Question

我试图从数据库动态生成表单，以便我可以简单地登录和进行更改，以及跟踪投票。表单生成正常，但我不能为我的生活弄清楚如何从数组中获取id以发布数据。唯一成功的尝试将最后一个设置为他们的所有topic_id，因为我试图在while语句中定义它。

  <?php  $cat_set = fetch_categories();

  if (!isset($business_name)) {
  $business_name = ''; }

    while($categories = mysqli_fetch_assoc($cat_set)) {
                echo '<div class="category">'. "<h3>" . "Best of the Best"      
   .' ' . ucfirst($categories["cat_name"]) . ' ' . "</h3>";

                $topic_set = get_topics_for_cat($categories["cat_id"]);
                    while($topics = mysqli_fetch_assoc($topic_set)) {
                        $topic_name = $topics["topic_name"];
                    echo '<div class="field">' . '<label for=' . 
  $topics['topic_id'] . '">' . ucfirst($topic_name) . ":" . '</label>';
                        echo '<input type="text"' . 'id="' . 
  $topics['topic_id'] . '"' . 'name="'
                        . $topics["topic_id"] . '"' . 'value="' . 
  $business_name . '"></div>';
                    }
                    echo "</div>";
    }
    ?>

<input type="submit" value="Submit" id="submit" name="submitform"></form>
</div>
<pre>
<?php
$user_id = 3000;
if (isset($_POST['submitform'])) {
    print_r($_POST);

foreach ($_POST as $business_name) {
    $filtered_business_name = mysqli_real_escape_string($dbc,  
$business_name);
    $query = "INSERT INTO votes (";
    $query .=" business_name, topic_id, user_id";
    $query .= ") VALUES (";
    $query .=" '$filtered_business_name', '$topics['topic_id']',   
'$user_id'";
    $query .=")";

    $votes = mysqli_query($dbc,$query);

更具体地说，我想，在我的查询中获取$ topic [＆＃39; topic_id＆＃39;]的最佳方法是什么？我试图逃避它以及我能想到的一切。

Answer 1

用大括号括起变量：

$query .=" '$filtered_business_name', '{$topics['topic_id']}', '$user_id'";

Answer 2

请执行以下操作：

 $topic = $topics['topic_id'];
$query .=" '$filtered_business_name', '$topic', '$user_id'";

Answer 3

我一定是误用了它。我做了很多编辑和事情，我自己的东西没有意义，所以我扭转了一堆并尝试了key =＆gt;值的东西......它就可以了。现在查询是：

if (isset($_POST['submitform'])) {

foreach($_POST as $key=>$value) {
    $filtered_business_name = mysqli_real_escape_string($dbc, $value);
    $query = "INSERT INTO votes (";
    $query .=" business_name, topic_id, user_id";
    $query .= ") VALUES (";
    $query .=" '$filtered_business_name', '$key', '$user_id'";
    $query .=")";

    $votes = mysqli_query($dbc,$query);
    confirm_query($votes);

}
}

感谢大家的帮助，希望这将有助于将来制作糟糕表格的人。

以动态生成的形式使用数组中的id

3 个答案: