PHP Ranking....

[Verbat]

I have a system that i'm working on that lats say has 1000 members now.

Each member has his own display page and own rating points.

 

What i want to do (and dont know how ) is to show on each of the members page his rank out of all the 1000 members.

Like:

 

V-Tech Ranked 317 of 1000 members

 

 

Of course all the info is set on a SQL db.

 

 

Thanks.

[burnside]

what is your table like in the db ?

[adam291086]

Well if the ranking is stored in the database then all you will need to do is count how many records you have in your database, mysql_count_rows i think, then echo that result out with there rank

 

echo $rank;

echo "out of";

echo $total records;

 

 

[Verbat]

what is your table like in the db ?

I have the "rating" field that will be changed all the time depending on what the members will do and by that field i want it to be sorted.

 

adam291086:

It's not sorted out, each member has his rating points and i want it to be sorted and shown by that.

[adam291086]

well in you sql do something like

 

SELECT * FROM Foo

ORDER BY Rank DESC

 

Thank will give you all the ranking starting with the highest rank

[Verbat]

well in you sql do something like

 

SELECT * FROM Foo

ORDER BY Rank DESC

 

Thank will give you all the ranking starting with the highest rank

 

I know how to creat a page that will show me all the members and sort it, but what i'm trying to do is find what is a rating of a specific member out of everyone and show it on a page without the rest of the members.

[burnside]

well if the rating is saved in the db why cant you just echo that value out?

[adam291086]

Well how are you getting that specific member? like burnside and i said, why not count the number of records and echo it out along with their rank

 

Well if the ranking is stored in the database then all you will need to do is count how many records you have in your database, mysql_count_rows i think, then echo that result out with there rank

 

echo $rank;

echo "out of";

echo $total records;

 

[Verbat]

well if the rating is saved in the db why cant you just echo that value out?

The rating is saved but it's like number of points that the member has, and i want to make a code that will how much points the rest of the members has and tell me what is the rank of that member out of everyone.

[burnside]

this ill leave this one to you adam291086  haha

[Verbat]

Well how are you getting that specific member? like burnside and i said, why not count the number of records and echo it out along with their rank

 

Well if the ranking is stored in the database then all you will need to do is count how many records you have in your database, mysql_count_rows i think, then echo that result out with there rank

 

echo $rank;

echo "out of";

echo $total records;

 

The rank is not saved in the DB, the rating is saved and the rating is just number of points the member has.

[adam291086]

well you could try this

 

<?php

//work out the number of members
$query1 = "SELECT * FROM `table_name`"; 
$result1 = mysql_query($query);
$row1 = mysql_count_rows($result1)

//work out the rank
$query = "SELECT * FROM `table_name` ORDER BY `rating` DESC"; 
$result = mysql_query($query);
$ranknumber = 1;

while($row = mysql_fetch_array($result)){		

echo $ranknumber . ". " . $row['username'] . ", with " . $row['rating'] . " points out of" . $row1. "<br>";		

$ranknumber++; 
// increases $ranknumber by 1	
}?>

[Verbat]

well you could try this

 

<?php

//work out the number of members
$query1 = "SELECT * FROM `table_name`"; 
$result1 = mysql_query($query);
$row1 = mysql_count_rows($result1)

//work out the rank
$query = "SELECT * FROM `table_name` ORDER BY `rating` DESC"; 
$result = mysql_query($query);
$ranknumber = 1;

while($row = mysql_fetch_array($result)){		

echo $ranknumber . ". " . $row['username'] . ", with " . $row['rating'] . " points out of" . $row1. "<br>";		

$ranknumber++; 
// increases $ranknumber by 1	
}?>

 

It's not working... There is an error in the script. And i dont think that this is what i had in mind

[adam291086]

sorry i can see where i have gone wrong.

 

You want to add up the total rating points of all the member and display there ranking and there point out of a possible x amount

[Verbat]

Maybe i'll explain better -

In my DB table i have 3 feilds: ID,USERNAME,POINTS

Now i have 5 members in there:

 

ID USERNAME POINTS

1  Test1        10

2  Test2        56

3  Test3        1

4  Test4        34

5  Test5        15

 

Now i have a page that shows each of the members "profile" and in that page i want it to show me the rank of one member.

If for example i'm viewing Test3 members page i want it to tell me there that he is ranked 5 out of 5 members as he has the lowest points.

 

Maybe it's more clear now.

[burnside]

like adam said :

 

SELECT * FROM Foo WHERE id_foo = foo

ORDER BY Rank DESC

 

will only show that id just echo the id number and that will give him his rank  i think not to sure though.

[adam291086]

This is what you want

 

<?php

//work out the number of members
$query1 = "SELECT * FROM `table_name`"; 
$result1 = mysql_query($query);
$row1 = mysql_count_rows($result1)

//work out the total number of points

$sum = "SELECT *, SUM(POINTS) FROM table name "; 
$result3 = mysql_query($sum);

//work out the rank
$query = "SELECT * FROM `table_name` ORDER BY `rating` DESC"; 
$result = mysql_query($query);
$ranknumber = 1;

while($row = mysql_fetch_array($result)){		

echo $row['USERNAME']."is ranked".$ranknumber. "out of" . $row1. "members" . "with a point score of". $row['POINTS']."out of a possible".$result3. "<br />"

$ranknumber++; 
// increases $ranknumber by 1	
}?>

 

That will add a rank to the user depening on there point score.

Work out how many members there are and echo that result

works out the total number of points that have been issues

so the echo will look like this

 

adam is ranked 1  out of 5 members with a point score of 10 out of a possible 20

[burnside]

@adam like the adam is ranked 1st hahaha

[Verbat]

It tells me that there is an error on this like -

$query = "SELECT * FROM `table_name` ORDER BY `rating` DESC"; 

 

And i changed the details so it will fit my DB

[adam291086]

add that to the end of thw sql

 

or die('Query failed. ' . mysql_error());

 

for example

 

$result = mysql_query($query) or die('Query failed. ' . mysql_error());

 

tell me what the errors are so i can help

[Verbat]

Parse error: syntax error, unexpected T_VARIABLE in c:\wamp\www\fusion\test.php on line 29

 

That's the error.

 

Any chance you have MSN or something?

[burnside]

i think missed a ; out i could be wrong though.

[adam291086]

i am at work so no msn. HAHA i working really hard

 

What is on line 29. We are not mind readers you have to be specific with errors and line numbers.

[Verbat]

i am at work so no msn. HAHA i working really hard

 

What is on line 29. We are not mind readers you have to be specific with errors and line numbers.

 

I wrote the line before

 

$query = "SELECT * FROM `table_name` ORDER BY `rating` DESC"; 

[adam291086]

post you actual code

 

do you have a table in your database called table_name. How can i help when you don't post the code your using.