Question

Been trying for hours this php embedding in html but something is wrong as i get a error:

Parse error: syntax error, unexpected $end in C:\Program Files\xampp\htdocs\profile.php on line 678

<select value='$pays' name='pays' id='pays' style='width: 204px;margin-bottom: 5px; outline-width:0;'>
<?php
$result = mysql_query("SELECT pays_az, pays_zz, pays_or FROM in_lays WHERE pays_flush = '1' ORDER BY pays_nom ASC");
while ($row = mysql_fetch_array ($result) )
{  ?>
     <option value="<?php echo $row['dd']; ?>"

     <?php   if($row['dd_id'] == $pays)
     {
        echo 'selected="selected" ' ;
     }
     else
     {
           if($row['dd_id'] == "61")
        {
       echo 'selected="selected"' ;
        }
     }
 ?>><?php echo $row['lala']; ?></option>
<?php}?>

</select>

dont pay attention to the names in the sql request.

Answer 1

Маке

<?php}?>

to

<?php } ?>

And if in first row $pays a php variable. If yes – please echo it