Muat gambar ke C # dan kemudian masukkan ke tabel MySQL

Saya akan menawarkan dua solusi. Solusi pertama adalah menyimpan gambar mentah dalam byte dalam database secara langsung. Solusi kedua adalah apa yang saya rekomendasikan secara pribadi - yaitu menggunakan jalur file gambar di database sebagai gantinya.

Berikut kutipan dari artikel yang memunculkan beberapa poin bagus tentang perlu atau tidaknya BLOB.

Inilah cara Anda memilih file gambar:

using (var openFileDialog = new OpenFileDialog())
{
   openFileDialog.Title = "Choose Image File";
   openFileDialog.InitialDirectory =
                Environment.GetFolderPath(Environment.SpecialFolder.MyPictures);
   openFileDialog.Filter = "Image Files (*.bmp, *.jpg)|*.bmp;*.jpg";
   openFileDialog.Multiselect = false;
   if (openFileDialog.ShowDialog() == DialogResult.OK)
   {
       pictureBox1.Image = new Bitmap(openFileDialog.FileName);
   }
   // store file path in some field or textbox...
   textBox1.Text = openFileDialog.FileName;
}

Solusi 1:Pendekatan BLOB

// Write to database like this - image is LONGBLOB type
string sql = "INSERT INTO imagetable (image) VALUES (@file)";
// remember 'using' statements to efficiently release unmanaged resources
using (var conn = new MySqlConnection(cs))
{
    conn.Open();
    using (var cmd = new MySqlCommand(sql, conn))
    {
        // parameterize query to safeguard against sql injection attacks, etc. 
        cmd.Parameters.AddWithValue("@file", File.ReadAllBytes(textBox1.Text));
        cmd.ExecuteNonQuery();
    }
}

// read image from database like this
string sql = "SELECT image FROM imagetable WHERE ID = @ID";
using (var conn = new MySqlConnection(cs))
{
   conn.Open();
   using (var cmd = new MySqlCommand(sql, conn))
   {
      cmd.Parameters.AddWithValue("@ID", myInt);
      byte[] bytes = (byte[])cmd.ExecuteScalar();   
      using (var byteStream = new MemoryStream(bytes))
      {
         pictureBox1.Image = new Bitmap(byteStream);
      }
   }
}

Solusi 2:Menyimpan Jalur File di Sistem File

// Some file movement to the desired project folder
string fileName = Path.GetFileName(this.textBox1.Text);
string projectFilePath = Path.Combine(projectDir, fileName);
File.Copy(this.textBox1.Text, projectFilePath);

// Write to database like this - imagepath is VARCHAR type
string sql = "INSERT INTO imagepathtable (imagepath) VALUES (@filepath)";
using (var conn = new MySqlConnection(cs))
{
    conn.Open();
    using (var cmd = new MySqlCommand(sql, conn))
    {
        cmd.Parameters.AddWithValue("@filepath", projectFilePath);
        cmd.ExecuteNonQuery();
    }
}

// read from database like this
string sql = "SELECT imagepath FROM imagepathtable WHERE ID = @ID";
using (var conn = new MySqlConnection(cs))
{
    conn.Open();
    using (var cmd = new MySqlCommand(sql, conn))
    {
        cmd.Parameters.AddWithValue("@ID", myInt);
        pictureBox1.Image = new Bitmap(cmd.ExecuteScalar().ToString());
    }
}