displaying Pictrure

[franknu]

ok I want to display a picture from my database i difines the variable but now it doesnt even show the frame for the picture , i also want to create a link or more info to another page.

my error message is Notice: Undefined index: includefiles/bizwebpage2.php in c:\program files\easyphp1-8\home\townsfinder\business_display.php on line 79
i tried difining it at the top but

[code=php:0] ((( Online 79 I have echo '<a href="'. $row['includefiles/bizwebpage2.php'].'">' . $row['BusinessName'] . '</a>'; //link to an)))  [/code] 

Can anyone please tell me how to do this i dont know what to do anymore

here is my full code:
[code=php:0]

<?php

$host = "localhost";
$username = "localhost";
$password = "abc123";
$database = "contacts";

$Categories = addslashes ($_POST['Categories']);
$make = addslashes ($_POST['make']);
$type = addslashes ($_POST['type']);

if(!$Categories ||!$make||!$type){
  echo "You have not enter all fields";
  exit;
  }
$db = mysql_connect($host, $username, $password) or die(mysql_error());
mysql_select_db($database) or die(mysql_error());


$bizwebpage2 = (isset($_FILES['bizwebpage2']) ? $_FILES['bizwebpage2']:'');
$Picture1 =  (isset($_FILES['Picture1']) ? $_FILES['Picture1']:'');


if(isset($_GET['Categories'])){
    $Categories = addslashes($_GET['Categories']);
    }
    else {
    echo("No Categories found.");
}
$query = "SELECT * FROM `business_info` WHERE `Categories` LIKE '%".$Categories."%'";
$result = mysql_query($query) or die (mysql_error());
$num_result = mysql_num_rows($result);
  echo "<table>";
if($num_result <= 0){
    echo("No Results found.");
    exit;
}
//include login row
  echo "<tr>";
  echo "<td>";
echo"login";
  echo "</td>";
  echo "</tr>";
  //inlude topsearch
  echo "<tr>";
  echo "<td>";
include("includefiles/topsearch.php");
  echo "</td>";
  echo "</tr>";
 
  //business found row
  echo"<tr>";
  echo"<td div align='right' bgcolor='DFDFDF'>";
  echo"<p>Number of business found: ".$num_result."</p>";
  echo"</td>";
  echo"</tr>";
 
while($row = mysql_fetch_array($result)){
echo "<tr>";
//picture colums
echo "<td>";
echo"<table>";
echo"<tr>";
echo"<td>";

echo "<img src='$Picture1' width=\'203\' height=\'152\'>"; // picture display
//business colums
echo"</td>";
echo"<td>";
echo '<a href="'. $row['includefiles/bizwebpage2.php'].'">' . $row['BusinessName'] . '</a>'; //link to another page
echo "<br>";
echo ($row["Slogan"]);
    echo "</strong><br>Address: ";
    echo (stripslashes($row["Business_Address"]));
    echo "<br>State: ";
    echo (stripslashes($row["make"]));
    echo "<br>City: ";
    echo (stripslashes($row["type"]));
    echo "<br>Tel: ";
    echo (stripslashes($row["Tel"]));
    echo "</p>";
echo"<hr>";
echo "</td>";
//include banner colums
echo "<td>";
include("includefiles/side_business_banner.php");
echo "</td>";
echo "</tr>";
echo "</table>";
echo "</td>";
echo "</tr>";

echo "</table>";
}

?>
[/code]

[franknu]

i made this change to the picture line. but still, i am having the same problem
[code=php:0]

echo "<img src=\'[$Picture1]\' width=\'203\' height=\'152\'>";

[/code]