I have found a great PHP script to rank my scores and have played around with some data in myPHPAdmin and it works as I hoped it would.

My next challenge is using it in PHP to extract the result. Heres the example code;

PHP Code:

$query="SELECT Sum(Score) As Rank, id FROM table WHERE id='2'"; 


The code only produces one result - as RANK - but I can't figure out how to get that now into a PHP variable. Any ideas much appreciated.

BTW - already tried mysql_fetch_array

maybe try mysql_result(): http://us2.php.net/mysql_result

true!
mysql_result()
should work.

$query="SELECT Sum(Score) As Rank, id FROM table WHERE id='2'"; 
$result = (mysql_query($query));
$row = (mysql_fetch_array($result));

This will run the query and put the records into a variable called $row

then use $row['fieldname'] to get the information from the field.

fieldname being the name of the column in the table.

Ok. The query is a little more complex than first stated so maybe that has something to do with it so here it is in its entirety;

PHP Code:

$query="SELECT count(*)+1 as Rank FROM (SELECT * FROM fav_blogs WHERE favScore > (SELECT sum(favScore) from fav_blogs where blogID='2') order by favScore desc) as Rankings";
$result=mysql_result($query,0) or die mysql_error(); 


But then I still get this unexpected T_STRING error occurring.

Am I missing some syntax in this?

Your sql is either a little more advanced than what I am used to or a cluster of stuff that doesn't work, but I don't believe you need the single quotes around the number 2.

An unexpected T_STRING error means that you have a terminating string somewhere, you forgot a semi-colon in your code.

die is a function. die(mysql_error()); not die mysql_error();

Originally Posted by Creadiv

Your sql is either a little more advanced than what I am used to or a cluster of stuff that doesn't work, but I don't believe you need the single quotes around the number 2.

For integers you don't need quotes, but it doesn't hurt to have them.

Cheers NullPointer. Changed the function syntax.

The error I get now is;

mysql_result(): supplied argument is not a valid MySQL result resource

Most of the time that error means that you did not connect to the db or you search didn't have any results.