Question

Am new to php… I have been battling on my dynamic checkboxes in such a way that if none is checked the form is return, also I need to retain what was checked when the form postback due to other invalid inputs.

$result = mysql_query("SELECT * FROM course") or die(mysql_error()); 
if ($result) 
{ 
  while ($row = mysql_fetch_array($result)){
    if (isset($_POST['courses']) and $_POST['courses'] == $row['cid']) {echo $row['cid'];} 
    print "<input type=\"checkbox\" name=\"courses[]\" value=\"$row[cid]\">$row[cname]\n"; 
  }
}

Help needed purely on php codes. Thanks in advance

Answer 1

Do this where the checkbox appears in the HTML on your php page:

<?php

   $checked = isset($_POST["checkboxname"]) ? " checked" : '' ;
   echo "<input type='checkbox' name='checkboxname' value='yes'" . $checked . ">";

?>

This will retain the checkbox state after the form has been posted.

UPDATE:

For your code, just do it like this, I think:

$result = mysql_query("SELECT * FROM course") or die(mysql_error()); 
if ($result) {
    while ($row = mysql_fetch_array($result)) {
        $checked = '';
/* ERROR:  if (isset($_POST['courses']) and $_POST['courses'] == $row['cid']) { */
        if (isset($_POST['courses']) {
           if (in_array($row['cid'], $_POST['courses']) {
               echo $row['cid'];
               $checked = " checked";
           }
        }
        echo "<input type=\"checkbox\" name=\"courses[]\"  value=\"$row[cid]\"" . $checked . ">$row['cname']\n"; 
    }
}

EDIT: The condition needs to change too, I think, as I show in the code above.