Question

我需要从我的数据库输出问题，以便用户可以在注册过程中回答这些问题。每个问题都有2个与之相关的选项，现在每个问题都输出到网站两次，每个问题下面有一个选项。我需要每个问题出现一次，其相关选项出现在它下面。请参阅下面的代码。

   <?php
       $sql = "SELECT Question_ID, Question FROM Questions;";

       $result = mysqli_query($conn, $sql);
       $resultCheck = mysqli_num_rows($result);

       if($resultCheck > 0):
          while($row = mysqli_fetch_assoc($result)):
             $questionid = (int)$row['Question_ID'];
            ?>
            <label><?php echo $row['Question'];?></label><input type="hidden" value="<?php echo $row['Question_ID'];?>"/>
   <?php
             $query = "SELECT Choice_ID, Choice FROM Choices WHERE Question_ID = '$questionid';";
             $results = mysqli_query($conn, $query);
             $resultsCheck = mysqli_num_rows($results);
               if($resultsCheck > 0):
                  while($rows = mysqli_fetch_assoc($results)):
        ?>
         <br><select id="choice">
            <option value="<?php echo $rows['Choice_ID']; ?>"><?php echo $rows['Choice']; ?></option>
        </select><br/>
        <?php
        endwhile;
        endif;
        endwhile;
        endif;
        ?>

Answer 1

您的问题中没有<form>，我想您知道自己在做什么，并且知道如何处理自<input>和<select>以来发布的数据元素没有任何name属性。
正如Strawberry建议的那样，建议使用准备好的陈述。

我已经更改了您的代码以满足您的需求。在您的代码中，您必须从<select>循环中取出while代码，然后只拥有<option>代码在循环中。

所以代码将是这样的：

<?php
$sql = "SELECT Question_ID, Question FROM Questions;";

$result = mysqli_query($conn, $sql);
$resultCheck = mysqli_num_rows($result);

if($resultCheck > 0):
    while($row = mysqli_fetch_assoc($result)):
        $questionid = (int)$row['Question_ID'];
        echo '<label>'.$row['Question'].'</label><input type="hidden" value="'.$row['Question_ID'].'">';

        $query = "SELECT Choice_ID, Choice FROM Choices WHERE Question_ID = '$questionid';";
        $results = mysqli_query($conn, $query);
        $resultsCheck = mysqli_num_rows($results);
        if($resultsCheck > 0):
            $string = '<br><select id="choice">';
            while($rows = mysqli_fetch_assoc($results)):
                $string .= '<option value="'.$rows['Choice_ID'].'">'.$rows['Choice'].'</option>';
            endwhile;
            $string .= '</select><br>';
            echo $string;
        endif;
    endwhile;
endif;
?>

Answer 2

像这样组织你的代码（为便于阅读而简化）

while($question = mysqli_fetch_assoc($result))
{
   echo "<label for='q'>...; // print question
   echo "<select name='answer'>"  
   while ($answer = mysqli_fetch_assoc($results))
   {
       echo '<option>'.$answer.</option>';
   }
   echo '</select>';
}

如果数据已经输出到网站，如何通过ID检查以避免重复

2 个答案: