Upload Image into database

Question

0.00/5 (No votes)

See more:

Hi I want to upload a image into database

Please help me out

I create a method

C#

public void insert_blog(ref string connection_string, ref Int64 blog_id, ref string title, ref string intro, ref string text, ref string keyword, ref byte categotry, out byte affected_rows)
   {
       affected_rows = 0;

       SqlConnection conn = new SqlConnection(connection_string);
       SqlCommand cmd = new SqlCommand();

       if (conn.State == ConnectionState.Closed)
       {
           conn.Open();
       }
       cmd.CommandText = "insert_blog";
       cmd.Connection = conn;
       cmd.CommandType = CommandType.StoredProcedure;

       cmd.Parameters.Add("@blogid", SqlDbType.BigInt).Value = blog_id;
       cmd.Parameters.Add("@intro", SqlDbType.VarChar).Value = intro;
       cmd.Parameters.Add("@title", SqlDbType.VarChar).Value = title;
       cmd.Parameters.Add("@text", SqlDbType.NVarChar).Value = text;
       cmd.Parameters.Add("@keyword", SqlDbType.VarChar).Value = keyword;
       cmd.Parameters.Add("@category", SqlDbType.TinyInt).Value = categotry;

       affected_rows = Convert.ToByte(cmd.ExecuteNonQuery());


       conn.Close();
       conn.Dispose();
   }

and

C#

string connection = System.Configuration.ConfigurationManager.ConnectionStrings["blog"].ToString();
        blog_class cla = new blog_class();
        string title = TextBox1_title.Text;
        string intro = TextBox1_intro.Text;
        string meta = TextBox1_meta_tags.Text;
        string emp = "";
        Int64 blog_id = 0;
        order_by_tag_photo_ids_wall(ref connection, out blog_id);
        blog_id = blog_id + 1;
        byte cat_id = Convert.ToByte(dropdown1.SelectedValue.ToString());
        byte n;
        cla.insert_blog(ref connection, ref blog_id, ref title, ref intro, ref emp, ref meta, ref cat_id, out n);
        if (n > 0)
        {
            Response.Redirect("~/user/redirect/edit_blog.aspx?blogid=" + blog_id);
        }
        else
        {
            Label1.Text = "Error Occured";
        }

How I upload Image I create a field in database

img varchar(300)

Posted 22-Dec-14 22:25pm

Arora1992

Updated 22-Dec-14 23:50pm

Thanks7872

v2

Add a Solution

Comments

Sinisa Hajnal 23-Dec-14 5:20am

Why all ref parameters? You're not using them for anything.

Arora1992 23-Dec-14 7:32am

I try to upload Image
Please check this code
protected void Button1_Click(object sender, EventArgs e)
{
if (FileUpload1.HasFile)
{
string str = FileUpload1.FileName;
FileUpload1.PostedFile.SaveAs(Server.MapPath(".") + "//uploads//" + str);
string path = "~//uploads//" + str.ToString();
SqlConnection con = new SqlConnection(@"Data Source=LUVLY;Initial Catalog=practice;User ID=sa;Password=lenovo@345");
con.Open();
SqlCommand cmd = new SqlCommand("insert into testing values('" + TextBox1.Text + "','" + path + "')", con);
cmd.ExecuteNonQuery();
con.Close();
Label1.Text = "completed";
SqlDataAdapter da = new SqlDataAdapter("select * from testing",con);
DataTable dt = new DataTable();
da.Fill(dt);
GridView1.DataSource = dt;
DataBind();
}
else
{
Label1.Text = "Please try again";
}

}

And Html Code

<html xmlns="http://www.w3.org/1999/xhtml">
<head runat="server">
<title></title>
</head>
<body>
<form id="form1" runat="server">
<div>
<asp:FileUpload ID="FileUpload1" runat="server" />
<asp:TextBox ID="TextBox1" runat="server">
<asp:Button ID="Button1" runat="server" Text="Button" OnClick="Button1_Click" />
</div>
<asp:Label ID="Label1" runat="server" Text="Label">
<asp:GridView ID="GridView1" runat="server" AutoGenerateColumns="False">
<columns>
<asp:TemplateField HeaderText="Images">
<itemtemplate>
<asp:Image ID="image" runat="server" ImageUrl='<%Eval("image") %>' Height="1250px" Width="850px" />

</form>
</body>
</html>

Image save into database But when i retrieve It doesnot show the image

4 solutions

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Prasad Khandekar · Answer 1 · 2014-12-22T22:54:00

Hello,

Please have a look at these Code Project articles. They should help you resolve the issue. Using varchar(300) is not a good idea unless your images are pretty small & that field definition will not allow you to store images in binary format.

Regards,

Sid_Joshi · Answer 2 · 2014-12-23T00:26:00

use image as a data type in sql

protected void Button1_Click(object sender, EventArgs e)
{
    if (FileUpload1.PostedFile != null && FileUpload1.PostedFile.FileName != "")
    {
        byte[] img = new byte[FileUpload1.PostedFile.ContentLength];
        HttpPostedFile Image = FileUpload1.PostedFile;
        Image.InputStream.Read(img, 0, (int)FileUpload1.PostedFile.ContentLength);
        string str = "insert img values(@img)";
        SqlConnection cn = new SqlConnection("data source=localhost;initial catalog=aadarsh;user id=sa;password=cos123");
        cn.Open();
        SqlCommand cmd = new SqlCommand(str, cn);
        cmd.Parameters.AddWithValue("@img", img);
        SqlDataReader dr = cmd.ExecuteReader();
        cn.Close();
    }
}

Arora1992 · Answer 3 · 2014-12-23T21:03:00

SqlConnection con = new SqlConnection(@"Data Source=LUVLY;Initial Catalog=practice;User ID=sa;Password=lenovo@345");
con.Open();
string cmdstr = "select * from testing";
SqlCommand cmd = new SqlCommand(cmdstr, con);
SqlDataAdapter da = new SqlDataAdapter(cmd);
DataTable dt = new DataTable();
da.Fill(dt);
GridView1.DataSource = dt;

GridView1.DataBind();

MohamedEliyas · Answer 4 · 2015-01-06T19:22:00

C#

if (FileUpload1.HasFile)
     {
         try
         {
             filename = System.IO.Path.GetFileName(FileUpload1.FileName);
             FileUpload1.SaveAs(Server.MapPath("images/") + filename);
             photo = "images/" + filename;
             //Label8.Text = "Upload status: File uploaded!";
         }
         catch (Exception ex)
         {
             photo = "Upload status: The file could not be uploaded. The following error occured: " + ex.Message;
             //Label8.Text
         }

Upload Image into database

4 solutions

Solution 1

Solution 2

Solution 3

Solution 4

Add your solution here

Preview 0