我会提供两种解决方案。 第一种解决方案是直接将原始图像以字节存储在数据库中。 第二个解决方案是我个人推荐的 – 而不是使用数据库中的图像文件的路径。

这里摘录一篇文章 ，提出一些关于是否BLOB的优秀观点。

那么，有几个原因导致您不应该在数据库中存储二进制数据：

您可能希望将二进制数据存储在数据库中有一个很好的理由：

以下是选择图像文件的方法：

 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; } 

解决方案1：BLOB方法

 // 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); } } } 

解决方案2：在文件系统上存储文件路径

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 // 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()); } }