PHP For() Loop Issue - php

Question

I have a question that I have been stuck on for several hours now. I have played around with numerous types of for() and while() loops. I put them in different locations with different variables and ran different things, nothing worked..
My question:
Why is my program giving all users below the first one the same level? You can clearly see in the picture that Nicolas has much more XP. (5,000 xp is level 20, and if "Nic5" was first in the database then it would change the "Skill Level" to 20.
I know that the returned variable $lvl isn't changing for each player that loads and this is why each player is getting the first player's level.
Can anybody help me with this please?
Notes:
0 = a column that holds experience for a players skill level.
class calculatelevel:
class calculatelevel {
function level($skillnum)
{
$host = "*";
$user = "*";
$pass = "*";
$db = "*";
$con = new mysqli($host, $user, $pass, $db) or die($con->error);
$res = $con->query("SELECT `0` FROM hiscores");
$max = 99;
while($row = $res->fetch_assoc()) {
$xp = $row['0'];
// Find the appropriate level
for ($lvl = 1; $lvl < $max; $lvl++) //this for loop runs 99 times
{
if ($xp < $this->experience($lvl))//if players xp in skill is less than experience(level 1-99)
{
// Level found
$lvl -= 1;
break;
}
}
}
return $lvl;
}
public function experience($lvl)
{
$xp = 0;
for($x = 1; $x < $lvl; $x++)
{
$xp += floor($x + 300 * pow(2, ($x / 7)));
}
return floor($xp / 4);
}
}
Method that writes database information to page.
if($res->num_rows > 0) {
echo "<table>
<tr>
<td>Rank</td>
<td>Username</td>
<td>Skill Level</td>
<td>Total Exp</td> </tr>";
while($row = $res->fetch_assoc()) {
echo 'ran';
$calc = new calculatelevel();
$level = $calc->level(0);
echo '<tr>
<td>'.($count+1).'</td>
<td>'. htmlspecialchars($row['username']) .'</td>
<td>'.number_format($level).'</td>
<td>'.number_format($row['0']).'</td>
</tr>';
$count++;
}
}

Answer 1

Your while loop and for loop are structured incorrectly. You are looking at only the first row and returning the level for that row every time. You break the for loop when you find the level for that row's player, then immediately return that level. Result: it looks like everyone has the same level.
EDIT: Okay, here are a few more thoughts.
First, a column named 0 is asking for trouble, as Mike W pointed out. You say 0 is a table, but if that's the case, your SELECT statement doesn't make sense. The first thing I would try is changing the column name to something that isn't a number, like xp.
Second, you really should make only one database connection and use it throughout the entire request, if possible. Opening a new connection each time a particular function runs will tie up a server quickly.
Third, the obvious problem in your current code is this:
function level($skillnum) {
// other code here....
// Okay, you load a row's data into $row, with the idea that you will repeat this.
while($row = $res->fetch_assoc()) {
// $xp is set to the experience points for the user you just loaded
$xp = $row['0'];
// You now look at each level, starting at 1, to see if the
// user's xp is greater than the cutoff for that level
for ($lvl = 1; $lvl < $max; $lvl++) //this for loop runs 99 times
{
// If the user's xp is less than the cutoff...
if ($xp < $this->experience($lvl))//if players xp in skill is less than experience(level 1-99)
{
// ... then you go back down one level...
// Level found
$lvl -= 1;
// ... and quit the for loop!
break;
}
}
// okay, you're out of the for loop, so you go back to the while loop...
// a new row is loaded...
// and $xp and $lvl are both overwritten with that user's values
}
// So, the while loop has run once for each player...
// ... but you are only returning one player's level!
return $lvl;
}
Also, you have defined calculatelevel::level to require a parameter $skillnum, but you never use it in the code posted here.
I suspect there is another glitch in your real code causing it to return the first player's level, rather than the last player's level. It could be a problem with the 0 column name; that really should change.