如何将文本框值添加到 Access 数据库？答案

【问题标题】：How do i add textbox values to Access database?如何将文本框值添加到 Access 数据库？
【发布时间】：2018-04-17 00:38:50
【问题描述】：

我想将文本框值添加到 access 数据库中的相关列，连接已经建立，但是当我单击提交按钮时，没有添加值。这是我尝试过的代码，感谢任何帮助

protected void Button1_Click(object sender, EventArgs e)
{
    string EmailAddress = TextBox1.Text;
    string UserName = TextBox2.Text;
    string Password = TextBox3.Text;

    try
    {
        OleDbConnection con = new OleDbConnection(@"Provider = Microsoft.ACE.OLEDB.12.0; Data Source = C:\Users\Bheki Ndhlovu\source\WebSites\WebSite8\App_Data\UserDatabase.accdb; Persist Security Info = False;");

        OleDbCommand cmd = new OleDbCommand();
        cmd = new OleDbCommand("INSERT INTO User(EmailAddress, UserName, Password) VALUES(@EmailAddress, @UserName, @Password)");
        con.Open();


         if (con.State == ConnectionState.Open)
         {
             TextBox1.Text = "sssss";

             cmd.Parameters.Add("@EmailAddress", OleDbType.VarChar).Value = TextBox1.Text;
             cmd.Parameters.Add("@UserName", OleDbType.VarChar).Value = TextBox2.Text;
             cmd.Parameters.Add("@Password", OleDbType.VarChar).Value = TextBox3.Text;

            cmd.ExecuteNonQuery();
            con.Close();

         }


    }
    catch (Exception error)
    {
        //Show error message as    error.Message
    }

}

【问题讨论】：

标签： c# asp.net .net vb.net ms-access

【解决方案1】：

尝试添加connection string 和OleDbCommand。

cmd = new OleDbCommand("INSERT INTO User(EmailAddress, UserName, Password) VALUES(@EmailAddress, @UserName, @Password)",con);

【讨论】：

【解决方案2】：

这是一个示例，所有数据操作都驻留在一个类中。如果添加新记录成功，则返回新的主键。失败时，您可以查询引发失败问题的异常。

using System;
using System.Windows.Forms;
using System.Data.OleDb;
using System.IO;

namespace MS_AccessAddNewRecord_cs
{
    public partial class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();
        }

        private void addRecordButton_Click(object sender, EventArgs e)
        {
            var ops = new Operations();
            var newId = 0;
            if (ops.AddNewRow(companyTextBox.Text, contactNameTextBox.Text, contactTitleTextBox.Text, ref newId))
            {
                newIdentifierTextBox.Text = $"{newId}";
            }
            else
            {
                MessageBox.Show($"{ops.Exception.Message}");
            }
        }
    }
    /// <summary>
    /// This class should be in a separate class file, I placed it here for easy of learning
    /// </summary>
    public class Operations
    {
        private OleDbConnectionStringBuilder Builder = new OleDbConnectionStringBuilder
        {
            Provider = "Microsoft.ACE.OLEDB.12.0",
            DataSource = Path.Combine(AppDomain.CurrentDomain.BaseDirectory, "Database1.accdb")
        };

        private Exception mExceptiom;
        public Exception Exception
        {
            get
            {
                return mExceptiom;
            }
        }
        /// <summary>
        /// Add a new record, upon success return the new primary key for the record in pIdentifier parameter
        /// </summary>
        /// <param name="pName"></param>
        /// <param name="pContactName"></param>
        /// <param name="pContactTitle"></param>
        /// <param name="pIdentfier"></param>
        /// <returns></returns>
        public bool AddNewRow(string pName, string pContactName, string pContactTitle, ref int pIdentfier)
        {
            bool Success = true;

            try
            {
                using (OleDbConnection cn = new OleDbConnection { ConnectionString = Builder.ConnectionString })
                {
                    using (OleDbCommand cmd = new OleDbCommand { Connection = cn })
                    {
                        cmd.CommandText = "INSERT INTO Customers (CompanyName,ContactName, ContactTitle) " + 
                                                          "Values(@CompanyName,@ContactName, @ContactTitle)";

                        cmd.Parameters.AddWithValue("@CompanyName", pName);
                        cmd.Parameters.AddWithValue("@ContactName", pContactName);
                        cmd.Parameters.AddWithValue("@ContactTitle", pContactTitle);

                        cn.Open();

                        int Affected = cmd.ExecuteNonQuery();

                        if (Affected == 1)
                        {
                            cmd.CommandText = "Select @@Identity";
                            pIdentfier = Convert.ToInt32(cmd.ExecuteScalar());
                        }
                    }
                }
            }
            catch (Exception ex)
            {
                Success = false;
                mExceptiom = ex;
            }

            return Success;

        }
    }
}

【讨论】：

【解决方案3】：

也许在 Page_Load 方法中您没有 if(!isPostback)，因此在执行 Button1_Click 方法之前，TextBoxes 的值会在 postback 上重置。

【讨论】：

【解决方案4】：

如果 EmptyWaterHole 的回答不是问题，是不是连接出错了？

确保“VarChar”是每个字段的正确数据类型。

此外，请确保值不超过大小（即：如果您将字段设置为最多允许 25 个字符并且您的值超过 25 个字符，则不会添加该值）。

此外，如果您不允许空值并且其中一个值超出限制，则不会添加整条记录。

【讨论】：

【解决方案5】：

先生。饥饿的。像这样试试。

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;
using System.Data.OleDb;

namespace WindowsFormsApplication2
{
    public partial class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();
        }

        private void button1_Click(object sender, EventArgs e)
        {

            OleDbConnection conn;
            conn = new OleDbConnection(@"Provider=Microsoft.Jet.OleDb.4.0;Data Source=C:\your_path_here\Northwind.mdb");

            conn.Open();

            OleDbCommand cmd = conn.CreateCommand();

            cmd.CommandText = @"INSERT INTO MyExcelTable([Fname], [Lname],  [Address])VALUES('" + textBox1.Text + "', '" + textBox2.Text + "','" + textBox3.Text + "')";
            cmd.ExecuteNonQuery();
            conn.Close();

        }

        public OleDbConnection myCon { get; set; }

        private void button2_Click(object sender, EventArgs e)
        {

            OleDbConnection conn = new OleDbConnection();
            conn.ConnectionString = @"Provider=Microsoft.ACE.OLEDB.12.0;Data Source=C:\Users\Ryan\Desktop\Coding\Microsoft Access\Northwind.mdb";

            string fstName  = textBox1.Text.Trim();
            string lstName  = textBox2.Text.Trim();
            string adres = textBox3.Text.Trim();
            OleDbCommand cmd = new OleDbCommand(@"INSERT INTO MyExcelTable (FName, LName, Address) VALUES (@FName, @LName, @Address)")
            {
                Connection = conn
            };

            conn.Open();

            if (conn.State == ConnectionState.Open)
            {
                // you should always use parameterized queries to avoid SQL Injection
                cmd.Parameters.Add("@FName", OleDbType.VarChar).Value = fstName;
                cmd.Parameters.Add("@LName", OleDbType.VarChar).Value = lstName;
                cmd.Parameters.Add("@Address", OleDbType.VarChar).Value = adres;

                try
                {
                    cmd.ExecuteNonQuery();
                    MessageBox.Show(@"Data Added");
                    conn.Close();
                }
                catch (OleDbException ex)
                {
                    MessageBox.Show(ex.Source + "\n" + ex.Message);
                    conn.Close();
                }
            }
            else
            {
                MessageBox.Show(@"Connection Failed");
            }
        }
        }
    }

【讨论】：

【解决方案6】：

如果你使用 access 作为你的数据库，试试这个它会起作用

try
{
    OleDbCommand command = new OleDbCommand();
    command.Connection = connection;
    command.CommandText = "INSERT INTO REPORT (patientName,tel,hostel,id no,department,diagnose,gender) values(@patientName,@tel,@hostel,@id no,@department,@diagnose,@gender)";
    connection.Open();
    command.Parameters.AddWithValue("@patientName", textBox1.Text);
    command.Parameters.AddWithValue("@tel", textBox2.Text);
    command.Parameters.AddWithValue("@hostel", textBox3.Text);
    command.Parameters.AddWithValue("@id no", textBox4.Text);
    command.Parameters.AddWithValue("@department", textBox5.Text);
    command.Parameters.AddWithValue("@diagnose", richTextBox1.Text);
    command.Parameters.AddWithValue("@gender", textBox6.Text);
    command.ExecuteNonQuery();
    connection.Close();
    MessageBox.Show("Patient record Have been save successfully....");

}
catch (Exception ex)
{
    MessageBox.Show("error" + ex);
}

【讨论】：