SQL calculate row number - php

Question

I doing league table football and There is a profile of Team (Like : Team.php?team=XXX)
In this page I want to show, What position of TeamXXX in League Table
Page League Table
<?php
$number = 0;
$sql = "SELECT * FROM `leaguetable` WHERE `league` = 'leaguename' ORDER BY pts DESC";
$query = mysql_query($sql);
while($rs=mysql_fetch_assoc($query)){
$number++;
?>
<table>
<thead>
<tr>
<th>Position</th>
<th>Team</th>
<th>Points</th>
</tr>
</thead>
<tbody>
<td><?php echo $number; ?></td>
<td><?php echo $rs['team']; ?></td>
<td><?php echo $rs['pts']; ?></td>
</tbody>
</table>
<?php } ?>
Data in Table leaguetable
id team pts
In team.php I want to show position of TeamXXX
<?php
$getTeam = mysql_fetch_assoc(mysql_query("SELECT * FROM `leaguetable` WHERE `team`='"$_GET['team']"'");
?>
<table>
<thead>
<tr>
<th>Position</th>
<th>Team</th>
<th>Points</th>
</tr>
</thead>
<tbody>
<td>#########</td>
<td><? echo $getTeam['team']; ?></td>
<td><? echo $getTeam['pts']; ?></td>
</tbody>
</table>
How can I know what position of teamXXX in leaguetable?
Any help will be greatly appreciated. Thank you very much

Answer 1

query
select id, team, pts, rnk
from
(
select leag.id, leag.team, leag.pts,
#rnk := if(leag.pts = #lag, #rnk,
if(#lag := leag.pts, #rnk + 1, #rnk + 1)) as rnk
from leaguetable leag
cross join ( select #rnk := 0, #lag := null ) params
where league = 'FA Cup'
order by leag.pts desc
) rankings
where team = 'Chelsea'
;
example.php
<?php
/**
* Mysqli initial code
*
* User permissions of database
* Create, Alter and Index table, Create view, and Select, Insert, Update, Delete table data
*
* #package PhpFiddle
* #link http://phpfiddle.org
* #since 2012
*/
require_once "dBug!.php";
require "util/public_db_info.php";
$short_connect = new mysqli($host_name, $user_name, $pass_word, $database_name, $port);
if (mysqli_connect_errno())
{
die("Failed to connect to MySQL: " . mysqli_connect_error());
}
/*
$sql = "create table leaguetable"
. "("
. " id integer primary key not null,"
. " team varchar(33) not null,"
. " pts integer not null default 0"
. ");";
$result = $short_connect->query($sql);
if(!$result)
{
die("Create table failed : " . mysqli_error($short_connect));
}
$sql = "insert into leaguetable"
. "( id, team, pts )"
. "values"
. "( 1, 'Liverpool', 22 ),"
. "( 2, 'Arsenal', 29 ),"
. "( 3, 'Chelsea', 23 ),"
. "( 4, 'Tottenham', 23)";
$result = $short_connect->query($sql);
if(!$result)
{
die("insert failed : " . mysqli_error($short_connect));
}
*/
//get all tables in the database
//$sql = "SHOW TABLES";
//get column information from a table in the database
//$sql="SELECT COLUMN_KEY, COLUMN_NAME, COLUMN_TYPE FROM information_schema.COLUMNS WHERE TABLE_NAME = 'books'";
//SQL statement for a table in the database
$sql = "select id, team, pts, rnk "
. "from"
. "("
. "select leag.id, leag.team, leag.pts,"
. "#rnk := if(leag.pts = #lag, #rnk,"
. " if(#lag := leag.pts, #rnk + 1, #rnk + 1)) as rnk "
. "from leaguetable leag "
. "cross join ( select #rnk := 0, #lag := null ) params "
. " where league = 'FA Cup' "
. "order by leag.pts desc;"
. ") rankings "
. where team = 'Chelsea';";
//result is boolean for query other than SELECT, SHOW, DESCRIBE and EXPLAIN
$result = $short_connect->query($sql);
if (($result) && ($result->num_rows > 0))
{
echo "<table>" . "<thead>" . "<tr>" . "<th>Position</th>" . "<th>Team</th>" . "<th>Points</th>" . "</tr>" . "</thead>" . "<tbody>";
//convert query result into an associative array
echo "<tr><td>" . $row['rnk'] . "</td><td>" . $row['team'] . "</td><td>" . $row['pts'] . "</td></tr>";
echo "</tbody></table>";
}
else
{
die("select failed : " . mysqli_error($short_connect));
}
$short_connect->close();
?>
output
<table>
<thead>
<tr>
<th>Position</th>
<th>Team</th>
<th>Points</th>
</tr>
</thead>
<tbody>
<tr>
<td>2</td>
<td>Chelsea</td>
<td>23</td>
</tr>
</tbody>
</table>
sqlfiddle

Answer 2

Hope this will help
$number = 0;
$points=0;
$sql = "SELECT * FROM `leaguetable` WHERE `league` = 'leaguename' ORDER BY pts DESC";
$query = mysql_query($sql);
while($rs=mysql_fetch_assoc($query)){
if($points!=$rs['pts'])
$number++;
if($rs['team']==$_GET['team']){
if($points==$rs['pts'])
$position=$number-1;
else
$position=$number;
}
$poins=$rs['pts'];
}

Answer 3

If you don't mind using two queries (I wouldn't mind it), it's pretty easy: Just count how many teams have more points, and add one (so that if three teams are tied in second position, they all get position 2).
$result = mysql_query("SELECT count(*)+1 AS POSTN FROM leaguetable WHERE
league = 'leaguename' AND pts > $points");
$row = mysql_fetch_assoc($result);
$position = $row["POSTN"];
Doing it in one query is a bit messier, since you need to embed this query in your original one:
"SELECT *, (SELECT count(*)+1 FROM leaguetable table2
WHERE league = 'leaguename' AND table2.pts > leaguetable.pts) AS POSTN
FROM leaguetable WHERE team = '$currentteam'"
But why are you using the mysql_* API for new code? Haven't you noticed all the dire warnings in pink boxes in the documentation? Do yourself a favor and switch to mysqli today, starting with this program.
Also: Never just inject $_GET[param] into your query string! You're giving yourself an SQL injection attack waiting to happen... and brittle, error-prone code until then.