Re: [PHP-DB] no temp var for mysql_fetch_array results

[jeffrey_n_Dyke]

>Hi, all --

>I have some code which looks about like

>  if ( $_POST['signup']['referrer'] )                       # a friend?
 > {
 >   $q = "select InvitedBy from customer where EMail
= '{$_POST['signup']['referrer']}'" ;
 >   $r = mysql_query($q,$dbro) ;
 >   $row = mysql_fetch_array($r) ;
 >   $i = $row[0] ;
 > }
 > else
 # no referrer supplied
 >   { $i = '' ; }

>and I would really prefer not to have to use $row just to set $i.  Since
>mysql_fetch_array() returns an array, and I know I want a field from just
>this one row (not looping), could I just

>  $i = mysql_fetch_array($r)[0] ;

Hi David....
      How about:
      $i = mysql_result(mysql_query($q,$dbro),0);
HTH
Jeff

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