Question

question table
========================
question_id
1
2
3
4


user_answer table
========================
user_id  question_id
33       2
44       4
33       1
44       3

This code will return the question id (2 and 1)
what I want is to retrieve the other question id from table question so I want
the result to be (3 and 4)

   $fadi = mysql_query("SELECT * FROM question
         LEFT OUTER JOIN user_answer
           ON user_answer.question_id = question.question_id
              WHERE user_answer.user_id = 33");

   Print "<table border cellpadding=3>";
   while($info = mysql_fetch_array($fadi))
   { Print "<tr>"; 
    Print "<th>question </th> <td>".$info['question_id'] . "</td></tr>";
   }  Print "</table>"; }

Answer 1

Edit: Improved version.

  $fadi = mysql_query("SELECT * FROM question WHERE  question.question_id NOT IN (SELECT user_answer.question_id FROM user_answer WHERE user_answer.user_id = 33)");

   Print "<table border cellpadding=3>";
   while($info = mysql_fetch_array($fadi))
   { Print "<tr>"; 
    Print "<th>question </th> <td>".$info['question_id'] . "</td></tr>";
   }  Print "</table>"; }

MySQLi version:

  $link = mysqli_connect($hostname, $username, $password, $database);
  if (!$link){ 
  echo('Unable to connect to database');
  }
  else{
  $fadi = mysqli_query("SELECT * FROM question WHERE  question.question_id NOT IN (SELECT user_answer.question_id FROM user_answer WHERE user_answer.user_id = 33)", $link);

   Print "<table border cellpadding=3>";
   while($info = mysqli_fetch_array($fadi,MYSQL_BOTH))
   { Print "<tr>"; 
    Print "<th>question </th> <td>".$info['question_id'] . "</td></tr>";
   }  Print "</table>"; }

  }
  mysqli_close($link);

See in action: http://www.sqlfiddle.com/#!2/27b6f/21