how can I higlight the correct cell in PHP - php

Question

I am trying to make a function in PHP which writes out a table, and looks in the database to find what cells should have info. the grid will always be the same size, but the content may be in different places.
I've gotten it to be able to look in the database, though it seems to only highlight the first cell, rather than the correct coordinates.
require("sql.php");
$sql = <<<SQL
SELECT *
FROM `maps`
WHERE `objpresent` = 1
SQL;
if(!$result = $db->query($sql)){
die('There was an error running the query [' . $db->error . ']');
} // ran the query
$xobj = array();
$yobj = array();
while($row = $result->fetch_assoc()){
//echo $row['x'] . $row['y'] . $row['object'] . '<br />';
$xobj[] += $row['x'];
$yobj[] += $row['y'];
}// get the rows
//find whether the row is obstructed
for($a=0; $a<=20-1; $a++) //rows (y)
{
for($i=0; $i<=25-1; $i++) //cols (x)
{
echo "<td>"; //between these write the apt content
// if (empty($xobj[$i]) || empty($yobj[$a]) ){
// echo '0';
//} //detect whether there is even a record for this space
if(!empty($xobj[$i]))
{
if(!empty($yobj[$a]))
{
echo $xobj[$i]; //debug
if($xobj[$i] == $i)
{
//echo $xobj[$i];
echo "A";
}
}
}
//echo "<td><img src='emptysym.png'></img></td>";
echo "</td>"; //add textual descriptions for now, add icons later
}
echo "</tr>";
}
this is my current(though rather messy) code.
if there is a row with the column x saying 2, and the column y saying 3, then it should put a letter at (2,3.
is it possible to fix this, or is there a better method for this?

Answer 1

Use a 2-dimensional array whose indexes are the x and y values from the database:
$xyobj = array();
while($row = $result->fetch_assoc()){
$xyobj[$row['x']][$row['y']] = true;
}
Then your output loop should be:
for ($y = 0; $y < 20; $y++) {
echo '<tr>';
for ($x = 0; $x < 25; $x++) {
echo '<td>';
if (isset($xyobj[$x][$y])) {
echo 'A';
}
echo '</td>';
}
echo '</tr>';
}