A simple php script error result with tables - php

Question

Hello guys i'm in need of php experts advice.
Here's my problem. A user has 2 inputs the start and the end this are all integers. this will able to identify odd and even. i have solve already odd and even and almost done. main problem ex. start value 1 end 5. 1 is odd it should be displayed on odd in table. but the problem is it is found in even table. initial value is the problem. the rest was good.
here's my code
<?php
$firstnum = $_POST['first_input'];
$secondnum = $_POST['second_input'];
$counter = 0;
echo "<table border='1'>";
if ($firstnum < $secondnum) {
echo "<tr>"; //first tr
echo "<th>"; echo "Even numbers"; echo "</th>";
echo "<th>"; echo "Odd numbers"; echo "</th>";
echo "</tr>";
for ($counter=$firstnum; $counter <= $secondnum ; $counter++) {
if ($counter % 2 == 0){
echo "<tr>";
echo "<td>";
echo $counter;
echo "</td>";
} else {
echo "<td>";
echo $counter;
echo "</td>";
echo "</tr>";
}
}
} elseif ($firstnum > $secondnum) {
# code...
//first num is < second num
echo "<tr>"; //first tr
echo "<th>"; echo "Even numbers"; echo "</th>";
echo "<th>"; echo "Odd numbers"; echo "</th>";
echo "</tr>";
for ($counter=$firstnum; $counter >= $secondnum ; $counter--) {
if ($counter % 2 == 0){
echo "<tr>";
echo "<td>";
echo $counter;
echo "</td>";
} else {
echo "<td>";
echo $counter;
echo "</td>";
echo "</tr>";
}
}
}
echo "</table>";
?>

Answer 1

Your issue is you have invalid html resulting from your if/else blocks.
If your if you have
<tr>
<td><td>
and in your else you have
<td></td>
</tr>
Both of these need full row/cell tags
<tr>
<td></td>
<td></td>
</tr>
So your code should look like
if ($counter % 2 == 0){
echo "<tr>";
echo "<td>";
echo $counter;
echo "</td>";
echo "<td>";
echo "</td>";
echo "</tr>";
} else {
echo "<tr>";
echo "<td>";
echo "</td>";
echo "<td>";
echo $counter;
echo "</td>";
echo "</tr>";
}