Select 填充但不显示值 (php/msqli)

Question

所以我正在开发一个下拉菜单，显示我网站上的用户，当我加载页面时，我可以看出它正在连接到数据库并选择正确的 table、列和所有内容，因为它正在显示正确数量的选项，但所有选项都是空白...我做错了什么？

<?php $users = $conn->query("SELECT username FROM users"); ?>

<label for="ref" class="ref">If you were referred to RDR PLEASE select your referrer's Username:</label>
<select name="referrer" style="width:150px">
        <option value="None"></option>
        <?php while($rows = $users->fetch_array())
        {
                $user = $rows['$users'];
                echo "<option value='$user'>$user</option>";
        } ?> </select>

Answer 1

您的列名似乎是您 sql 查询中的“用户名”。

        <?php while($rows = $users->fetch_array())
        {
                $user = $rows['username'];
                echo "<option value='$user'>$user</option>";
        } ?> </select>

Select 填充但不显示值 (php/msqli)

Select Populating but not showing values (php/msqli)

php

database

forms

mysqli