Browse an image file and upload into database

Charith Liyanagamage

Hi.....

I need a help again...........

I have a one TextBox and two Buttons in my application. One is for Browse and other is for Upload.
I want to browse an image file using Browse button (then the path file will appear in the text box) and upload that file into database.

Then i want to load this stored button into PictureBox using another button called View.

I found below code for browsing.....please support me to do the rest.

    private void button1_Click(object sender, EventArgs e)
        {
            OpenFileDialog fdlg = new OpenFileDialog();
            fdlg.Title = "Open File";
            fdlg.InitialDirectory = @"c:\";
            fdlg.Filter = "All files (*.*)|*.*|All files (*.*)|*.*";
            fdlg.FilterIndex = 2;
            fdlg.RestoreDirectory = true;
            if (fdlg.ShowDialog() == DialogResult.OK)
            {
                textBox1.Text = fdlg.FileName;
            }
        }

If u could please give me a hand to do this.

Thanks,
CharithMax

7 Replies

Know the answer? Post it — somebody with the same question will find it here.

Kirtan Patel17y ago
Accepted answer
Hi charith Here is Precedure i have written to Inset Image Data into SQl Database and Again Read it into Picture box

Here i Have defined two Columns in Database "ID int and "photo" with data type var-binary in database

// Procedure How to Insert Image into Database

using System.Data.SqlClient;
using System.IO;

private void btnInsertImage_Click(object sender, EventArgs e)
        {
            OpenFileDialog ofd = new OpenFileDialog();

            ofd.RestoreDirectory = true;

            if (ofd.ShowDialog() == DialogResult.OK)
            {
               textBox1.Text = ofd.FileName;
               byte[] photo = File.ReadAllBytes(textBox1.Text);
               SqlConnection con = new SqlConnection(@"Data Source=.\SQLEXPRESS;AttachDbFilename=|DataDirectory|\Database1.mdf;Integrated Security=True;User Instance=True");
               con.Open();
               SqlCommand comm = new SqlCommand("insert into ImageData(photo) values(@photo)", con);
               comm.Parameters.AddWithValue("@photo", photo);
               comm.ExecuteNonQuery();
               con.Close();
               MessageBox.Show("Data inserted");
            }
        }

// Read Image into PictureBox From database

        private void btnReadImageData_Click(object sender, EventArgs e)
        {
            byte[] photo;
            MemoryStream m = null;
            SqlConnection con = new SqlConnection(@"Data Source=.\SQLEXPRESS;AttachDbFilename=|DataDirectory|\Database1.mdf;Integrated Security=True;User Instance=True");
             con.Open();
             SqlCommand comm = new SqlCommand("select * from ImageData where id=1", con);
             SqlDataReader reader = comm.ExecuteReader();
             while ( reader.Read())
             {
                 photo= (byte[])(reader["photo"]);
                 m = new MemoryStream(photo);
             }
             pictureBox1.Image = Image.FromStream(m);
        }

Happy Coding :)
Charith Liyanagamage17y ago
Hello friend,

please send the same solution for MS Access 2007 db. And please mention what is the data type that i should select when creating access data table.

Thanks.
Kirtan Patel17y ago
You can Get photo Binary both Way

By using Index or using Column name

i n my code i called it with column name "photo" that is in my database

you can use both :)
Charith Liyanagamage17y ago
Hello Dude Kitran...

Ur code works well. But I got some Indexexception @ this point(Bold green coloured line) when retrieving the image.

while ( reader.Read())
             {
               // photo= (byte[])(reader["photo"]);
                 m = new MemoryStream(photo);
             }

so I modified the code line as below

   while (reader.Read())
            {
                //photo = (byte[])(reader["photo"]);

                photo = (byte[])(reader[1]); -------Modified line

                m = new MemoryStream(photo);

            }

Now it works properly....

Anyway Thank you for ur big help.............
Charith Liyanagamage17y ago
Thank You very much dude...
Charith Liyanagamage17y ago
Thank you very much
Master Billa17y ago
Hi

I think some more information i need provide here

1. your image store column database as "Image" DataType.

2. after you need read your image as byte, then you need send to db.

Save C# code

protected void btnSubmit_Click(object sender, EventArgs e)

    {

        SqlConnection connection = null;

        try

        {



            Byte[] imgByte = null;

            if ( textBox1.Text!="")

            {



                byte[] imgByte = File.ReadAllBytes(textBox1.Text);

            }

            // Insert the employee name and image into db

string conn = ConfigurationManager.ConnectionStrings ["EmployeeConnString"].ConnectionString;

            connection = new SqlConnection(conn);

            connection.Open();

string sql = "INSERT INTO EmpDetails(empimg) VALUES(@eimg) SELECT @@IDENTITY";

            SqlCommand cmd = new SqlCommand(sql, connection);

            cmd.Parameters.AddWithValue("@eimg", imgByte);

            int id = Convert.ToInt32(cmd.ExecuteScalar());



        }

        catch

        {

            lblResult.Text = "There was an error";

        }

        finally

        {

            connection.Close();

        }

    }
private void btnview_Click(object sender, EventArgs e)

        {

            byte[] imgeData;

            MemoryStream datastream = null;

            using (SqlConnection con = new SqlConnection(ConnectionString))
{

             con.Open();

             using(SqlCommand comm = new SqlCommand("select ImageColumn from EmpDetails where ID='"+YourID+"'", con)
{

             using(SqlDataReader reader = comm.ExecuteReader())
{

             while ( reader.Read())

             {

                 imgeData = (byte[])(reader["ImageColumn"]);

                 datastream = new MemoryStream(imgeData);

             }
pictureBox1.Image = Image.FromStream(datastream );
}
}
}



        }
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