PHP Forum

**bennyboywonder** · 06-22-2005, 08:13 AM

If anyone can tell me why this form is not passing the post values from the first page to the 'formactions.php' I would be HUGELY grateful. the switch statement IS going to the right place, because the text at the bottom prints out, just for some reason the POST values are not being passed to this PHP page.

HTML Code:

					   <div class="commentsform">
                            <form action="formactions.php" method="post">
                                <div class="hidden">
                                    <input type="hidden" name="formaction" value="addcomment" />
                                    <input type="hidden" name="postid" value="<?php echo $postid; ?>" />
                                    <input type="hidden" name="commentlogid" value="<?php echo $logid; ?>" />
                                    <input type="hidden" name="returnpage" value="<?php echo $page; ?>" />
                                </div>

                                <div ><label for="name">name</label><input type="text" name="newcommentname" /></div>
                                <div ><label for="email">email</label><input type="text" name="commentemail" /></div>
                               <div><label for="body" cols="3">body</label><textarea name="commentbody"></textarea></div>
                                 <div><label></label><input type="submit" class="submitbutton" value="submit comment" /></div
                            </form>
                        </div>

PHP Code:


			
switch($_POST['formaction']) {

        case "addcomment":

        $commentpostidsend = $_POST['postid'];

        echo $commentpostidsend;

        $commentnamesend = $_POST['newcommentname'];

        echo $commentnamesend;

        if ($commentnamesend == "") $commentnamesend = "an anonymous reader";

        var $commentemailsend = $_POST['commentemail'];

        echo $commentemailsend;

        var $commentbodysend = $_POST['commentbody'];

        echo $commentbodysend;

        $commentlogid = $_POST['commentlogid'];

        $commenttimesend = date('H:i:s');

        $commentdatesend = date('Y-m-d');

        if ($_POST['admincomment'] == "on")    $commentadminsend = 1;

        else $commentadminsend = 0;

        $query3 = "INSERT INTO comments (EntryID, CommentDate, CommentTime, EntryName, EntryEmail, CommentBody, Admin, logid) VALUES ('$commentpostidsend', '$commentdatesend', '$commenttimesend', '$commentnamesend', '$commentemailsend', '$commentbodysend', $commentadminsend, $commentlogid)";

        mysql_query($query3) or die('error, insert query (query 3) failed');

        echo "<h4>Adding comment...</h4>";

        break;

}

**Kyrnt** · 06-22-2005, 08:22 AM

I don't know if it is related, but the </div after your submit button is not closed (may have been a cut and paste oversight).

Are you trying to print out the value of $POST[formaction] in formactions.php before you send it into the switch statement? It would be helpful to know if the value is there at all as opposed to not coming through the switch statement as you expect.

**bennyboywonder** · 06-22-2005, 10:51 AM

Quote:

Originally Posted by Kyrnt

I don't know if it is related, but the </div after your submit button is not closed (may have been a cut and paste oversight).

Are you trying to print out the value of $POST[formaction] in formactions.php before you send it into the switch statement? It would be helpful to know if the value is there at all as opposed to not coming through the switch statement as you expect.

how embarrassing. Yes closing that </div> tag seemed to solve my problem. Thanks very much :P

**Kyrnt** · 06-22-2005, 11:22 AM

No problem -- happy coding.