我該如何啓動它?我知道我需要做一個mysql查詢,但如何轉換數據,在下拉列表中的選項。請記住,它是在一個表單內,從MySQL表發送結果。以MySQL數據爲表格提取下拉列表的數組
(EDIT)
我printf的,這就是我想裏面工作:
<?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();
?>
感謝!我認爲這會對我有幫助。我想你應該改變這個問題,如果可以的話,以反映真正的問題。即,「爲下拉列表獲取數組結果」。我出於好奇,偶然讀過這篇文章,但可能找到了我在一段時間內遇到的問題的答案。 – motorbaby
沒關係;我編輯了這個問題。 (PHPMyAdmin與該問題無關。) – motorbaby