如何獲取顯示的數據庫QUICKFIX上所有選中框的ID？

Question

我在數據庫中顯示了我的數據表格，左邊是複選框。我想找到一種方法將複選框鏈接到問題編號（ID）。當我點擊提交時，我想要選中的ID被回顯。幾乎我希望有人能夠選擇他們想要的問題然後顯示出來。

<?php 

    $con=mysqli_connect("####","####","#####","###"); 

    $result = mysqli_query($con,"SELECT * FROM q_and_a "); 

echo "<table border='1'> 
    <tr> 
<th>Add</th> 
<th>#</th> 
<th>Question</th> 
<th>A</th> 
<th>B</th> 
<th>C</th> 
<th>D</th> 
<th>Answer</th> 

</tr>"; 

while($row = mysqli_fetch_array($result)) 
{ 
echo "<tr>"; 

echo '<td><input type="checkbox" name="questions[]" value="$id"></td>'; 

echo "<td>" . $row['id'] . "</td>"; 
echo "<td>" . $row['question'] . "</td>"; 
echo "<td>" . $row['choiceA'] . "</td>"; 
echo "<td>" . $row['choiceB'] . "</td>"; 
echo "<td>" . $row['choiceC'] . "</td>"; 
echo "<td>" . $row['choiceD'] . "</td>"; 
echo "<td>" . $row['answer'] . "</td>"; 
echo "</tr>"; 
} 
echo "</table>"; 

mysqli_close($con); 

?> 

提交按鈕

<form method="POST" action="makeTest.php"> 
<input type="submit" name="make" value="Make Test"> 
</form> 

Answer 1

趕快做些編輯您的代碼，然後它會工作。 1.改變您的查詢通過添加字段不*（以確保性能和顯示順序）

$result = mysqli_query($con,"SELECT id,question,choiceA,choiceB,choiceC,choiceD,answer FROM q_and_a "); 

然後之前而塊開放的形式標記（HTML）

<?php 
//above codes will be there as you show before 
echo '<form method="POST" action="makeTest.php">'; 
while($row = mysqli_fetch_array($result)){ 
{ $id=$row['id']; // initialize your id here, so as to pass it in checkbox too 
// then continue with your code 
} 
?> 
<input type="submit" name="make" value="Make Test"> 
</form> 

在maketest.php喲可以喊得ckeckbox使用的foreach，見下文

foreach($_POST['questions'] as $questions){ 
          //do your job 

    }