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Home/ Questions/Q 9156225
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Editorial Team
  • 0
Editorial Team
Asked: June 17, 20262026-06-17T12:43:52+00:00 2026-06-17T12:43:52+00:00

How can i start it? I know that i need to make an mysql

  • 0

How can i start it? I know that i need to make an mysql query, but how to transform data, in options in the dropdown list. And remember, it’s inside a form, to send the result from a MySQL table.

(EDIT)

I am working inside a printf That is what i want:

<?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();
?>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-17T12:43:52+00:00Added an answer on June 17, 2026 at 12:43 pm

    I’ve discovered the problem. My first $dl = mysql_fetch_array($resultdl, MYSQL_ASSOC); pulls the first row from the table. When I start while loop the mysql_fetch_array picks up with the next row. So, the solution is remove the first call. This is the final code:

    <?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>
            ");
    
    ?>
    
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