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PHP Script Forums » Web Developer Tutorials » Discuss our Web Development Tutorials

(20 posts)

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Started 6 months ago by websmartinc
Latest reply from Olaf

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Olaf
PHP Coder

Just tested the script with your database table:

there is one warning:
Notice: Undefined index: count in /home/user/httpdocs/dom.com/test.php on line...

This is because count doesn't exists the first time (you can ignore this one, you will never see that on a production server)

And the values for your parent menu items should link to pages where this script is used as well otherwise you get a 404 error check this screen:

Posted 6 months ago #
websmartinc
Member
okay, great... here's what I get now
```
<ul id="nav">

<li><a>Home</a></li>
<li><a>Products</a></li>
```
Posted 6 months ago #
websmartinc
Member

in the code I am seeing there are no hyperlinks and no sub items showing up...

Posted 6 months ago #
Olaf
PHP Coder

put your code on some URL and post it here

Posted 6 months ago #
websmartinc
Member

http://www.iowensound.com/testing/test.php

tx!!!

Posted 6 months ago #
Olaf
PHP Coder

soory but this is not possible with the script I posted before. Rename the script to test.txt and let me know (remove the database password)

Posted 6 months ago #
websmartinc
Member

tx

as per your instructions, i have changed it to
http://www.iowensound.com/testing/test.txt and removed the password...

Posted 6 months ago #

Olaf
PHP Coder

compare this function from the tutorial with you own code:

function dyn_menu($parent_array, $sub_array, $qs_val = "menu", $main_id = "nav", $sub_id = "subnav", $extra_style = "foldout") {
    $menu = "<ul id=\"".$main_id."\">\n";
    foreach ($parent_array as $pkey => $pval) {
        if (!empty($pval['count'])) {
            $menu .= "  <li><a class=\"".$extra_style."\" href=\"".$pval['link']."?".$qs_val."=".$pkey."\">".$pval['label']."</a></li>\n";
        } else {
            $menu .= "  <li><a href=\"".$pval['link']."\">".$pval['label']."</a></li>\n";
        }
        if (!empty($_REQUEST[$qs_val])) {
            $menu .= "<ul id=\"".$sub_id."\">\n";
            foreach ($sub_array as $sval) {
                if ($pkey == $_REQUEST[$qs_val] && $pkey == $sval['parent']) {
                    $menu .= "<li><a href=\"".rebuild_link($sval['link'], $qs_val, $sval['parent'])."\">".$sval['label']."</a></li>\n";
                }
            }
            $menu .= "</ul>\n";
        }
    }
    $menu .= "</ul>\n";
    return $menu;
}

EDIT: looks like a problem with the new parser script on the forum (code is changed)

Posted 6 months ago #

websmartinc
Member

this one has closing ul tags whereas the one I had before did not. I tried this code and the list shows the main level items and they are linked, however the sub items do not show. also, the Products does not link to Products.php it links to products.php?menu=2... not sure if that is supposed to do that or not.
i have reloaded the text file above with the changed code.

Posted 6 months ago #
Olaf
PHP Coder

The previous code was wrong (parse error by the forums plugin).
About the links: you need to add REAL links, create 3 pages, use them for the links. If you click one of the "parent" links you will see that sub-links will show up.
This is how this menu works, there is no option to keep the menu unfolded on initial load

Posted 6 months ago #

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