PHP - 如何选择由mysqli_fetch_array（）给出的行的特定索引？

Question

请参考下面的图片：

http://i.stack.imgur.com/6hBPC.png

例如，如果用户点击该行的按钮，说：“你有数学测验”时，认为“问答ID”值行然后会被传递给另一个PHP文件。

这里是我当前的代码：

<?php 
    $con=mysqli_connect("127.0.0.1", "root", "", "quizmaker"); 
    if (mysqli_connect_errno($con)) 
    { 
     echo "MySqli Error: " . mysqli_connect_error(); 
     } 

    $now=date("m/d/Y"); 
    $sql=mysqli_query($con,"SELECT * FROM quiz_query WHERE quiz_date='$now'"); 
    $count=mysqli_num_rows($sql); 

    if($count>=1) 
    { 
    echo "<table border='1' width='50%'>"; 
    echo "<form action='answer_quiz.php' method='post'>"; 
    echo "<tr> 
    <td>You have a pending quiz!</td><td>&nbsp;</td><td>&nbsp;</td> 
    </tr>"; 
    $number=1; 

    while($result=mysqli_fetch_array($sql)) 
    { 
     echo "<tr>"; 
     echo "<td>You have a quiz for " . $result['subject'] . "</td>"; 
     echo "<td>Quiz ID: " .$result['quiz_ID']. "</td>"; 
     echo "<td><input type='submit' name='button' id='button' value='Take Quiz'>"; 
     echo "<input type='hidden' name='quiz[$number]' value='$result[quiz_ID]'>"; 
     echo "</td>"; 
     echo "</tr>"; 
     $number++; 
     } 


    echo "</form>"; 
    echo "</table>";  
    } 
    else 
    { 
     "You have no quiz! :D"; 
     } 

    mysqli_close($con); 
    ?> 

Answer 1

移动这一行：

echo "<form action='answer_quiz.php' method='post'>"; 

里面的while循环。

---

而且，变化

echo "<input type='hidden' name='quiz[$number]' value='$result[quiz_ID]'>" 

与

echo "<input type='hidden' name='quizId' value='$result[quiz_ID]'>" 

现在，在answer_quiz.php您会收到$_POST['quizId']与您需要的值。

Answer 2

变化的，同时：

while($row = $result->fetch_array(MYSQLI_ASSOC)){ 

echo $row['subject']; 

} 

Answer 3

你忘了在你的变量报价： 相反

echo "<input type='hidden' name='quiz[$number]' value='$result[quiz_ID]'>"; 

的

应该

echo "<input type='hidden' name='quiz[$number]' value='$result[\"quiz_ID\"]'>";