Question

对于我的项目，我试图通过使用我数据库中过去学生的数据来预测一个过去学生的A级数学成绩。所以，paststudent1获得了数学A级的A。在第一年的考试中，他们获得了计算机A，物理B，化学A，男性和97％的出勤率。如果当前1年级的学生得到相同的成绩，相同的科目，男性和相同的出勤率，我想说他们将获得相同的成绩。

我的代码中的错误说：＆＃34;致命错误：未捕获的异常＆＃39; PDOException＆＃39; with message＆＃39; SQLSTATE [42000]：语法错误或访问冲突：1064 SQL语法中有错误;查看与您的MariaDB服务器版本对应的手册，以获得正确的语法，以便在附近使用＃39;等级1 =＆＃39;数学＆＃39;，＆＃39; A＆＃39;＆＃39;在第1行＆＃39;在/home/u717042829/public_html/predicter.php:70堆栈跟踪：＃0 /home/u717042829/public_html/predicter.php(70）：PDOStatement-＆gt;执行（）＃1 {main}抛出/ home / u717042829 /public_html/predicter.php在第70行＆＃34;

第70行是：

＆＃13;

			$query->execute();

＆＃13;

。

＆＃13;

<?php 
error_reporting(E_ALL); ini_set('display_errors', 1);
require("includes/config.php");
//if not logged in redirect to login page
if(!$user->is_logged_in()){ header('Location: login.php'); } 

//define page title
$title = "Predict a Student's Grade";

?>
<!DOCTYPE html>
<html>



<head>

    <meta charset="utf-8">
    <meta http-equiv="X-UA-Compatible" content="IE=edge">
    <meta name="viewport" content="width=device-width, initial-scale=1">
    <meta name="description" content="">
    <meta name="author" content="">

    <title>Past Student</title>

    <!-- Bootstrap Core CSS -->
    <link href="css/bootstrap.min.css" rel="stylesheet">

  


    <!-- Custom Fonts -->
    <link href="font-awesome/css/font-awesome.min.css" rel="stylesheet" type="text/css">

    <!-- HTML5 Shim and Respond.js IE8 support of HTML5 elements and media queries -->
    <!-- WARNING: Respond.js doesn't work if you view the page via file:// -->
    <!--[if lt IE 9]>
        <script src="https://oss.maxcdn.com/libs/html5shiv/3.7.0/html5shiv.js"></script>
        <script src="https://oss.maxcdn.com/libs/respond.js/1.4.2/respond.min.js"></script>
    <![endif]-->

</head>

<body>




<?php 

	if (isset($_POST['name'])) {


		$name     = $_POST['name'];
        $subject1 = $_POST['subject1'];
        $grade1   = $_POST['grade1']; 
        $subject2 = $_POST['subject2'];
        $grade2   = $_POST['grade2'];
        $subject3 = $_POST['subject3'];
        $grade3   = $_POST['grade3'];
        $subject4 = $_POST['subject4'];
        $grade4   = $_POST['grade4'];
        $attendance = $_POST['attendance'];
        $gender     = $_POST['gender'];


			$query = $db->prepare("SELECT * FROM paststudent WHERE subject1, grade1 = :subject1, :grade1");
			$query->bindParam(':subject1', $subject1);
			$query->bindParam(':grade1', $grade1);
			$query->execute();

          
          	




	}
	?>
	<h1> Your predicted grade is <?php echo $query ?> </h1>

</body>
</html>

＆＃13;

现在我按小步骤将主题1与1级相匹配，然后进行2级2级，因为我太干了。

Answer 1

您的查询的部分看起来不正确：

WHERE subject1, grade1 = :subject1, :grade1

你很可能意味着要改变它看起来像这样：

WHERE subject1 = :subject1 AND grade1 = :grade1

所以你的最终版本应为：

$query = $db->prepare(  " SELECT * FROM paststudent "
                      . " WHERE subject1 = :subject1 AND grade1 = :grade1");

我建议阅读一些关于SQL基础知识的教程。

Answer 2

这不是您从PDO获得结果的方法，请尝试

<body>
<table>
....


$query->execute();

          }
while($row = $query->fetch(PDO::FETCH_ASSOC)){

      echo "<tr><td>".$row['user']."</td></tr>"; --change user to what the column is actually called
}

    ?>
</table>  
</body>
</html>

我的pdo语句中的错误

2 个答案: