如何将表名作为参数传递给使用Gridview更新数据库中的表 [英] How to pass table name as parameter to update a table in database using Gridview

问题描述

嗨朋友

Hi Friends

I have Multiple Tables in my Access Database. And I want to Update a Particular Table through passing a parameter for table name in Gridview.

For Example if i want to update a table with name of Satya. i will type satya in textbox nd click on Search button particular table is updating to gridview. now i want to update the same table.

I tried my code as below.

private void btnSaveGrid_Click(object sender, EventArgs e)
       {
           using (OleDbConnection conn = new OleDbConnection(@"Provider=Microsoft.ACE.OLEDB.12.0;Data Source=C:\Users\Administrator\Documents\Database3.accdb"))
           {

               conn.Open();
               try
               {

                   foreach (DataGridViewRow row in dataGridView1.Rows)
                   {

                       if (!row.IsNewRow)
                       {
                           string tabl1 = textBox1.Text.ToString();
                           using (OleDbCommand cmd = new OleDbCommand("INSERT INTO tabl1 (EmpName, Sname) VALUES(@EmpName,@Sname)", conn))
                           {

                               cmd.Parameters.AddWithValue("@EmpName", row.Cells[1].Value);
                               cmd.Parameters.AddWithValue("@Sname", row.Cells[2].Value);


                               cmd.ExecuteNonQuery();
                               MessageBox.Show("success");
                           }
                       }
                   }
               }
               catch(Exception ex)
               {
                   MessageBox.Show(ex.ToString());
               }
               conn.Close();
           }

// below table Search Code is working good.

private void btnSearch_Click(object sender, EventArgs e)
        {
            string connectionString = @"Provider=Microsoft.ACE.OLEDB.12.0;Data Source=C:\Users\Administrator\Documents\Database3.accdb";
            OleDbConnection con = new OleDbConnection(connectionString);
            con.Open();

            try
            {
                OleDbCommand cmd = new OleDbCommand();
                cmd.Connection = con;
                String query = String.Format("Select * from {0}", textBox1.Text);
                cmd.CommandText = query;

                OleDbDataAdapter da = new OleDbDataAdapter(cmd);
                DataTable dt = new DataTable();
                da.Fill(dt);
                dataGridView1.DataSource = dt;


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

       }

推荐答案

你可以使用方法String.Format来解决这个问题。

You can use the method String.Format to solve this problem.

string queary = String.Format("INSERT INTO {0} (EmpName, Sname) VALUES (@EmpName, @Sname)", tabl1);

非常感谢老板。



按要求工作。

Thank You so much Boss.

It is Working as per requirement.

private void btnSaveGrid_Click(object sender, EventArgs e)
        {
            using (OleDbConnection conn = new OleDbConnection(@"Provider=Microsoft.ACE.OLEDB.12.0;Data Source=C:\Users\Administrator\Documents\Database3.accdb"))
            {

                conn.Open();
                try
                {

                    foreach (DataGridViewRow row in dataGridView1.Rows)
                    {

                        if (!row.IsNewRow)
                        {
                            string tabl1 = textBox1.Text.ToString();
                            string queary = String.Format("INSERT INTO {0} (EmpName, Sname) VALUES (@EmpName, @Sname)", tabl1);
                            using (OleDbCommand cmd = new OleDbCommand(queary, conn))
                            {

                                cmd.Parameters.AddWithValue("@EmpName", row.Cells[1].Value);
                                cmd.Parameters.AddWithValue("@Sname", row.Cells[2].Value);


                                cmd.ExecuteNonQuery();
                                MessageBox.Show("success");
                            }
                        }
                    }
                }
                catch(Exception ex)
                {
                    MessageBox.Show(ex.ToString());
                }
                conn.Close();
            }
        }

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