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displaying Pictrure

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 fileseasyphp1-8hometownsfinderbusiness_display.php on line 79
i tried difining it at the top but it is not working maybe because it a link when i click on the link it takes me to the root not the actual bizwebpage2.php file

[code=php] ((( 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]

<?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]

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PHP

2 Comments(s)

Copy linkTweet thisAlerts:
@so_is_thisNov 13.2006 — That should be either this:

echo '<a href="includefiles/bizwebpage2.php">' . $row['BusinessName'] . '</a>';

or this:

echo '<a href="'. $row['LinkPage'].'">' . $row['BusinessName'] . '</a>';

Where "LinkPage" must be the name of a column in your database result set.
Copy linkTweet thisAlerts:
@franknuauthorNov 13.2006 — ohh, there is going to be more info so what should be like a loop them,

ok let;s say, the colum would be id how would i tell it to show just certains colums after they click on BusinessName or how i would i set the colors and format
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