bumdeal2 Posted April 3, 2010 Share Posted April 3, 2010 I have two forms that pass info between each other (quesid),the problem i am having is when i pass he value from one form to another, the vaue being passed in the url is $quesie, i dont know where the $quesie is coming froma it is not in my code could u please help. My first form is 1. <?php include('auth.php'); include('database_access_param.php'); include('auth.php'); $qid=$_POST['qid']; $qzid=$_POST['qzid']; $quesid=$_POST['quesid']; if(isset($_GET['quesid'])) { $quesid=$_GET['quesid']; } if(isset($_GET['qid'])) { $qzid=$_GET['qid']; } echo $quesid; echo $qzid; mysql_connect( $hostname, $dbuser, $dbpassword) or die ( 'Unable to connect to server' ); mysql_select_db($dbname ) or die ( 'Unable to select database' ); $sql = "SELECT * FROM questiona WHERE qzid = '$qzid' && quesid = '$quesid' "; print("</head>"); print('<body>'); print('<form action="grad.php?quesid=$quesid" method="post" id="quiz">'); print('<table class="report" align=left width=100%>'); print('<tr><td> </td></tr>'); $result=mysql_query($sql) or die ('Unable execute the query'); ; if(mysql_numrows($result)) { $row = mysql_fetch_row($result); { $questid=$row[0]; $quesid=$row[1]; $question=$row[2]; $answer=$row[3]; $ranswer=$row[4]; $ranswer1=$row[5]; $ranswer2=$row[6]; print('<tr><td >'.$quesid.' '.$question.'</td>'.$Quiz_Topic.'</tr>'); print('<tr><td><input type="radio" name="answer" id="answer" value="'.$ranswer .'"> '.$ranswer.'</td></tr>'); print('<tr><td><input type="radio" name="answer" id="answer" value="'.$ranswer1 .'"> '.$ranswer1 .'</td></tr>'); print('<tr><td><input type="radio" name="answer" id="answer" value="'.$ranswer2 .'"> '.$ranswer2 .'</td></tr>'); print('<tr><td><input type="radio" name="answer" id="answer" value="answer"> '.$answer .'</td></tr>'); print('<tr><td> <input type=hidden name=answer value="'.$answer.'"><input type= hidden name=quesid value="$quesid"></tr>'); } print('<tr><td colspan=2><input type=submit name=submit value="Submit" ><input type=reset name=reset value="Clear" ></td></tr></table>'); } else { echo "<div id='results'>Your result is: $subTotal%</div>"; print('<input type =hidden name=score value="'.$score.'">'); print('</form>'); } ?> My second page is 1. < ?php include('auth.php'); include('database_access_param.php'); import_request_variables("pgc",""); //echo" act_save_answer.php<br>"; $quesid = $_POST['quesid']; $qzid= $_POST['qzid']; $answer = $_POST['answer']; $exist=0; $score=0; $cday= date('d'); $cmonth= date('m'); $cyear= date('Y'); $dbcurrent_date=$cyear.'-'.$cmonth.'-'.$cday; $db_link=mysql_connect($hostname, $dbuser, $dbpassword) or die("Unable to connect to the server!"); mysql_select_db($dbname) or die("Unable to connect to the database."); $ssql = "SELECT * FROM scores WHERE username = '$user' AND qid = '$qid'"; if($quesid<11) { $totalCorrect = 0; if ($answer == "answer") { $totalCorrect++; } $num = 10; $subTotal = $totalCorrect * $num; $subTotal; $quesid++; header("location: PlayGame.php?quesid=$quesid "); if($quesid=11) { $isql="insert into scores(scoreid,qid, username, score, qdate) values (0, '".$user."','".$qid."',".$subTotal.",'".$dbcurrent_date."')"; //echo $isql."<br>"; if(!mysql_query($isql)) { echo mysql_errno() . ": "; echo mysql_error() . "<BR>"; } } } else { echo "Sorry. You cannot take this quiz second time!"; print('<a href="index.php?boxaction=home">Home</a>'); } ?> Link to comment https://forums.phpfreaks.com/topic/197467-error-in-code/ Share on other sites More sharing options...
ialsoagree Posted April 3, 2010 Share Posted April 3, 2010 It's coming from right here: print('<form action="grad.php?quesid=$quesid" I believe you're looking for: print('<form action="grad.php?quesid='.$quesid.'" Keep in mind, you're using single quotes for this string, so variables inside single quotes are NOT parsed: $example_string = 'stuff'; echo 'This is the first example: '.$example_string; // This is the first example: stuff echo 'This is the second example: $example_string'; // This is the second example: $example_string echo "This is the third example: $example_string"; // This is the third example: stuff Link to comment https://forums.phpfreaks.com/topic/197467-error-in-code/#findComment-1036514 Share on other sites More sharing options...
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