錯誤獲取PHP變量嵌入了

-1

<?php 
include('lib/db.php'); 
if(isset($_GET['id'])) 
{ 
$cid=mysql_real_escape_string($_GET['id']); 
//echo $sub; 
$rs=mysql_query("select * from category where id='$cid'"); 
$r=mysql_fetch_array($rs); 
$q=rand(1,2); 
$rs1=mysql_query("select * from questions where qid='$q'"); 

$r1=mysql_fetch_array($rs1); 
} 
?> 
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> 
<html xmlns="http://www.w3.org/1999/xhtml"> 
<head> 
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> 
<title>Test</title> 
<script type="text/javascript" src="js/jquery-2.1.0.js"></script> 
<script type="text/javascript" src="js/tsubmit.js"></script> 

</head> 
<body> 
<header> 
<?php echo $r['cat_name'];?> Test 
</header> 
<form method="post" id="test" role="form"> 
<label for="question">Question:</label> 
<?php echo $r1['question']; ?><br/> 
<input type="radio" value="<?php echo $r1['ans1']; ?>" name="ans1" /><?php echo $r1['ans1']; ?> 
<br/><input type="radio" value="<?php echo $r1['ans2']; ?>" name="ans1" /><?php echo $r1['ans2']; ?> 
<br/><input type="radio" value="<?php echo $r1['ans3']; ?>" name="ans1" /><?php echo $r1['ans3']; ?> 
<br/><input type="radio" value="<?php echo $r1['ans4']; ?>" name="ans1" /><?php echo $r1['ans4']; ?> 
<br/><input type="submit" value="submit" name="submit" id="submit"/> 
</form> 


</body> 
</html>

和嵌入式tsubmit.js獲取數據的問題網頁是

$(document).ready(function() 
          { 
var start; 
start= new Date()/1000; 
function submit(x){ 
     var end=new Date()/1000; 
      var timespent=end-x; 
      alert(timespent); 
      var ans=$('input[name=ans1]:checked').val(); 
      alert(ans); 
      var cans="<?php echo $r['ans']; ?>"; 
      //alert('<?php echo $r["ans"]; ?>'); 
      alert(cans); 
      // var qid="<?php echo $q; ?>"; 
      alert(qid); 
      //var datastring='ans='+ans+'&timespent='+timespent+'cans='+cans+'qid='+qid; 
      $.ajax({ 
        type:"POST", 
        url:"result.php", 
        data:'ans='+ans+'&timespent='+timespent+'&cans='+cans+'&qid='+qid           //success:success() 
        }); 

     } 


     $("#test").submit(function(event){ 
     alert('user clicked submit');     
      submit(start); 
      //event.preventDefault(); 
     }); 
    });

但我不是罐頭的價值，也result.php沒有得到任何

<?php 
    include('lib/db.php'); 
    echo $_POST['ans']; 
    echo "<script> alert('ans')</script>"; 
    ?>

echo $_post['ans']不獲取任何東西

來源

2015-08-20 Ravi Kumar

請勿將技術從PHP直接轉換爲js環境。您可能會因語法錯誤而殺死腳本。總是使用'json_encode（）'。 –

，並且您正在通過'

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