double image insert

See more:

C#

WPF

MySQL

I use WPF and C# in this project.
There is only an Insert button and a datagrid.
First I choose an image from datagrid,
then I press Insert button.
In my case when I press insert once, two images are inserted to my DB at the same time.
Where is the problem in my code?:

<window x:class="PupilReg.MainWindow" xmlns:x="#unknown">
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
Title="Image Insert" Height="674" Width="1009" WindowStartupLocation="CenterScreen" FontSize="15" FontWeight="Bold" ResizeMode="CanResizeWithGrip">
<grid margin="0,10,-8,-50">
<datagrid name="dataGrid1" height="565" width="435" margin="474,53,100,65">
<datagrid.itembindinggroup>
<bindinggroup>

<datagrid.columns>
<datagridtextcolumn header="Image_ID" binding="{Binding Path=Image_ID}" width="80">
<datagridtemplatecolumn header="Image_BLOB" width="200" isreadonly="True">
<datagridtemplatecolumn.celltemplate>
<datatemplate>
<Image Source="{Binding Path=Image_BLOB}" Width="160" Height="160" />





<Button x:Name="btnInsert" Content="Insert" HorizontalAlignment="Left" Margin="93,117,0,0" VerticalAlignment="Top" Width="75" Click="btnInsert_Click"/>



using MySql.Data.MySqlClient;
using System;
using System.Data;
using System.Windows;

namespace PupilReg
{
public partial class MainWindow : Window
{
string constr = "Data Source=localhost;port=3306;Initial Catalog=test;User Id=root;password=2525";
public MainWindow()
{
InitializeComponent();
BindGrid1();
}
public void BindGrid1()
{
MySqlConnection con = new MySqlConnection(constr);
con.Open();
MySqlCommand cmd = new MySqlCommand("SELECT * FROM images", con);
MySqlDataAdapter da = new MySqlDataAdapter(cmd);
DataTable dt = new DataTable();
da.Fill(dt);
dataGrid1.ItemsSource = dt.DefaultView;
}
private void btnInsert_Click(object sender, RoutedEventArgs e)
{
DataRowView SelectedRowValue = (DataRowView)dataGrid1.SelectedValue;
byte[] ImageBytes = (byte[])SelectedRowValue.Row.ItemArray[1];
MySqlConnection con = new MySqlConnection(constr);
MySqlCommand cmd = new MySqlCommand("INSERT INTO images (Image_BLOB) VALUES (@ImageSource)", con);
cmd.Parameters.Add("@ImageSource", MySqlDbType.Blob, ImageBytes.Length).Value = ImageBytes;
con.Open();
int i = cmd.ExecuteNonQuery();
if (i > 0)
{
cmd.ExecuteNonQuery();
MessageBox.Show("Image Inserted");
}
con.Close();
BindGrid1();
}
}
}