은 가치 숨겨진 필드를 PHP를 얻으려고 노력 - 가져 오기 오류

Question

HTML/PHP :

  <?php if(!empty($_GET['pID'])) $the_pID = mysql_real_escape_string($_GET['pID']); 
      #echo $the_pID; 
      ?> 

      <form action="inc/q/prof.php?pID=<?php echo $the_pID; ?>" method="post">    
      <select id="courseInfoDD" name="courseInfoDD" tabindex="1"><?php while($row3 = $sth3->fetch(PDO::FETCH_ASSOC)) { 
        echo "<input type='hidden' name='cID' value='$_POST['cID']'"; 
        echo "<option><?php".$row3['prefix']." ".$row3['code']."</option>"; }echo "</select>"; ?> 
      <input type="text" id="addComment" name="addComment" tabindex="3" value="Enter comment" /> 

     <input type="hidden" name="pID" value="<?php echo $the_pID; ?>"> 
     <input type="submit" name="submit" id="submit" /> 
     </form> 

PHP/MYSQL을 나는 HTML/PHP와 함께 다음과 같은 오류가

<?php // Get select box options 
$pID3 = filter_input(INPUT_GET, 'pID', FILTER_SANITIZE_NUMBER_INT); 
$username = "###"; 
$password = "#####"; 
     $pdo3 = new PDO('mysql:host=#####;dbname=####', $username, $password); 
     $pdo3->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); 
$sth3 = $pdo3->prepare(' 
    SELECT pID, C.cID, C.prefix, C.code 
    FROM Department D, Course C, Professor P 
    WHERE pID = ? 
    AND D.dID = C.dID 
    AND D.dID = P.dID; 
'); 
     $sth3->execute(array(
      $pID3 
     )); 
?> 

<?php 

$connect = mysql_connect("#####", $username, $password) or die ("Error , check your server connection."); 
mysql_select_db("####"); 

//Get data in local variable 
if(!empty($_POST['addComment'])) 
    $INFOO=mysql_real_escape_string($_POST['addComment']); 
if(!empty($_POST['pID'])) 
    $PIDD=mysql_real_escape_string($_POST['pID']); 
if(!empty($_POST['courseInfoDD'])) 
    $COURSEE=mysql_real_escape_string($_POST['courseInfoDD']); 

#print_r($_POST); 
echo $the_pID; 

// check for null values 
if (isset($_POST['submit'])) { 
$query="INSERT INTO Comment (info, pID, CName) values('$INFOO','$PIDD','$COURSEE')"; 
mysql_query($query) or die(mysql_error()); 
echo "Your message has been received"; 
} 
#else if(!isset($_POST['submit'])){echo "No blank entries";} 
#else{echo "Error!";} 
?> 

위 : Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING

나는이 줄을 다룰 생각이다 : echo <input type='hidden' name='cID' value='<?php echo '$_POST['cID']';?><option>".$row3['prefix']." ".$row3['code']."</option>"; }echo "</select>"; ?>

누구에게도 문제가 발생합니까? 당신은 PHP에서 옵션 태그를 인쇄하고 싶었지만 외부 나중에 왼쪽처럼

<?php echo '$_POST['cID']';?><option>".$row3['prefix'] 

보인다 "옵션";-) 후 다른 PHP 태그를 열어 시작할 수

Answer 1

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