PHP Calculate rank from database - php

Question

I got a little problem, I've got a database, in that database are different names, id, and coins. I want to show people their rank, so your rank has to be 1 if you have the most coins, and 78172 as example when your number 78172 with coins.
I know I can do something like this:
SELECT `naam` , `coins`
FROM `gebruikers`
ORDER BY `coins` DESC
But how can I get the rank you are, in PHP :S ?

Answer 1

You can use a loop and a counter. The first row from MySql is going the first rank,I.e first in the list.

Answer 2

I presume you want something like:
1st - John Doe
2nd - Jane Doe
..
..
right?
See: http://www.if-not-true-then-false.com/2010/php-1st-2nd-3rd-4th-5th-6th-php-add-ordinal-number-suffix
Helped me a while ago.

Answer 3

You could use a new varariable
$i = "1";
pe care o poti folosi in structura ta foreach,while,for,repeat si o incrementezi mereu.
and you use it in structures like foreach,while,for,repeat and increment it
$i++;
this is the simplest way

Answer 4

No code samples above... so here it is in PHP
// Your SQL query above, with limits, in this case it starts from the 11th ranking (0 is the starting index) up to the 20th
$start = 10; // 0-based index
$page_size = 10;
$stmt = $pdo->query("SELECT `naam` , `coins` FROM `gebruikers` ORDER BY `coins` DESC LIMIT {$start}, {$page_size}");
$data = $stmt->fetchAll();
// In your template or whatever you use to output
foreach ($data as $rank => $row) {
// array index is 0-based, so add 1 and where you wanted to started to get rank
echo ($rank + 1 + $start) . ": {$row['naam']}<br />";
}
Note: I'm too lazy to put in a prepared statement, but please look it up and use prepared statements.

Answer 5

If you have a session table, you would pull the records from that, then use those values to get the coin values, and sort descending.
If we assume your Session table is sessions(session_id int not null auto_increment, user_id int not null, session_time,...) and we assume that only users who are logged in would have a session value, then your SQL would look something like this: (Note:I am assuming that you also have a user_id column on your gebruikers table)
SELECT g.*
FROM gebruikers as g, sessions as s WHERE s.user_id = g.user_id
ORDER BY g.coins DESC
You would then use a row iterator to loop through the results and display "1", "2", "3", etc. The short version of which would look like
//Connect to database using whatever method you like, I will assume mysql_connect()
$sql = "SELECT g.* FROM gebruikers as g, sessions as s WHERE s.user_id = g.user_id ORDER BY g.coins DESC";
$result = mysql_query($sql,$con); //Where $con is your mysql_connect() variable;
$i = 0;
while($row = mysql_fetch_assoc($result,$con)){
$row['rank'] = $i;
$i++;
//Whatever else you need to do;
}
EDIT
In messing around with a SQLFiddle found at http://sqlfiddle.com/#!2/8faa9/6
I came accross something that works there; I don't know if it will work when given in php, but I figured I would show it to you either way
SET #rank = 0; SELECT *,(#rank := #rank+1) as rank FROM something order by coins DESC
EDIT 2
This works in a php query from a file.
SELECT #rank:=#rank as rank,
g.*
FROM
(SELECT #rank:=0) as z,
gebruikers as g
ORDER BY coins DESC

Answer 6

If you want to get the rank of one specific user, you can do that in mysql directly by counting the number of users that have more coins that the user you want to rank:
SELECT COUNT(*)
FROM `gebruikers`
WHERE `coins` > (SELECT `coins` FROM `gebruikers` WHERE `naam` = :some_name)
(assuming a search by name)
Now the rank will be the count returned + 1.
Or you do SELECT COUNT(*) + 1 in mysql...