Re: Building a query to fetch an array

[Barry Galbraith]

>echo "$clli = $row['$pointcode']<br>";

Nope! pointcode here is the name of the field you return in your SELECT sql statement.
$row['pointcode'] is an element of the $row array. It turns out there is one element in that array, and it's name is 'pointcode'.

This is a real good case of confusing your variables.

DON'T DO THIS!

$pointcode = mysql_result($pointcode, 0);

The left side $pointcode is a (possibly) a string. The right side $pointcode is (possibly) a resource.

I don't think you substituted the code correctly, so here's the full code.

<?php

mysql_connect("localhost", "root", "");

$clli = mysql_real_escape_string($_POST['clli']);

if ($clli==NULL)
echo "Please enter an 11 digit CLLI code!";
else
{
$pointcode = mysql_query("SELECT pointcode FROM database1.tester1 WHERE clli='$clli'" );
// $pointcode will be false, or a resource

$pointcode_num_rows = mysql_num_rows($pointcode);

if ($pointcode_num_rows==0)
echo "CLLI code does not exist!";
else
{
while ($row = mysql_fetch_array($pointcode)) {
    echo "$clli = $row['pointcode']<br>";
    }
}

}

?>

Good luck,
Barry.