Question

嗨，任何帮助将不胜感激，因为我似乎无法找到导致我的数据被复制的错误，并且向用户回复3次而不是一次。如果有人能帮我解决这个问题，谢谢。

<?php

session_start();
//Connect to database 

if(!isset($_SESSION['username'])) { //check if user-name is not given

    echo('you are not authorised to veiw this page, log in with admin account to view it');

    //show login page
    include 'login.html';
}
$con=mysqli_connect("localhost:3307","root","usbw","degreeproject");

//check if query sucessful else report error
$classid = $_GET['classid'];
$studentid = $_GET['userid'];
$sql3 = "SELECT usernames.username, quizzes.QuizID, results.score, results.QuizID,enrolledclasses.UserID, quizzes.ClassID FROM `results`, enrolledclasses, usernames, quizzes 
                WHERE usernames.username = results.username
                AND quizzes.QuizID = results.QuizID             
                AND enrolledclasses.UserID = $studentid
                AND quizzes.ClassID = $classid;";


if (!$result3=mysqli_query($con,$sql3)) {
    die('Error: ' . mysqli_error($con));
}       
else {
    //count the number of records from the query result
    $count=mysqli_num_rows($result3);
}   

if($count>=1) {
    $result3 = mysqli_query($con,$sql3); 
    while(list($Username, $QuizID, $Score) = mysqli_fetch_row($result3)) {
            $sql = "SELECT COUNT(Question) as total FROM questions WHERE QuizID = $QuizID";
            $result = mysqli_query($con, $sql);
            $numquestions;
            if($result->num_rows > 0) {
                $numquestions = $result->fetch_assoc()['total'];
            }

            echo"<table>";
            echo "<tr>";
                echo "<th>Username: </th>";
                echo "<th>QuizID: </th>";
                echo "<th>Score: </th>";
                echo "<th>Go Back: </th>";
            echo "</tr>";

            echo "<tr>";
            echo("<td>$Username</td>");
            echo("<td>$QuizID</td>");
            echo("<td>$Score / $numquestions</td>");
            echo("<td></td>");
            echo "<td><a href ='../CurrentlyBeingWorkedOn/index.php'> Go Back Home &nbsp; &nbsp;</a></td>";
            echo"</tr>";
            echo"</table>";
        }
    }


$con->query($sql3);

echo("<p>Successfully Retrieved The Results For The Quizzes Students Have Taken.</p>");
echo("<p><a href='javascript:history.back(1);'>Return To Previous Page</a></p>");



?>

这是代码的结果以及输出给用户的内容。

[产生的结果] [1] [1]：https://i.stack.imgur.com/hn5jm.png

Answer 1

从我看到你的加盟条件有问题。用户名和结果，测验和结果之间存在关联。但是注册类与任何其他表之间没有关系。在where子句中添加关系，您将找到所需的输出。

SELECT usernames.username, quizzes.QuizID, results.score, results.QuizID,enrolledclasses.UserID, quizzes.ClassID FROM `results`, enrolledclasses, usernames, quizzes 
WHERE usernames.username = results.username
AND quizzes.QuizID = results.QuizID
AND enrolledclasses.UserID = usernames.UserID          
AND enrolledclasses.ClassID= quizzes.ClassID          
AND enrolledclasses.UserID = $studentid
AND quizzes.ClassID = $classid;

Answer 2

尝试使用显式JOIN语句重写SQL查询：

"SELECT usernames.username, quizzes.QuizID, results.score,
    results.QuizID,enrolledclasses.UserID, quizzes.ClassID 
FROM usernames u 
    JOIN results r ON u.username=r.username 
    JOIN quizzes q ON q.QuizID=r.QuizID 
    JOIN enrolledclasses ec ON ec.?=?.? 
WHERE q.ClassID = $classid AND ec.UserID=$studentid"

您需要为您的注册类表提供一个连接条件，我无法从您给定的代码中看到 - 也许您可以将其加入users表上的用户名或类似内容？只需替换上面给出的代码中的问号。如果您没有任何可以加入的标准，则必须在enrolledclasses表中引入一个。

数据正在回响3次而不是一次

2 个答案: