Upload and Download Files From Database in Web API

Mudita Rathore
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Article

Introduction

This article explains how to upload files in the database and download files from the database in the ASP.NET Web API. This file can be a text file or an image file.

The procedure for creating the application is as in the following.

Step 1

First we create a database in SQL.

Open SQL Server 2012.
Select "New Query" and create the database and table.

The following commands are used for creating the database and table:

create database Demo
use Demo
create table Datafile(ID int IDENTITY NOT NULL,Filerecord image NOT NULL ,Filetype varchar(50) NOT NULL,Name varchar(50) NOT NULL)

Step 2

Create the Web API application:

Start Visual Studio 2012.
From the start window select "New Project".
In the Template window select "Installed" -> "Visual C#" -> "Web".
Select "ASP.NET MVC4 Web Application" and click the "OK" button.

From the "MVC4 Project" window select "Web API".

Click the "OK" button.

Step 3

Open the "HomeController" file and write the code for uploading the file in the database. This file exists:

In the "Solution Explorer".
Select "Controller" -> "HomeController".

Add the following code:

using System;
using System.Collections.Generic;
using System.Data.SqlClient;
using System.IO;
using System.Linq;
using System.Web;
using System.Web.Mvc;
namespace FileUploadDatabase.Controllers
{
public class HomeController : Controller
{
public bool Infile(HttpPostedFileBase imgfile)
{
return (imgfile != null && imgfile.ContentLength > 0) ? true : false;
}
public ActionResult Index()
{
foreach (string Save in Request.Files)
{
if (!Infile(Request.Files[Save])) continue;
string fileType = Request.Files[Save].ContentType;
Stream file_Strm = Request.Files[Save].InputStream;
string file_Name = Path.GetFileName(Request.Files[Save].FileName);
int fileSize = Request.Files[Save].ContentLength;
byte[] fileRcrd = new byte[fileSize];
file_Strm.Read(fileRcrd, 0, fileSize);
const string connect = @"Server=.;Database=Demo; User Id=sa; password=wintellect;";
using (var conn = new SqlConnection(connect))
{
var qry = "INSERT INTO Datafile (Filerecord, Filetype, Name)VALUES (@Filerecord, @Filetype, @Name)";
var cmd = new SqlCommand(qry, conn);
cmd.Parameters.AddWithValue("@Filerecord", fileRcrd);
cmd.Parameters.AddWithValue("@Filetype", fileType);
cmd.Parameters.AddWithValue("@Name", file_Name);
conn.Open();
cmd.ExecuteNonQuery();
}
}
return View();
}
public ActionResult DownloadImage()
{
const string connect = @"Server=.;Database=Demo;User id=sa;password=wintellect;";
List<string> fileList = new List<string>();
using (var con = new SqlConnection(connect))
{
var query = "SELECT Filerecord, Filetype,Name FROM Datafile";
var cmd = new SqlCommand(query, con);
con.Open();
SqlDataReader rdr = cmd.ExecuteReader();
while (rdr.Read())
{
fileList.Add(rdr["Name"].ToString());
}
}
ViewBag.Images = fileList;
return View();
}
public FileContentResult GetFile(int id)
{
SqlDataReader rdr;
byte[] fileContent = null;
string fileType = "";
string file_Name = "";
const string connect = @"Server=.;Database=Demo;User id=sa;password=wintellect;";
using (var con = new SqlConnection(connect))
{
var query = "SELECT Filerecord, Filetype, Name FROM Datafile WHERE ID = @ID";
var cmd = new SqlCommand(query, con);
cmd.Parameters.AddWithValue("@ID", id);
con.Open();
rdr = cmd.ExecuteReader();
if (rdr.HasRows)
{
rdr.Read();
fileContent = (byte[])rdr["Filerecord"];
fileType = rdr["Filetype"].ToString();
file_Name = rdr["Name"].ToString();
}
}
return File(fileContent, fileType, file_Name);
}
}
}