Ok I have this code:

//this is the hi.php//
<?php
//highlight items//
$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("juliver", $con);

$result = mysql_query("SELECT * FROM hi WHERE pp='2'");

$hi = "";

while($row = mysql_fetch_array($result)) //<--- this is the line 13//
  {
  $hi .= "<div id='hicontainer'><a id='download_now' class='imgx' href='#?w=700' rel='popup'><img src='".$row['name']."' />";
  $hi .= "<p>".$row['title']."</p>";
  $hi .= "<a href='#?w=700' id='".$row['id']."' rel='popup' class='imgx'>View full</a>    </div>";
  }

//Lots of lots of code here I just specified those code which in error.//

mysql_close($con);

?>

and heres where im going to display the output actually.

<td>
<!--highlight items-->
<div id="tbtitle">
<img src="Images/galleryicon.png"/><p>Highlight items</p>
</div>
<div id="tblgal">
<? echo $hi; ?>
</div>
</td>

But i get an error saying: "Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\madeinusa\hi.php on line 13."

please, im stuck with this. Thank you in advance.

• php
• html

You probably have an error in your query "SELECT * FROM hi WHERE pp='2'"
Try executing that query directly in mysql and see what happens. Make sure you have the table names and columns spelt right.

Also, try quoting the table names and columns in backtick ` characters to avoid accidentally using MySQL reserved words.

$result has a boolean value - probably because your query failed.

You should validate the result, before trying to use it.

After your query, you could do something like the following to see why your query fails:

if(!$result)
  exit(mysql_error());

or, more simply, you can use

$result = mysql_query("SELECT * FROM hi WHERE pp='2'") or die(mysql_error());

check to this:

mysql_connect("localhost","username","password");

First you should use proper exception handling in the code. After the line

$result = mysql_query("SELECT * FROM hi WHERE pp='2'");

you should check whether it returns resource id or any error in the way

if(!$result){
    // if you use try-catch, then use
   throw new Exception(mysql_error());
   // Or you can use
   die(mysql_error());
}else{
     while($row = mysql_fetch_array($result)) //<--- this is the line 13//
     {
        $hi .= "<div id='hicontainer'><a id='download_now' class='imgx' href='#?w=700'     rel='popup'><img src='".$row['name']."' />";
        $hi .= "<p>".$row['title']."</p>";
        $hi .= "<a href='#?w=700' id='".$row['id']."' rel='popup' class='imgx'>View    full</a>    </div>";
     }

}    

"SELECT * FROM hi WHERE pp='2'" query may return false means there are no records that full fill query.Try echo mysql_num_rows($result); in order to know no of rows return for query.