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我在我的错误报告中收到一条 mysqli 警告，指出Warning: mysqli_stmt::bind_result() [mysqli-stmt.bind-result]: Number of bind variables doesn't match number of fields in prepared statement in ... on line 36 有谁知道我如何正确绑定结果以便此警告消失？我真的看不出问题出在哪里，但是在使用 mysqli 时，我是一个初学者。

下面是代码：

<?php
  // PHP code
  session_start(); 

//connected to db

  // required variables (make them explciit no need for foreach loop)
  $teacherusername = (isset($_POST['teacherusername'])) ? $_POST['teacherusername'] : '';
  $teacherpassword = (isset($_POST['teacherpassword'])) ? $_POST['teacherpassword'] : '';
  $loggedIn = false;

  if (isset($_POST['submit'])) {

    $teacherpassword = md5(md5("j3Jf92".$teacherpassword."D203djS"));  

    // don't use $mysqli->prepare here
    $query = "SELECT * FROM Teacher WHERE TeacherUsername = ? AND TeacherPassword = ? LIMIT 1";
    // prepare query
    $stmt=$mysqli->prepare($query);
    // You only need to call bind_param once
    $stmt->bind_param("ss",$teacherusername,$teacherpassword);
    // execute query
    $stmt->execute(); 
    // get result and assign variables (prefix with db)
    $stmt->bind_result($dbTeacherForename,$dbTeacherSurname,$dbTeacherUsername,$dbTeacherPassword);

    while($stmt->fetch()) {
      if ($teacherusername == $dbTeacherUsername && $teacherpassword == $dbTeacherPassword) {
        $loggedIn = true;
      }
    }

    if ($loggedIn == true){
      // left your session code as is - but think wisely about using
      $_SESSION['teacherforename'] = $dbTeacherForename;
      $_SESSION['teachersurname'] = $dbTeacherSurname;
      header( 'Location: menu.php' ) ;
      die();
    }

       /* close statement */
    $stmt->close();

    /* close connection */
    $mysqli->close();
  }
?>
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2 回答 2

1

不推荐使用通配符 *。表中的列可能比您需要的 4 列更多？

我会去

SELECT TeacherForname, TeacherSurname, TeacherUsername, TeacherPassword FROM Teacher WHERE TeacherUsername = ? AND TeacherPassword = ? LIMIT 1
于 2012-08-10T14:01:06.733 回答
0

发生错误是因为您要求 mysqli 将x列绑定到y变量 where x != y。表中有多少列Teacher？

于 2012-08-10T14:11:29.543 回答