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Hi

 

Just trying to save on some php by trying this query:

 

  $query = "SELECT @rank := @rank +1 AS rank, userid, percent FROM ranking ORDER BY percent ASC";

 

Which I have tried directly against the mysql instance with total success.  Trying from within PHP the rank column doesn't seem to be there...  Is this even allowed, do the @ need escaping, any suggestions welcome... even if I have to use a view, please let me know.

 

Heres the code fragment:

 

    $conn = get_db_conn();

    $query = "SELECT @rank := @rank +1 AS rank, userid, percent FROM ranking ORDER BY percent ASC";

    $res = mysql_query($query,$conn);

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

        echo $row['rank'], " ";

        echo $row['userid'], " ";

        echo $row['percent'], " <br>";

    }

 

The userid and percent are echoed the rank is simply missing... 

 

Thanks for any hep.

 

PHP newbie :(

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that's helpful for me as a newbie, thank you.  I also checked the SQL itself Here's the some resultset:

 

SELECT @rank := @rank +1 AS rank, userid, percent FROM ranking ORDER BY percent DESC

Array

(

    [0] =>

    [rank] =>

    [1] => 573522302

    [userid] => 573522302

    [2] => 11.36

    [percent] => 11.36

)

 

As you can see the rank "virtual column" contains no value...  Do I conclude such virtual columns as unsupported?

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i remember now... that bit of the query wasn't working so i took it off and left just the it from the SELECT.  Seems I have to set the rank firstly on this connection and then select the rank

 

    $query = "SET @rank := 0";
    mysql_query($query,$conn);
    $query = "SELECT @rank := @rank +1 AS rank, userid, percent FROM ranking ORDER BY percent DESC";
    echo "<pre>",$query,"</pre>";

 

Could these I combine these into one mysql_query() call, seemingly it doesn't like me using a ; to concatenate these statements?

 

... but i'm helped and many thanks

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