php - 获取表单中包含 MySQL 数据的下拉列表数组

Question

我该如何启动它？我知道我需要进行 mysql 查询，但是如何转换数据，在下拉列表中的选项中。请记住，它位于表单内，用于发送 MySQL 表的结果。

(编辑)

我正在 printf 中工作，这就是我想要的:

<?php ob_start();
include('/../../config.php'); 

if(isset($_POST['edit_id']) && !empty($_POST['edit_id'])) { 

$edit_id = mysql_real_escape_string($_POST['edit_id']); 
$result = mysql_query("SELECT username, password, nome, cidade, pais, base, isactive,   admin, dov, checador, dinheiro, email, datanascimento, profissao, idivao, idvatsim, horas, rank FROM acars_users WHERE `id`='".$edit_id."'");
$resultdl = mysql_query("SELECT * FROM acars_hubs");

$data = mysql_fetch_array($result);
$dl = mysql_fetch_array($resultdl); 


printf("<div  align=\"center\">
<br><form method=\"post\" action=\"editar2.php\">
<p><font size=\"2\" face=\"Segoe UI, Arial, Helvetica, sans-serif\"   align=\"center\">Modifique os campos que deseja para <strong>editar este membro.</font><br>
<br>

<table width=\"700\" border=\"0\" align=\"center\" >
<tr>
    <td>Base Operacional:</td>
    <td><label for=\"hub\"></label>
      <select name=\"hub\">
         <option>".$dl['name']."</option>
       </select>
    </td>
    </td>
</tr>
</table></br></br>
<input name=\"edit_id\" value=\"$edit_id\" type=\"hidden\">
<input type=\"image\" src=\"img/Editar.PNG\" width='85' height='30'></form>
</form>

</table> 
</div>

");
while ($data = mysql_fetch_array($result));
while ($dl = mysql_fetch_array($resultdl));
ob_end_flush();
?>

Answer 1

你的意思是这样吗？

<select>
<?php while($row = mysql_query("SELECT * FROM table")){ ?>
<option><?=$row['column']; ?></option>
<?php } ?>
</select>