I have been struggling to set this up for months and months!
I need help with setting rank to my database.
This is how my current code looks like:
$db->queryNoReturn("SET #a:=0");
return $db->query("
SELECT * FROM
(SELECT
`FFA_Stats`.`id`,
`FFA_Stats`.`player_uuid`,
`FFA_Stats`.`points`,
`FFA_Stats`.`hits`,
`FFA_Stats`.`shots`,
`FFA_Stats`.`wins`,
`FFA_Stats`.`tkills`,
`FFA_Stats`.`tdeaths`,
(`FFA_Stats`.`tkills`/`FFA_Stats`.`tdeaths`) as `KDR`,
`player`.`name`,
`player`.`uuid`,
`player`.`online`,
(#a:=#a+1) AS rank
FROM `FFA_Stats`
INNER JOIN `player` ON `FFA_Stats`.`player_uuid`=`player`.`uuid`
ORDER BY `points` DESC
) AS `sub`
");
Basically its sorting it by points and you can check how it looks like here: http://filipvlaisavljevic.com/clash/ffa.php
All I want to do is add rank to the sorted table so the player with the most points would be #1 etc.
Does anyone know what to do?
Usually a rank number would be an integer that you could generate from iterating through the rows of the query result. eg. echo $count++;
If you have calculated or attributed a rank in your database then you can add 'order by' statements separated by commas. eg.
FROM `FFA_Stats`
INNER JOIN `player`
ON `FFA_Stats`.`player_uuid`=`player`.`uuid`
ORDER BY `rank` DESC, `points` DESC) AS `sub`
");
Related
i have a problem my database looks like this DATABASE STRUCTURE
i want to sort the lboard and ignore the repeated values
like this REQUIRED OUTPUT
can anyone help me?
You could use GROUP BY to get single result for one user and MAX function to get its maximum score
the query will be like this
SELECT id, exam_id, user_id, MAX(lboard) FROM `leaderb` GROUP BY user_id ORDER BY lboard DESC
SELECT id, exam_id, user_id, MAX( lboard )
FROM wp_watu_takings WHERE exam_id = 3
GROUP BY user_id
ORDER BY MAX( lboard ) DESC
Look at my code, I want the select statement order by the count percentage after I fetch the data from this select statement, obviously, it's not logical. What can I do? Help, appreciate.
<?php
//myslq connection code, remove it because it's not relate to this question
$stm =$db->prepare("SELECT id ,term_count, COUNT(user_id) as count FROM sign WHERE term IN (:term_0,:term_1) GROUP BY user_id ORDER by count DESC");
//trying replace order by count with $combine_count, but it's wrong
$term_0="$term[0]";
$term_1="$term[1]";
$stm->bindParam(":term_0", $term_0);
$stm->bindParam(":term_1", $term_1);
stm->execute();
$rows = $stm->fetchALL(PDO::FETCH_ASSOC);
foreach ($rows as rows) {
$count=$rows['count'];
$term_count_number=$rows['term_count'];
$count_percentage=round(($count/$count_user_diff)*100);
$count_key_match=round(($count/$term_count_number)*100);
$combine_count=round(($count_percentage+$count_key_match)/2);
//issue is here, I want the select statement order by $combine_count
}
?>
SELECT id ,term_count, COUNT(user_id) as `count`
FROM sign
WHERE term IN (:term_0,:term_1)
GROUP BY user_id
ORDER by `count` DESC");
Since "count" is a function, it would be better to put backtics around the non-function "counts", as done above.
GROUP BY should list the field not aggregated. Otherwise, it does not know which id and term_count to fetch. So, depending on what you are looking for,
Either do
SELECT user_id, COUNT(*) as `count` -- I changed this line
FROM sign
WHERE term IN (:term_0,:term_1)
GROUP BY user_id
ORDER by `count` DESC");
or do
SELECT id ,term_count, COUNT(*) as `count`
FROM sign
WHERE term IN (:term_0,:term_1)
GROUP BY id ,term_count -- I changed this line
ORDER by `count` DESC");
SQL Syntax Logic
SELECT column1, count(column1) AS amount
FROM table_name
GROUP BY column1
ORDER BY amount DESC
LIMIT 12
I need to order my query by date first...
So I used this:
SELECT * FROM `mfw_navnode` order by `id` DESC
I wanted to order my results from last to first.
Then what I am trying to do
is to add a query over it, which would group my results by node_name..
The result should be..all the top nodes grouped by "category/node name type", while the first node that I see is was ordered the highest for its category in the first query..
I thought to do something like this:
SELECT * FROM(
SELECT * FROM `mfw_navnode` order by `id` DESC) AS DD
WHERE (node_name='Eby' OR node_name='Laa' OR node_name='MIF' OR node_name='Amaur' OR node_name='Asn' )
GROUP BY DD.node_name
I get no result..or any response from phpmyadmin when I input that result..
Where do I get wrong?
Note , I dont want to group my results and then order them..
I want them to be ordered, and then grouped. After being grouped..I want the result of each group to have the highest value ..from the other rows in the group
It is not sufficient to perform the ordering first, as even then MySQL makes no guarantee over which record it will select for each group. From the manual:
The server is free to choose any value from each group, so unless they are the same, the values chosen are indeterminate.
You must instead identify the records of interest with a subquery, then join the result with your table again in order to obtain the related values:
SELECT *
FROM mfw_navnode NATURAL JOIN (
SELECT node_name, MAX(id) AS id FROM mfw_navnode GROUP BY node_name
) AS DD
WHERE node_name IN ('Eby', 'Laa', 'MIF', 'Amaur', 'Asn')
Ordered by ID and group by node_name
SELECT * FROM `mfw_navnode`
WHERE (node_name='Eby' OR node_name='Laa' OR node_name='MIF' OR node_name='Amaur' OR node_name='Asn' )
GROUP BY DD.node_name
ORDER BY `id` DESC
Grouping is used commonly when You are using some aggregate function (sum, max, min, count, etc). If You don't use such function in Your query then why do You want to group the results?
Anyway, this should do the trick:
SELECT *
FROM mfw_navnode
WHERE id IN (SELECT id
FROM mfw_navnode
WHERE node_name IN ('Eby', 'Laa', 'MIF', 'Amaur', 'Asn')
GROUP BY node_name)
ORDER BY id
The following SQL may yield you the required output:
SELECT node_name, MAX(id)
FROM mfw_navnode
GROUP BY node_name
ORDER BY node_name
I see two problems with your SQL.
1) placing the order by in the inline select does nothing (and is probably causing an error)
2) you are grouping on node_name but you are not aggregating anything
SELECT COUNT(id) as row_count, node_name FROM( SELECT * FROM mfw_navnode ) AS DD
WHERE (node_name='Eby' OR node_name='Laa' OR node_name='MIF' OR node_name='Amaur' OR node_name='Asn' )
GROUP BY DD.node_name
order by node_name desc
further I am not sure why you need the inline select as the where could simply be on the original select ( perhaps you have something more complex going on that you didn't show )
SELECT COUNT(id) as row_count, node_name
from mfw_navnode
WHERE node_name='Eby' OR node_name='Laa' OR node_name='MIF' OR node_name='Amaur' OR node_name='Asn'
GROUP BY node_name
order by node_name desc
I have a database with users and points (in fact it's a percentage, but that doesn't matter). The user(s) with the highest number of points is on the first rank, the second on the second rank ...
I could get the rank of a $searchedUserID if I did somethink like this:
SELECT `user_id`, `points` FROM `usertable` ORDER BY `points` DESC
/** This function returns the rank of a user. The rank nr. 1 is the best.
* It is possible that some users share a rank.
*
* #param int $searchedUserID the ID of the user whose rank you would like to
* know
*
* #return int rank
*/
function getUserRank($searchedUserID)
{
$userArray = getAllUsersOrderedByPoints();
$rank = 0;
$lastPoints = -1; // Never happens
foreach ( $userArray as $user) {
if ($user['point'] != $lastPoints) $rank++;
if ($user['user_id'] == $searchedUserID) break;
}
return $rank;
}
Isn't there a more direct way to get this with (My)SQL?
If not: Can the PHP-part be improved?
(edit: I could store the rank calculated by PHP directly in the database ... but this would mean I had to make quite a lot of UPDATEs.)
edit2: Perhaps GROUP BY could be used? Something like:
SELECT `user_id`, `points` FROM `usertable` GROUP BY `points` ORDER BY `points` DESC
The problem with this query is the possibility, that I don't get the searched user_id. It would be necessary to send a second query:
SELECT `user_id` FROM `usertable` WHERE `points` = $pointsOfTheUser
You ask:
Isn't there a more direct way to get this with (My)SQL?
Yes. In SQL:2003, you could use the DENSE_RANK() window function. In MySQL, you can simulate this, as the (dense) rank of a record by some score is simply the count + 1 of distinct better scores:
SELECT u.user_id,
u.points,
1 + COUNT(DISTINCT others.points) AS `dense_rank`
FROM users u
LEFT JOIN users others
ON u.points < others.points -- Which other users have more points?
WHERE user_id = ?
GROUP BY 1, 2;
perhaps an inner join with group by and sort would do the trick?
SELECT * FROM
INNER JOIN
(
SELECT user_id AS uid, max(points) AS score
FROM usertable GROUP BY user_id
)
AS ds ON usertable.user_id = ds.uid AND usertable.points = ds.score
ORDER BY score DESC
Just thinking on paper (pixels) .. wouldn't that give you the list in order from highest points to lowest with a unique record per user ... or are you looking to have these sorted where you can clarify a tie as a single "place" in the rankings?
so I'm trying to create a ranking system for my website, however as a lot of the records have same number of points, they all have same rank, is there a way to avoid this?
currently have
$conn = $db->query("SELECT COUNT( * ) +1 AS 'position' FROM tv WHERE points > ( SELECT points FROM tv WHERE id ={$data['id']} )");
$d = $db->fetch_array($conn);
echo $d['position'];
And DB structure
`id` int(11) NOT NULL,
`name` varchar(150) NOT NULL,
`points` int(11) NOT NULL,
Edited below,
What I'm doing right now is getting records by lets say
SELECT * FROM tv WHERE type = 1
Now I run a while loop, and I need to make myself a function that will get the rank, but it would make sure that the ranks aren't duplicate
How would I go about making a ranking system that doesn't have same ranking for two records? lets say if the points count is the same, it would order them by ID and get their position? or something like that? Thank you!
If you are using MS SQL Server 2008R2, you can use the RANK function.
http://msdn.microsoft.com/en-us/library/ms176102.aspx
If you are using MySQL, you can look at one of the below options:
http://thinkdiff.net/mysql/how-to-get-rank-using-mysql-query/
http://www.fromdual.ch/ranking-mysql-results
select #rnk:=#rnk+1 as rnk,id,name,points
from table,(select #rnk:=0) as r order by points desc,id
You want to use ORDER BY. Applying on multiple columns is as simple as comma delimiting them: ORDER BY points, id DESC will sort by points and if the points are the same, it will sort by id.
Here's your SELECT query:
SELECT * FROM tv WHERE points > ( SELECT points FROM tv WHERE id ={$data['id']} ) ORDER BY points, id DESC
Documentation to support this: http://dev.mysql.com/doc/refman/5.0/en/sorting-rows.html
Many Database vendors have added special functions to their products to do this, but you can also do it with straight SQL:
Select *, 1 +
(Select Count(*) From myTable
Where ColName < t.ColName) Rank
From MyTable t
or to avoid giving records with the same value of colName the same rank, (This requires a key)
Select *, 1 +
(Select Count(Distinct KeyCol)
From myTable
Where ColName < t.ColName or
(ColName = t.ColName And KeyCol < t.KeyCol)) Rank
From MyTable t