数据库中插入图片的问题

Question

private void button_click(object sender,System.EventArgs e)
{
FileStream fs;
string PathImage;
OleDbConnection OleConn=new OleDbConnection(@"Provide=Microsoft.Jet.OLEDB.4.0;Data Source=E:\db1.mdb");
OleConn.Open();
if(DialogResult.OK==this.openFileDialog1.ShowDialog())
{
 PathImage=this.openFileDialog1.FileName;
fs=new FileStream(PathImage,System.IO.FileMode.Open,System.IO.FileAccess.Read);
byte[] ib=new byte[fs.Length];
fs.Read(ib,0,ib.Length);
fs.Close();OleDbCommand cmd=new OleDbCommand("Insert into[image1](Image1) values (@img)",OleConn);
((OleDbParameter)cmd.Parameters.Add("@img",OleDbType.Binary)).Vaule=ib;
cmd.ExecuteNonQuery();
OleConn.Close();
MessageBox.Show("写入成功！");
}
}

Answer 1

//添加和读取全给你搞定了。
private void button1_Click(object sender, System.EventArgs e)
{
//get file info
string filePath=Application.StartupPath+"\\it.jpg";
FileInfo fi=new FileInfo(filePath);
FileStream fileStream=fi.OpenRead();
int length=(int)fileStream.Length;
byte[] fileData=new byte[length];
fileStream.Read(fileData,0,length);

//Sql 
OleDbCommand command = new OleDbCommand ("INSERT INTO t1 (filename,fileData)" + "VALUES (@filename,@fileData)", myConnection);
//add para 1
System.Data.OleDb.OleDbParameter parafileName = new OleDbParameter("@filename", System.Data.OleDb.OleDbType.VarChar,50); 
parafileName.Value ="it.jpg";
command.Parameters.Add(parafileName); //add para2
System.Data.OleDb.OleDbParameter paramPersonImage = new OleDbParameter("@fileData", System.Data.OleDb.OleDbType.Binary); 
paramPersonImage.Value = fileData;
command.Parameters.Add(paramPersonImage); //Excute 
this.OpenDB();
command.ExecuteNonQuery();
myConnection.Close();
} private void button2_Click(object sender, System.EventArgs e)
{
this.OpenDB();
string strSel="select * from t1 order by id desc";
OleDbCommand cmd=new OleDbCommand(strSel,myConnection);
OleDbDataReader dr=cmd.ExecuteReader();
if(dr.Read())
{
byte[] by=(byte[])dr.GetValue(2);
MemoryStream ms=new MemoryStream(by);
Image img=Image.FromStream(ms);
pb.Image=img;
}
dr.Close();
cmd.Dispose();
myConnection.Close();
}
private void button1_Click(object sender, System.EventArgs e)
{
//get file info
string filePath=Application.StartupPath+"\\it.jpg";
FileInfo fi=new FileInfo(filePath);
FileStream fileStream=fi.OpenRead();
int length=(int)fileStream.Length;
byte[] fileData=new byte[length];
fileStream.Read(fileData,0,length);

//Sql 
OleDbCommand command = new OleDbCommand ("INSERT INTO t1 (filename,fileData)" + "VALUES (@filename,@fileData)", myConnection);
//add para 1
System.Data.OleDb.OleDbParameter parafileName = new OleDbParameter("@filename", System.Data.OleDb.OleDbType.VarChar,50); 
parafileName.Value ="it.jpg";
command.Parameters.Add(parafileName); //add para2
System.Data.OleDb.OleDbParameter paramPersonImage = new OleDbParameter("@fileData", System.Data.OleDb.OleDbType.Binary); 
paramPersonImage.Value = fileData;
command.Parameters.Add(paramPersonImage); //Excute 
this.OpenDB();
command.ExecuteNonQuery();
myConnection.Close();
} private void button2_Click(object sender, System.EventArgs e)
{
this.OpenDB();
string strSel="select * from t1 order by id desc";
OleDbCommand cmd=new OleDbCommand(strSel,myConnection);
OleDbDataReader dr=cmd.ExecuteReader();
if(dr.Read())
{
byte[] by=(byte[])dr.GetValue(2);
MemoryStream ms=new MemoryStream(by);
Image img=Image.FromStream(ms);
pb.Image=img;
}
dr.Close();
cmd.Dispose();
myConnection.Close();
}