Insert multiple values from textbox and show them into a datagridview

Satyapriya Nayak

7 Comments

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Virender Kumar8y ago
I got error : - column 'LoonNo.' doesn't belong to table. in below code
Virender Kumar8y ago
Namespace IncentiveCalculator{ public partial class frmInCal : Form { public frmInCal() { InitializeComponent(); } DataTable dt = new DataTable(); DataRow dr; private void Form1_Load(object sender, EventArgs e) { comboBox1.Items.Add("A"); comboBox1.Items.Add("B"); comboBox1.Items.Add("C"); comboBox1.Items.Add("GEN"); comboBox1.SelectedItem = "A"; cbLoomType.Items.Add("Small"); cbLoomType.Items.Add("Large"); cbLoomType.SelectedItem = "Small"; } private void btnExit_Click(object sender, EventArgs e) { DialogResult dialogResult = MessageBox.Show("Are you sure?", "Incentive Calculator", MessageBoxButtons.YesNo); if (dialogResult == DialogResult.Yes) { this.Close(); } else { } } private void btnAdd_Click(object sender, EventArgs e) { dr = dt.NewRow(); dr["LoomNo."]=txtLoomNo.Text; dr["LoomType"]=cbLoomType.SelectedItem; dr["WeaverID"]=txtWId.Text; dr["QualityCode"]=txtQCode.Text; dr["Points"]=txtPoints.Text; dt.Rows.Add(dr); DataGrid1.DataSource = dt; txtLoomNo.Text = ""; txtWId.Text = ""; txtQCode.Text = ""; txtWId.Text = ""; } } }
Virender Kumar8y ago
I got below error:-
Divya Diya8y ago
Its quite easy for beginners to understand
Nithin Varghese9y ago
Its good and very helpful
Gokul10y ago
it's good
Nilesh Jadav11y ago
Nice one

In this blog we will know how to insert multiple values from textbox and show them into a datagridview.

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;

namespace Add_multiple_rows_to_datagridview
{

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

DataTable dt = new DataTable();
DataRow dr;

private void btn_add_Click(object sender, EventArgs e)
{

dr = dt.NewRow();
dr["Name"] = txt_name.Text;
dr["Address"] = txt_address.Text;
dr["Age"] = txt_age.Text;
dt.Rows.Add(dr);
dataGridView1.DataSource = dt;
clear();
}

private void Form1_Load(object sender, EventArgs e)
{
dt.Columns.Add("Name");
dt.Columns.Add("Address");
dt.Columns.Add("Age");
}

private void clear()
{
txt_name.Text = "";
txt_address.Text = "";
txt_age.Text = "";
}

}
}