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  1. POFetch array for dropdown list with MySQL data in a form
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    copied!<p>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. </p> <p>(EDIT)</p> <p>I am working inside a printf That is what i want:</p> <pre><code>&lt;?php ob_start(); include('/../../config.php'); if(isset($_POST['edit_id']) &amp;&amp; !empty($_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("&lt;div align=\"center\"&gt; &lt;br&gt;&lt;form method=\"post\" action=\"editar2.php\"&gt; &lt;p&gt;&lt;font size=\"2\" face=\"Segoe UI, Arial, Helvetica, sans-serif\" align=\"center\"&gt;Modifique os campos que deseja para &lt;strong&gt;editar este membro.&lt;/font&gt;&lt;br&gt; &lt;br&gt; &lt;table width=\"700\" border=\"0\" align=\"center\" &gt; &lt;tr&gt; &lt;td&gt;Base Operacional:&lt;/td&gt; &lt;td&gt;&lt;label for=\"hub\"&gt;&lt;/label&gt; &lt;select name=\"hub\"&gt; &lt;option&gt;".$dl['name']."&lt;/option&gt; &lt;/select&gt; &lt;/td&gt; &lt;/td&gt; &lt;/tr&gt; &lt;/table&gt;&lt;/br&gt;&lt;/br&gt; &lt;input name=\"edit_id\" value=\"$edit_id\" type=\"hidden\"&gt; &lt;input type=\"image\" src=\"img/Editar.PNG\" width='85' height='30'&gt;&lt;/form&gt; &lt;/form&gt; &lt;/table&gt; &lt;/div&gt; "); while ($data = mysql_fetch_array($result)); while ($dl = mysql_fetch_array($resultdl)); ob_end_flush(); ?&gt; </code></pre>
 

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