0

我该如何启动它？我知道我需要在下拉列表中的选项中进行 mysql 查询，但是如何转换数据。请记住，它在一个表单中，用于从 MySQL 表中发送结果。

（编辑）

我在 printf 里面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();
?>
4

2 回答 2

2

你的意思是这样的吗？

<select>
<?php while($row = mysql_query("SELECT * FROM table")){ ?>
<option><?=$row['column']; ?></option>
<?php } ?>
</select>
于 2013-01-06T02:16:47.230 回答
0

我发现了问题。我的第$dl = mysql_fetch_array($resultdl, MYSQL_ASSOC);一个从表中拉出第一行。当我开始while循环mysql_fetch_array时，接下一行。因此，解决方案是删除第一个电话。这是最终代码：

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

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

    $edit_id = mysql_real_escape_string($_POST['edit_id']); 

    $result = mysql_query("SELECT * FROM acars_users WHERE `id`='".$edit_id."'");
    $data = mysql_fetch_array($result);
    $resultdl = mysql_query("SELECT * FROM acars_hubs");

    printf("<div  align=\"center\">
    <form method=\"post\" action=\"actions/actions_editar.php\">
    ");

           while ($dl = mysql_fetch_array($resultdl, MYSQL_ASSOC)){

        printf(" 
        <option value=".$dl["id"].">".$dl["name"]."</option>;
        ");

           }

    printf("
        <input name=\"edit_id\" value=\"$edit_id\" type=\"hidden\">
        <input type=\"image\" src=\"../../images/botao_editar.PNG\" width='85' height='30'></form>
        ");

?>
于 2013-01-20T15:35:12.657 回答