I want to create form design like attached file.

Question

Hi All I want to create form design same like attached file. please help me about this can i create.

Feroz Khan · Answer

Hi Bikesh Srivastava i want to create in windows form so how can i ..

Bikesh Srivastava · Answer

Hi If you using window form hen its very easy . other wise use css javascriptjquerybootsrap in web application.

Delpin Susai Raj · Answer

https://support.office.com/en-us/article/Insert-a-file-attachment-control-8e8e102f-fad5-4440-bc28-d28944f85287

Dhananjay Kumar · Answer

Step 1 Create your database and table Step 2 And design your form according to your requirement Here is your complete code Source code Default.aspx @ Page Language =C# AutoEventWireup =true CodeFile =Default.aspx.cs Inherits =_Default %> DOCTYPE html > html xmlns =http://www.w3.org/1999/xhtml> head runat =server> title > title > head > body > form id =form1 runat =server> div > table class =auto-style1 align =center> tr > td colspan =2> h1 > Attech file h1 > td > tr > tr > td > file name td > td > asp : TextBox ID =TextBox1 runat =server> asp : TextBox > td > tr > tr > td > Upload file td > td > asp : FileUpload ID =FileUpload1 runat =server /> td > tr > tr > td > asp : Button ID =Button1 runat =server Text =Upload OnClick =Button1_Click /> td > tr > table > div > form > body > html > Step 3 : Add a folder in your solution Explorer Step 4: Code file Default.aspx.cs using System; using System.Data.SqlClient; using System.Configuration; using System.IO; public partial class _Default : System.Web.UI. Page { SqlConnection con = new SqlConnection ( ConfigurationManager .ConnectionStrings[ Connectionstring ].ConnectionString); SqlCommand cmd; protected void Page_Load( object sender EventArgs e) { } protected void Button1_Click( object sender EventArgs e) { string myfile; if (FileUpload1.HasFile) { string sExtension = Path .GetExtension(FileUpload1.FileName); FileUpload1.SaveAs(Request.PhysicalApplicationPath + \NewFolder1\ + FileUpload1.FileName); myfile = ~\NewFolder1\ + FileUpload1.FileName; } else { myfile =  ; } con.Open(); cmd = new SqlCommand ( insert into file values(' + TextBox1.Text + '' +myfile + ')  con); cmd.ExecuteNonQuery(); con.Close(); Response.Write(  alert('File Uploaded successfully')  ); TextBox1.Text =  ; } }