(simple?) Problem out putting MySQL query results

Hi,

I have an example query that was supplied with a post (zip) code database. It is supposed to calculated which suppliers are closest to the entered postcode.

After validating the postcode format the function below is called

function getDistanceUK($fromPcode)
{
	$_from_postcode = splitpostcode($fromPcode); // parse postcode to match database search criteria
	$_from_postcode = substr(md5($_from_postcode), 0, 8); // encrypts received value using PHP md5
	$_to_postcode = splitpostcode($targetPcode); // parse postcode to match database search criteria
	$_to_postcode = substr(md5($_to_postcode), 0, 8); // encrypts received value using PHP md5
	
$db_conn = db_connect(); // connect to database

$query = "SELECT sup_name, sup_postcode_md5, (SQRT(POW((b.coord1 - a.coord1), 2) + POW((b.coord2 - a.coord2), 2))/1000) * 0.621 AS distance
    FROM uk_code a, uk_code b, suppliers
    WHERE a.code = '".$_from_postcode."' AND b.code = suppliers.sup_postcode_md5
    HAVING (distance < 5)
    ORDER BY distance asc ";

	if (!$query) {
  echo '</table>';
  exit('<p>Error retrieving data from database!<br />'.
      'Error: ' . mysql_error() . '</p>');
	}

/// MY EFFORTS TO PRINTS THE RESULTS

while ($list = mysql_fetch_array($query)) {
  echo "<tr valign='top'>\n";
   $sup_postcode_md5 = $list['sup_postcode_md5'];
  echo "<td>$sup_postcode_md5</td>\n";
  echo "</tr>\n";
}

	}

PHP:

However my efforts to display the results show the error "mysql_fetch_array(): supplied argument is not a valid MySQL result resource "

Any one see where im going wrong?

 

dude, you create the query, but didn't run it in mysql.

add this after the "$query = SELECT sup_name,".....

$result=mysql_query($query) or die(mysql_error());

remove line 16-20

then replace
while ($list = mysql_fetch_array($query)) {
with
#
while ($list = mysql_fetch_array($result)) {

 

Thank you dude! Caouldn't see the wood for the trees!
Your a star!

 

You'll probably get this straight away too!

How can i add an "if returns no records 'echo 'this message'"?

if (!$result){
echo 'none1';
}

seems to print everytime

 

TRY

$count=mysql_num_rows($result) ;
if($count==0){echo "none1";}

 

Thanks, but

"mysql_fetch_array(): supplied argument is not a valid MySQL result resource"

Am i putting it in the right place?

	$db_conn = db_connect(); // connect to database

$query = "SELECT sup_name, sup_postcode, sup_address, sup_postcode_md5, (SQRT(POW((b.coord1 - a.coord1), 2) + POW((b.coord2 - a.coord2), 2))/1000) * 0.621 AS distance
    FROM uk_code a, uk_code b, suppliers
    WHERE a.code = '".$_from_postcode."' AND b.code = suppliers.sup_postcode_md5
    HAVING (distance < '".$area."')
    ORDER BY distance asc ";

$result=mysql_query($query) or die(mysql_error());

	echo "<table width=\"100%\"  align=\"left\" border=\"0\" cellspacing=\"0\" cellpadding=\"0\" ><tr>";

	while ($list = mysql_fetch_array($result)) {
   		$sup_postcode_md5 = $list['sup_postcode_md5'];
     	$sup_name = $list['sup_name'];
	    $sup_postcode = $list['sup_postcode'];
		$distance = $list['distance'];
		$distance = round($distance);
		$sup_address = $list['sup_address'];

   echo "<td colspan=\"3\" align=\"left\"><p class=\"suppliersName\">$sup_name</p></td>";
   echo "<td align=\"right\"><p class=\"suppliersDistance\">$distance miles away</p></td></tr>";
   echo "<tr><td colspan=\"3\" align=\"left\" width=\"130px\"><span class=\"suppliersAddress\">$sup_address $sup_postcode</span></td>";
   echo "<td align=\"left\">&nbsp;</td>";
	echo "</tr>";
}

echo "</table>";

	}
$count=mysql_num_rows($result) ;
if($count==0){echo "none1";}

PHP:

 

Thanks dude. being thick again. i worked it out.

Thanks again for your help!