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Multiple file upload to SQL

I am looking for an example of uploading a file(s) to an SQL server. I want to be able to upload multiple files to SQL using the OpenFileDialog to select the files. This is a windows form app not a web client. I am looking for an example in c# if possible.
Thank You

ASKER CERTIFIED SOLUTION

Dmitry G

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SOLUTION

Anurag Thakur

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NoSleepForTheWeary

ASKER

Thank you for the responces. I am looking for how to parse the the value returned from the OpenFileDialog. http://www.shabdar.org/store-save-images-in-sql-server.html This one is close to what I am looking for but I can seem to figure out what is the delimeter between the file names in the return string when choosing more then one.

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Anurag Thakur

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NoSleepForTheWeary

ASKER

This is what I came up with thanks for the input.

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Data.Sql;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;
using System.Data.SqlClient;
using System.IO;
 
namespace WriteBlob
{
    public partial class Form1 : Form
    {
        SqlConnection sqlCnnUPDATE = new System.Data.SqlClient.SqlConnection();
            
        public Form1()
        {
            InitializeComponent();
            sqlCnnUPDATE.ConnectionString = ("data source=SQLSERVER;"+
                "initial catalog=FILEDB;"+
                "user id=sa;password=password;");
 
                     
 
        }
 
 
 
 
        public void UploadFile(
            string pFILENAMEDESC,
            string pFILEDESC,
            float pRUNNINGTIME,
            DateTime pFILEDATE,
            int FILETYPE,
            string FilePath)
        {
            try
            {
 
              
                byte[] file = GetFile(FilePath);
 
                // Construct INSERT Command
                SqlCommand addEmp = new SqlCommand(
                    "INSERT INTO TBLFILES (" +
                    "FILENAMEDESC,FILEDESC,RUNNINGTIME,FILEBLOB,FILEDATE,FILETYPE) " +
                    "VALUES(@FILENAMEDESC,@FILEDESC,@RUNNINGTIME,@FILEBLOB,@FILEDATE,@FILETYPE)", sqlCnnUPDATE);
 
                addEmp.Parameters.Add("@FILENAMEDESC", SqlDbType.NVarChar, 100).Value = pFILENAMEDESC;
                addEmp.Parameters.Add("@FILEDESC", SqlDbType.NVarChar, 300).Value = pFILEDESC;
                addEmp.Parameters.Add("@RUNNINGTIME", SqlDbType.Float).Value = pRUNNINGTIME;
                addEmp.Parameters.Add("@FILEBLOB", SqlDbType.Image, file.Length).Value = file;
                addEmp.Parameters.Add("@FILEDATE", SqlDbType.DateTime).Value = pFILEDATE;
                addEmp.Parameters.Add("@FILETYPE", SqlDbType.Int).Value = FILETYPE;
                
                // Open the Connection and INSERT the BLOB into the Database
                if (sqlCnnUPDATE.State != ConnectionState.Open)
                {
                    sqlCnnUPDATE.Open();
                } 
                addEmp.ExecuteNonQuery();
                sqlCnnUPDATE.Close();
 
                
            }
            catch
            {
                
            }
 
 
 
 
        }
        
 
        // **** Read Image into Byte Array from Filesystem
        public static byte[] GetFile(string filePath)
        {
            FileStream fs = new FileStream(filePath, FileMode.Open, FileAccess.Read);
            BinaryReader br = new BinaryReader(fs);
 
            byte[] file = br.ReadBytes((int)fs.Length);
 
            br.Close();
            fs.Close();
 
            return file;
        }
 
        private void button1_Click(object sender, EventArgs e)
        {
            if (ofd.ShowDialog() == DialogResult.OK)
            {
                string filename;
                string FilePath;
                
                foreach (string fileListItem in ofd.FileNames)
                {
                    filename = System.IO.Path.GetFileName(fileListItem);
                    FilePath = System.IO.Path.GetDirectoryName(fileListItem) + '\\' + filename;
                    listBox1.Items.Add((object)filename);
                    UploadFile(filename, filename, 60, DateTime.Now, 1, FilePath);
 
                }
 
            }
 
 
        }
 
     
        
    }
}

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