how could i solve this code problem in php [duplicate] - php

Question

This question already has answers here:
SET a variable in SELECT statement - MySQL
(2 answers)
Closed 4 years ago.
i want to rank my users by score from my 'scores' table with php using mysql . i wrote this query that is actually working in SQL Tab of phpMyAdmin :
SET #rank=0;
SELECT #rank:=#rank+1 AS rank, name, score
FROM scores
ORDER BY score DESC;
but i cant figure it out , how to use it in my php Code . i tried:
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo $row["score"]. $E ." ". $row["name"]. $N ;
echo PHP_EOL;
echo PHP_EOL;
}
} else {
echo "0 results";
}
i think my problem is with my while loop . but i dont know what should i do :(
PleaseHelp(); :(
guys problem was solved . Our Dear Friend #mallik1055 noticed that the problem was with query and im sending multiple request . then i noticed That phpMyAdmin automatically runs multiple query unlike PHP , so i tried mysqli_multi_query and it worked , thank you a lot <3

Answer 1

That should work.
$rank = 0;
$rank = $rank+1;
$sql = "SELECT rank, name, score FROM scores ORDER BY score DESC";
$result = mysqli_query($conn, $sql);
$row = mysqli_fetch_assoc($result);
if ($row > 0) {
while($row > 0) {
echo $row["score"].$E." ".$row["name"].$N.;
echo PHP_EOL;
echo PHP_EOL;
}
} else {
echo "0 results";
}

Answer 2

$row["rank"] is an element in your fetched rows, just like "score" and "name"