FileUpload Controle in ASP.NET Save Images in Database C# VB.NET

In this example i am Uploading Images using FileUpload Control and saving or storing them in SQL Server database in ASP.NET with C# and VB.NET.

Database is having a table named Images with three columns.
1. ID Numeric Primary key with Identity Increment.
2. ImageName Varchar to store Name of Image.
3. Image Image to store image in binary format.

After uploading and saving images in database, images are displayed in GridView.

To know how to Store and Retrieve Images in Data Base using Microsoft .NET read it.

Html markup of the page look like

<form id="form1" runat="server">
<div>
<asp:TextBox ID="txtName" runat="server" Width="95px">
</asp:TextBox>
<asp:FileUpload ID="FileUpload1" runat="server"/>
<asp:Label ID="lblMessage" runat="server">
</asp:Label>
<asp:Button ID="btnUpload" runat="server" 
            OnClick="btnUpload_Click" Text="Upload"/>
</div>
</form>

Write this code in Click Event of Upload Button

C# Code behind

View source     

01protected void btnUpload_Click(object sender, EventArgs e)

02{

03 string strImageName = txtName.Text.ToString();

04 if (FileUpload1.PostedFile != null && 

05     FileUpload1.PostedFile.FileName != "")

06  {

07   byte[] imageSize = new byte

08                 [FileUpload1.PostedFile.ContentLength];

09  HttpPostedFile uploadedImage = FileUpload1.PostedFile;

10  uploadedImage.InputStream.Read

11     (imageSize, 0, (int)FileUpload1.PostedFile.ContentLength);

12 

13                                                           // Create SQL Connection 

14  SqlConnection con = new SqlConnection();

15  con.ConnectionString = ConfigurationManager.ConnectionStrings

16                         ["ConnectionString"].ConnectionString;

17 

18                                                          // Create SQL Command 

19 

20 SqlCommand cmd = new SqlCommand();

21 cmd.CommandText = "INSERT INTO Images(ImageName,Image)" +

22                   " VALUES (@ImageName,@Image)";

23 cmd.CommandType = CommandType.Text;

24 cmd.Connection = con;

25 

26 SqlParameter ImageName = new SqlParameter

27                     ("@ImageName", SqlDbType.VarChar, 50);

28 ImageName.Value = strImageName.ToString();

29 cmd.Parameters.Add(ImageName);

30 

31 SqlParameter UploadedImage = new SqlParameter

32               ("@Image", SqlDbType.Image, imageSize.Length);

33 UploadedImage.Value = imageSize;

34 cmd.Parameters.Add(UploadedImage);

35 con.Open();

36 int result = cmd.ExecuteNonQuery();

37 con.Close();

38 if (result > 0)

39 lblMessage.Text = "File Uploaded";

40 GridView1.DataBind();

41 }

42}

VB.NET Code

View source     

01Protected Sub btnUpload_Click

02(ByVal sender As Object, ByVal e As EventArgs)

03 

04    Dim strImageName As String = txtName.Text.ToString()

05    If FileUpload1.PostedFile IsNot Nothing AndAlso 

06       FileUpload1.PostedFile.FileName <> "" Then

07 

08        Dim imageSize As Byte() = New Byte

09          (FileUpload1.PostedFile.ContentLength - 1) {}

10 

11        Dim uploadedImage__1 As HttpPostedFile = 

12                                 FileUpload1.PostedFile

13 

14        uploadedImage__1.InputStream.Read(imageSize, 0, 

15              CInt(FileUpload1.PostedFile.ContentLength))

16 

17                                                             ' Create SQL Connection 

18        Dim con As New SqlConnection()

19        con.ConnectionString = 

20                 ConfigurationManager.ConnectionStrings

21                  ("ConnectionString").ConnectionString

22 

23                                                                ' Create SQL Command 

24 

25        Dim cmd As New SqlCommand()

26        cmd.CommandText = "INSERT INTO Images

27           (ImageName,Image) VALUES (@ImageName,@Image)"

28        cmd.CommandType = CommandType.Text

29        cmd.Connection = con

30 

31        Dim ImageName As New SqlParameter

32                  ("@ImageName", SqlDbType.VarChar, 50)

33        ImageName.Value = strImageName.ToString()

34        cmd.Parameters.Add(ImageName)

35 

36        Dim UploadedImage__2 As New SqlParameter

37            ("@Image", SqlDbType.Image, imageSize.Length)

38        UploadedImage__2.Value = imageSize

39        cmd.Parameters.Add(UploadedImage__2)

40        con.Open()

41        Dim result As Integer = cmd.ExecuteNonQuery()

42        con.Close()

43        If result > 0 Then

44            lblMessage.Text = "File Uploaded"

45        End If

46        GridView1.DataBind()

47    End If

48End Sub

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Sep 28, 2011