Question

我是c＃编程的新手。我使用FileStream时遇到问题。我想通过搜索数据库中人员的ID来从数据库中检索图片。它工作!!但是当我尝试两次检索人的照片时（两次插入相同的ID）。它会给IOException

"The process cannot access the file 'C:\Users\dor\Documents\Visual Studio 2010\Projects\studbase\studbase\bin\Debug\foto.jpg' because it is being used by another process."

我有1个按钮 - ＆gt; buttonLoad | 1 pictureBox - ＆gt; pictureBoxPictADUStudent

这是buttonLoad上的代码

        string sql = "SELECT Size,File,Name FROM studgeninfo WHERE NIM = '"+textBoxNIM.Text+"'";
        MySqlConnection conn = new MySqlConnection(connectionSQL);
        MySqlCommand comm = new MySqlCommand(sql, conn);

        if (textBoxNIM.Text != "")
        {
            conn.Open();
            MySqlDataReader data = comm.ExecuteReader();

            while (data.Read())
            {

                int fileSize = data.GetInt32(data.GetOrdinal("size"));
                string name = data.GetString(data.GetOrdinal("name"));


                using (FileStream fs = new FileStream(name, FileMode.Create, FileAccess.Write))
                {
                        byte[] rawData = new byte[fileSize];
                        data.GetBytes(data.GetOrdinal("file"), 0, rawData, 0, fileSize);
                        fs.Write(rawData, 0, fileSize);
                        fs.Dispose();
                        pictureBoxPictADUStudent.BackgroundImage = new Bitmap(name);
                }
            }

            conn.Close();
        }

        else
        {
            MessageBox.Show("Please Input Student NIM ", "Warning!", MessageBoxButtons.OK, MessageBoxIcon.Warning);
        }

Answer 1

您正在此处打开文件：

using (FileStream fs = new FileStream(name, FileMode.Create, FileAccess.Write))
                                   // ^^^^ This is your filename..

..而且Bitmap也试图打开文件来阅读..在这里：

pictureBoxPictADUStudent.BackgroundImage = new Bitmap(name);
                                                   // ^^^^ You are using it again here

Bitmap在您写文件时无法从文件中读取..

编辑：

正如评论中所指出的，即使调用fs.Dispose()，也会发生这种情况。见这里：Does FileStream.Dispose close the file immediately?

Answer 2

此处的问题是文件被new Bitmap()锁定。所以你必须确保一旦加载了位图，它就会锁定文件：

        using (FileStream fs = new FileStream(name, FileMode.Create, FileAccess.Write))
        {
                byte[] rawData = new byte[fileSize];
                data.GetBytes(data.GetOrdinal("file"), 0, rawData, 0, fileSize);
                fs.Write(rawData, 0, fileSize);
        }

        using (var bmpTemp = new Bitmap(name))
        {
            pictureBoxPictADUStudent.BackgroundImage= new Bitmap(bmpTemp);
        }

更新：我更新了我的答案以反映您的最新答案。有关详细信息，请转到this post

Answer 3

这样写 using（FileStream fs = new FileStream（name，FileMode.Create，FileAccess.Write））

{

} pictureBoxPictADUStudent.BackgroundImage = new Bitmap（name）;

文件正由另一个进程使用

3 个答案:

{