很棒的一個程式 C# 寫入/讀取 Access 圖檔

2018-05-27

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Access

讀取Access 檔案資料 by Oledb

最近花了我數小時的時間,在網路世界裡,讓我看到一些不凡的高手,留下一些很棒的程式,

我將內容留下來,做為我的參考,怕未來這些網站消失,就看不到那麼好的程式

資料來源 : https://social.msdn.microsoft.com/Forums/windows/en-US/98d68c15-b28c-4f3e-92d7-e5b4aa6c2554/how-to-load-image-from-acces-database-and-put-it-in-picturebox-c?forum=winformsdatacontrols

Then the code to save the picture to database:

private void button1_Click(object sender, EventArgs e)
{
   string fileName = @"D:\Pictures\Pic.PNG";
   string strConn = @"Provider=Microsoft.ACE.OLEDB.12.0;Data Source=D:\Documents\Database1.accdb";
   FileInfo fi = new FileInfo(fileName);
   using (FileStream fs = fi.OpenRead())
   {
      byte[] bytes = new byte[fs.Length];
      fs.Read(bytes, 0, Convert.ToInt32(fs.Length));
      using (OleDbConnection cn = new OleDbConnection(strConn))
      {
          cn.Open();
          using (OleDbCommand cm = new OleDbCommand())
          {
             cm.Connection = cn;
             cm.CommandType = CommandType.Text;
             cm.CommandText = "insert into Table3 (Img) values(@file)";
             OleDbParameter spFile = new OleDbParameter("@file", SqlDbType.Image);
             spFile.Value = bytes;
             cm.Parameters.Add(spFile);
             cm.ExecuteNonQuery();
             MessageBox.Show("Saved successfully");
          }
      }
   }
}

Then the code to load it from database to PictureBox:

DataSet ds = new DataSet();

private void showImage(int index)
{
   byte[] bytes = (byte[])ds.Tables[0].Rows[index][1];
   MemoryStream memStream = new MemoryStream(bytes);
   try
   {
       Bitmap myImage = new Bitmap(memStream);
       pictureBox1.Image = myImage;
   }
   catch (Exception ex)
   {
       MessageBox.Show(ex.Message);
       pictureBox1.Image = null;
   }
}

private void button2_Click(object sender, EventArgs e)
{
   string strConn = @"Provider=Microsoft.ACE.OLEDB.12.0;Data Source=D:\Documents\Database1.accdb";
   using (OleDbConnection cnn = new OleDbConnection(strConn))
   {
      cnn.Open();
      using (OleDbDataAdapter da = new OleDbDataAdapter("select * from Table3", cnn))
      {
         da.Fill(ds);
         Random ran = new Random();
         int index = ran.Next(ds.Tables[0].Rows.Count);
         showImage(index);
      }
   }
}

Sure, also you can use OpenFileDialog control to let use choose the file from his computer:

private void button1_Click(object sender, EventArgs e)
{
   OpenFileDialog openFileDialog1 = new OpenFileDialog();
   openFileDialog1.Title = "Choose:";
   openFileDialog1.Filter = "All file(*.*)|*.*";
   if (openFileDialog1.ShowDialog() == DialogResult.OK)
   {
      string fileName = openFileDialog1.FileName;
      string strConn = @"Provider=Microsoft.ACE.OLEDB.12.0;Data Source=D:\Documents\Database1.accdb";
      //...
   }
}

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