Question

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


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

此代码将返回问题ID（2和1）我想要的是从表格问题中检索其他问题ID，以便我想要结果为（3和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

我相信你要找的是IS NULL：

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

您只需从问题表中检索问题ID即可更进一步：

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

Answer 2

编辑：改进版本。

  $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版本：

  $link = mysqli_connect($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);

请参阅操作：http://www.sqlfiddle.com/#!2/27b6f/21

Answer 3

可能的答案是采用您使用的相同mysql_query 然后从mysql_fetch_array检索'question_id'值，最后，如果你有问题的计数（如果不是你可以使用mysql计数），使用php函数array_diff()来检索（从增量排序） question_id的整数值数组或者来自'question'表中的question_id值数组的数组）正是你想要的

SQL左外连接选择不匹配的记录？

3 个答案: