Re: MySQL query problems

But what's your prob ? When you say the second one seems to never be executed ... Does the line: $row = mysql_fetch_array($result2); launches a Php Error ? And pay attention, because you're using mysql_fetch_array and mysql_fetch_row, be sure that you are not treating the result in the same way ;-) "Beauford.2005" <beauford.2005@rogers.com> a écrit dans le message de news: 000001c34bb5$5772a200$6401a8c0@p1... > Hi, > > Not sure if the problem here is PHP or MySQL, but here we go. I am > trying to do two queries on a database - one after the other, but the > second one never seems to get executed. The two queries are identical > except for two variables. I have checked my form and they are correct > and are being sent to the PHP script the way they should be. > > The variables are $playerto, $playerfrom, $nameto, $namefrom. The 'TO' > query is the one that doesn't work. > > Here is my code. Any help is appreciated. > > if ($namefrom != $nameto) { > if ($playerfrom != $playerto) { > > include("2004server.inc"); > if($error) { > include("trades-input.php"); > exit; > } > > $query1 = "select manager.idn, manager.total, > roster.idp, position, points from roster join reference > join manager where manager.idn=reference.idn and > reference.idp=roster.idp and manager.idn like '$namefrom' and > roster.idp like '$playerfrom'"; > > $result1 = mysql_query($query1) or $mysqlerror = > mysql_error(); > if ($mysqlerror) { > $error = "$d_base_error$email_error"; > include("trades-input.php"); > exit; > } > > $line = mysql_fetch_row($result1); > > mysql_free_result($result1); > > $query2 = "select manager.idn, manager.total, > roster.idp, position, points from roster join reference > join manager where manager.idn=reference.idn and > reference.idp=roster.idp and manager.idn like '$nameto' and > roster.idp like '$playerto'"; > > $result2 = mysql_query($query2) or $mysqlerror = > mysql_error(); > if ($mysqlerror) { > $error = "$d_base_error$email_error"; > include("trades-inputs.php"); > exit; > } > > $row = mysql_fetch_array($result2); > > mysql_free_result($result2); > } > } >