Use asp.net to upload pictures and save them to SQL Server, and then read and display them from SQL Server

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1, use asp.net to upload pictures and save them to SQL Server, then read and display them from SQL Server
One, upload and deposit SQL Server
Database structure
CREATE TABLE Test
{
ID identity (1,1),
Fimage image
}
Related stored procedures
Create proc UpdateImage
(
@UpdateImage Image
)
As
Insert into Test (fimage) VALUES (@UpdateImage)
Go
Add the following in the Upphoto.aspx file:
<input id= "Upphoto" name= "Upphoto" runat= "Server" type= "file" >
<asp:button id= "Btnadd" name= "Btnadd" runat= "server" text= "upload" ></asp:Button>

Then add the Click event handling code for the Btnadd button in the post code file UpPhoto.aspx.cs:
private void btnAdd_Click (object sender, System.EventArgs e)
{
Get the image and convert the image to byte[]
Httppostedfile Upphoto=upphoto.postedfile;
int upphotolength=upphoto.contentlength;
Byte[] Photoarray=new byte[upphotolength];
Stream Photostream=upphoto.inputstream;
Photostream.read (photoarray,0,upphotolength);

Connecting to a database
SqlConnection conn=new SqlConnection ();
Conn. connectionstring= "Data source=localhost;database=test; User Id=sa; Pwd=sa ";

SqlCommand cmd=new SqlCommand ("UpdateImage", conn);
Cmd.commandtype=commandtype.storedprocedure;

Cmd. Parameters.Add ("@UpdateImage", sqldbtype.image);
Cmd. parameters["@UpdateImage"]. Value=photoarray;

If you want to add a picture without using a stored procedure, change the above four code to read:
String strsql= "Insert into Test (fimage) VALUES (@FImage)";
SqlCommand cmd=new SqlCommand (strsql,conn);
Cmd. Parameters.Add ("@FImage", sqldbtype.image);
Cmd. parameters["@FImage"]. Value=photoarray;

Conn. Open ();
Cmd. ExecuteNonQuery ();
Conn. Close ();
}

Second, read and display from SQL Server
Add the following code where you want the picture to appear:
<asp:image id= "Imgphoto" runat= "Server" imageurl= "showphoto.aspx" ></asp:image>

Showphoto.aspx Body Code:
private void Page_Load (object sender, System.EventArgs e)
{
if (! Page.IsPostBack)
{
SqlConnection Conn=new SqlConnection ()
Conn. connectionstring= "Data source=localhost;database=test; User Id=sa; Pwd=sa ";

String Strsql= "select * from Test where id=2";//here suppose you get a picture with ID 2
SqlCommand cmd=new SqlCommand ()
Reader. Read ();
Response.contenttype= "Application/octet-stream";
Response.BinaryWrite ((byte[]) reader["Fimage");
Response.End ();
Reader. Close ();
}
}

3, save the picture in SQL Server in WinForm and read it from SQL Server and display it in PictureBox

1, stored in SQL Server
Database structure and stored procedures used, same as above
1.1, add a OpenFileDialog control to the form named Ofdselectpic
1.2, add an open File button on the form and add the following click event code:
Stream MS;
Byte[] Picbyte;
Ofdselectpic.showdialog ();
if (Ofdselectpic.showdialog () ==dialogresult.ok)
{
if ((Ms=ofdselectpic.openfile ())!=null)
{
MessageBox.Show ("OK");
Picbyte=new Byte[ms. Length];
Ms. position=0;
Ms. Read (Picbyte,0,convert.toint32 (MS). Length));
MessageBox.Show ("read finished!");

Connecting to a database
SqlConnection conn=new SqlConnection ();
Conn. connectionstring= "Data source=localhost;database=test; User Id=sa; Pwd=sa ";

SqlCommand cmd=new SqlCommand ("UpdateImage", conn);
Cmd.commandtype=commandtype.storedprocedure;

Cmd. Parameters.Add ("@UpdateImage", sqldbtype.image);
Cmd. parameters["@UpdateImage"]. Value=picbyte;

Conn. Open ();
Cmd. ExecuteNonQuery ();
Conn. Close ();

Ms. Close ();
}
}

2, read and display in PictureBox
2.1 Add a PictureBox, named Ptbshow
2.2 Add a button to add the following response event:
SqlConnection conn=new SqlConnection ();
Conn. connectionstring= "Data source=localhost;database=test; User Id=sa; Pwd=sa ";

String strsql= "Select Fimage from Test where id=1";

SqlCommand cmd=new SqlCommand (strsql,conn);

Conn. Open ();
SqlDataReader Reader=cmd. ExecuteReader ();
Reader. Read ();

MemoryStream ms=new MemoryStream ((byte[)) reader["Fimage");

Image Image=image.fromstream (ms,true);

Reader. Close ();
Conn. Close ();

Ptbshow.image=image;

Add the following in the Upphoto.aspx file:
<input id= "Upphoto" name= "Upphoto" runat= "Server" type= "file" >
<asp:button id= "Btnadd" name= "Btnadd" runat= "server" text= "upload" ></asp:Button>

Then add the Click event handling code for the Btnadd button in the post code file UpPhoto.aspx.cs:
private void btnAdd_Click (object sender, System.EventArgs e)
{
Get the image and convert the image to byte[]
Httppostedfile Upphoto=upphoto.postedfile;
int upphotolength=upphoto.contentlength;
Byte[] Photoarray=new byte[upphotolength];
Stream Photostream=upphoto.inputstream;
Photostream.read (photoarray,0,upphotolength);

Connecting to a database
SqlConnection conn=new SqlConnection ();
Conn. connectionstring= "Data source=localhost;database=test; User Id=sa; Pwd=sa ";

SqlCommand cmd=new SqlCommand ("UpdateImage", conn);
Cmd.commandtype=commandtype.storedprocedure;

Cmd. Parameters.Add ("@UpdateImage", sqldbtype.image);
Cmd. parameters["@UpdateImage"]. Value=photoarray;

If you want to add a picture without using a stored procedure, change the above four code to read:
String strsql= "Insert into Test (fimage) VALUES (@FImage)";
SqlCommand cmd=new SqlCommand (strsql,conn);
Cmd. Parameters.Add ("@FImage", sqldbtype.image);
Cmd. parameters["@FImage"]. Value=photoarray;

Conn. Open ();
Cmd. ExecuteNonQuery ();
Conn. Close ();
}

Second, read and display from SQL Server
Add the following code where you want the picture to appear:
<asp:image id= "Imgphoto" runat= "Server" imageurl= "showphoto.aspx" ></asp:image>

Showphoto.aspx Body Code:
private void Page_Load (object sender, System.EventArgs e)
{
if (! Page.IsPostBack)
{
SqlConnection Conn=new SqlConnection ()
Conn. connectionstring= "Data source=localhost;database=test; User Id=sa; Pwd=sa ";

String Strsql= "select * from Test where id=2";//here suppose you get a picture with ID 2
SqlCommand cmd=new SqlCommand ()
Reader. Read ();
Response.contenttype= "Application/octet-stream";
Response.BinaryWrite ((byte[]) reader["Fimage");
Response.End ();
Reader. Close ();
}
}

3, save the picture in SQL Server in WinForm and read it from SQL Server and display it in PictureBox

1, stored in SQL Server
Database structure and stored procedures used, same as above
1.1, add a OpenFileDialog control to the form named Ofdselectpic
1.2, add an open File button on the form and add the following click event code:
Stream MS;
Byte[] Picbyte;
Ofdselectpic.showdialog ();
if (Ofdselectpic.showdialog () ==dialogresult.ok)
{
if ((Ms=ofdselectpic.openfile ())!=null)
{
MessageBox.Show ("OK");
Picbyte=new Byte[ms. Length];
Ms. position=0;
Ms. Read (Picbyte,0,convert.toint32 (MS). Length));
MessageBox.Show ("read finished!");

Connecting to a database
SqlConnection conn=new SqlConnection ();
Conn. connectionstring= "Data source=localhost;database=test; User Id=sa; Pwd=sa ";

SqlCommand cmd=new SqlCommand ("UpdateImage", conn);
Cmd.commandtype=commandtype.storedprocedure;

Cmd. Parameters.Add ("@UpdateImage", sqldbtype.image);
Cmd. parameters["@UpdateImage"]. Value=picbyte;

Conn. Open ();
Cmd. ExecuteNonQuery ();
Conn. Close ();

Ms. Close ();
}
}

2, read and display in PictureBox
2.1 Add a PictureBox, named Ptbshow
2.2 Add a button to add the following response event:
SqlConnection conn=new SqlConnection ();
Conn. connectionstring= "Data source=localhost;database=test; User Id=sa; Pwd=sa ";

String strsql= "Select Fimage from Test where id=1";

SqlCommand cmd=new SqlCommand (strsql,conn);

Conn. Open ();
SqlDataReader Reader=cmd. ExecuteReader ();
Reader. Read ();

MemoryStream ms=new MemoryStream ((byte[)) reader["Fimage");

Image Image=image.fromstream (ms,true);

Reader. Close ();
Conn. Close ();

Ptbshow.image=image;