如何从数据库中检索图像并将其显示在图片框中 [英] how to retrieve images from database and displaying it in picturebox

问题描述

我有Access数据库。我在访问数据库中插入了员工数据和图像。当我点击Datagridview Click事件数据正在更新到文本框。用同样的点击照片即更新到图片框。但我无法在图片框中显示图像。请帮帮我。



我将图像转换成字节并存储在数据库中。喜欢

Hi,

I have Access Database. I inserted Employee Data and images in access Database. When i clicked on Datagridview Click Event data is updating to textboxes. With the same click Photo is to update to picturebox. But i am not able to display image in picture box. Please help me.

I converted image into bytes and stored in Database. like

MessageBox.Show("Saving Data at Index :" + rowPosition.ToString());

              OleDbCommand oledbinsert = new OleDbCommand("Insert INTO EmployeeData(EmpID,EmployeeName,PassportNo,PassportExpiry,AddressMalaysia,

OleDbParameter imageParameter = oledbinsert.Parameters.AddWithValue（@ Img，SqlDbType.Binary）;

imageParameter .Value = ImageAsBytes;

imageParameter.Size = ImageAsBytes.Length;



int rowsAffected = oledbinsert.ExecuteNonQuery（）;

MessageBox.Show（数据存储成功+ rowsAffected.ToString（）+Row）;

MessageBox.Show（数据成功添加到员工数据）;

OleDbParameter imageParameter = oledbinsert.Parameters.AddWithValue("@Img", SqlDbType.Binary);
imageParameter.Value = ImageAsBytes;
imageParameter.Size = ImageAsBytes.Length;

int rowsAffected = oledbinsert.ExecuteNonQuery();
MessageBox.Show("Data Stored Successfully in" + rowsAffected.ToString() + "Row");
MessageBox.Show("Data Successfully Added to Employee Data");

推荐答案

Your Solution Is very simple

void RetriveImage()
{
    OleDbConnection conn = new OleDbConnection(@"Your Connection String");
    conn.Open();
    using (conn)
    {

        Image img ;


        byte[] photo ;

        OleDbCommand cmd = new OleDbCommand("SELECT Pic From ATEST where ID=4", conn);
        OleDbDataAdapter reader;
        reader = cmd.ExecuteReader();
        reader.Read();
        photo =(byte []) reader.GetValue(0);

        MemoryStream ms = new MemoryStream(photo);

        img = Image.FromStream(ms);

        pictureBox1.Image = img;




    }
    conn.Close();

}

谢谢兄弟。但它无法正常工作。

只需查看gridview点击事件中的代码：

Thank You Bro. But its not working.
Just Look at my code in gridview click event :

private void dataGridView1_CellContentClick(object sender, DataGridViewCellEventArgs e)
        {
            if (e.RowIndex >= 0)
            {
                try
                {

                    DataGridViewRow row = dataGridView1.Rows[e.RowIndex];
                    Txt_EmpID.Text = row.Cells[0].Value.ToString();
                    Txt_EmpName.Text = row.Cells[1].Value.ToString();
                    Txt_Passport.Text = row.Cells[2].Value.ToString();
                    Dtp_PpExpiry.Text = row.Cells[3].Value.ToString();
                    txt_Address1.Text = row.Cells[4].Value.ToString();
                    Txt_Address2.Text = row.Cells[5].Value.ToString();
                    dtpDOB.Text = row.Cells[6].Value.ToString();
                    Txt_Emergency.Text = row.Cells[7].Value.ToString();
                    Txt_Emerg2.Text = row.Cells[8].Value.ToString();
                    Txt_Bank.Text = row.Cells[9].Value.ToString();
                    Txt_Dl.Text = row.Cells[10].Value.ToString()
    MessageBox.Show("Loading Data");
                }
                catch (Exception ex)
                {
                    MessageBox.Show("Error" + ex);
                }
               // pictureBox1.Image = row.Cells[18].Value as Image;
                //pictureBox1.Image = dataGridView1.CurrentRow.Cells[18].Value as Image;

               // LoadImage();
            }
            RetriveImage();
        }<pre lang="cs">

 <pre lang="cs">void RetriveImage()
        {
            OleDbConnection DBConnection = new OleDbConnection(@"Provider=Microsoft.ACE.OLEDB.12.0;Data Source=C:\Users\Administrator\Desktop\EmployeeData.accdb");
            DBConnection.Open();
            using (DBConnection)
            {

                Image img;


                byte[] photo;

                OleDbCommand cmd = new OleDbCommand("SELECT img From EmployeeData where EmpID=4", DBConnection);
                //OleDbDataAdapter reader = new OleDbDataAdapter();
              //  OleDbDataReader reader;
                OleDbDataReader reader = cmd.ExecuteReader();
               // reader = cmd.ExecuteReader();
                while (reader.Read())
                {
                    photo = (byte[])reader.GetValue(0);

                    MemoryStream ms = new MemoryStream(photo);

                    img = Image.FromStream(ms);


                    pictureBox1.Image = img;
                }
                

            }
            DBConnection.Close();

        }
<pre lang="cs">

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