Re: Getting Percentage of coloumn value

Here is my script: <? include "includes/required.php"; do_html_header('Name Here'); $query = "select @total_items=count(id) from tececo_stats"; $query1= "select referer,count(referer)/@total_items from tececo_stats group by referer"; $result = mysql_query($query) or die("Query failed: $query<br>" . mysql_error()); $result1 = mysql_query($query1) or die("Query failed: $query1<br>" . mysql_error()); $num_results = mysql_num_rows($result); ?> <table width="500"> <? for ($i=0; $i < $num_results; $i++) { $row = mysql_fetch_array($result1); echo '<tr><td>'.$row['referer'].'</td><td>'.$row['count(referer)'].'</td></tr>'; } ?> </table> <? do_html_footer(); ?> How can I echo count(referer) ? I currently get this error Warning: Undefined index: count(referer) in C:\Inetpub\TecEco_PHP\stats_interface\referer_base.php on line 16 -- JJ Harrison webmaster@tececo.com www.tececo.com "Jj Harrison" <webmaster@tececo.com> wrote in message news:20020709113437.34570.qmail@pb1.pair.com... > Could some one give me an idea as to what I SQL query could use to get the > percentages of each differant colomn value. > > ie if this was my table: > +-------+---+ > | name | id | > +-------+---+ > | foobar | 1 | > | foobar | 2 | > | foobar | 3 | > | barfoo | 4 | > +-------+---+ > > I would get this result(Then later do stuff with it in PHP): > > +-------+--------+ > | name | percent | > +-------+--------+ > | foobar | 75 | > | barfoo | 25 | > +-------+--------+ > > Thanks in advance, > > > -- > JJ Harrison > webmaster@tececo.com > www.tececo.com > > > >