如何获取列值并在select标记中显示它们

Question

我试过这段代码链接.但结果只显示了列的最后一个值.我想通过选择目的显示select标签中列的所有值.如何获得适当的价值？

Answer 1

避免在option标记中编写如此多的PHP代码.在数据库之前写入数据库获取代码并将其存储在数组中.

稍后,继续循环$array.

  <?php
     $dbc = mysqli_connect(DB_HOST,DB_USER,DB_PASSWORD,DB_NAME);
     $query = "SELECT id FROM done_add";
     $data = mysqli_query($dbc, $query);
     $array=[];
     while ($row = mysqli_fetch_array($data)) {
         $array[] = $row['id'];
     }
   ?>

    <select name="selectlink">
        <?php foreach ($array as $arr) { ?>
            <option value = ""> <?php print($arr); ?></option>
        <?php } ?>
    </select>