
using System.Web.Security;
using System.Web.UI;
using System.Web.UI.HtmlControls;
using System.Web.UI.WebControls;
using System.Web.UI.WebControls.WebParts;
using System.Xml.Linq;
using System.Data.OleDb;

public partial class Register : System.Web.UI.Page
{

    protected void Page_Load(object sender, EventArgs e)
    {
        string ArticleTitle = "", ArticleBody = "", ArticleAuthor = "", ArticleDate = "";
        if (IsPostBack)
        {
            try
            {
                ArticleTitle = Request.Form["ArticleTitle"].ToString();
                ArticleDate = Request.Form["ArticleDate"].ToString();
                ArticleAuthor = Request.Form["ArticleAuthor"].ToString();
                ArticleBody = Request.Form["ArticleBody"].ToString();


                string dpath = Server.MapPath(@"App_Data") + "/MySite.mdb";
                string connectionstring = @"Data source='" + dpath + "';Provider='Microsoft.Jet.OLEDB.4.0';";
                OleDbConnection con = new OleDbConnection(connectionstring);
                string QuaryString = string.Format("insert into tblArticles(ArticleTitle,ArticleBody, ArticleAuthor, PostDate) values ('{0}','{1}','{2}','{3}')", ArticleTitle, ArticleBody, ArticleAuthor, ArticleDate);
                OleDbCommand cmd = new OleDbCommand(QuaryString, con);
                con.Open();
                cmd.ExecuteNonQuery();

            }
            catch
            {
                Response.Redirect("Default.aspx");
            }
        }
    }
    }

我的问题是为什么当我点击“发送”按钮时，表单中的信息（标题，作者......）没有插入到数据库中？我该如何解决？希望得到帮助，谢谢。

4

2 回答 2

1

最好的方法是使用按钮单击事件，但在您的情况下。您可以为每个元素使用 AutoPostBack 属性：

<asp:TextBox ID="ArticalTitle" runat="server" Text="Button"  AutoPostBack="true"/>

当用户更改此元素的焦点时（按 TAB 等），代码将回发文本框中的值。但是在这种情况下，您的值会一一传递，因此您需要在尝试将它们写入数据库之前存储这些值，或者您可以使用面板控件并在面板控件之外放置一个虚拟按钮以更改焦点：

<asp:Panel ID="Panel1" runat="server" AutoPostBack="true">
    <asp:TextBox ID="TextBox1" Text="asd" runat="server" />
    <asp:TextBox ID="TextBox2" Text="asd" runat="server" />
</asp:Panel>
<asp:Button ID="Button1" runat="server" Text="Button" />

当用户单击按钮面板控件自动回发时，您可以捕获文本框中的值。

希望这可以帮助。

于 2013-01-05T10:51:18.633 回答
1

我认为您应该编写一个按钮单击事件来调用该函数：

。CS

  protected void Button1_Click(object sender, EventArgs e)
    {
        ArticleTitle = Request.Form["ArticleTitle"].ToString();
            ArticleDate = Request.Form["ArticleDate"].ToString();
            ArticleAuthor = Request.Form["ArticleAuthor"].ToString();
            ArticleBody = Request.Form["ArticleBody"].ToString();


            string dpath = Server.MapPath(@"App_Data") + "/MySite.mdb";
            string connectionstring = @"Data source='" + dpath + "';Provider='Microsoft.Jet.OLEDB.4.0';";
            OleDbConnection con = new OleDbConnection(connectionstring);
            string QuaryString = string.Format("insert into tblArticles(ArticleTitle, PostDate) values ('{0}','{1}','{2}','{3}')", ArticleTitle, ArticleBody, ArticleAuthor, ArticleDate);
            OleDbCommand cmd = new OleDbCommand(QuaryString, con);
            con.Open();
            cmd.ExecuteNonQuery();
    }

.aspx：

<asp:Button ID="Button1" runat="server" Text="Button" OnClick="Button1_Click" />

在 web.config 中使用 StoredProcedure： connectString：</p> 

<add name="CGB_GlobalInfoConnectionString" connectionString="Data Source=127.0.0.1;Initial Catalog=TestDB;Persist Security Info=True;User ID=sa;Password=11111" providerName="System.Data.SqlClient" />

。CS：

SqlConnection connection = new SqlConnection(DataManager.DBConnectString);
            try
            {
                SqlCommand cmd = new SqlCommand();
                cmd.CommandType = CommandType.StoredProcedure;
                cmd.Connection = connection;
                cmd.CommandText = "Insert_AddSomething";
                cmd.Parameters.Add("@UserID", SqlDbType.Int).Value = userid;

                connection.Open();
                cmd.ExecuteNonQuery();

                }
于 2013-01-05T10:05:57.577 回答