» Main » PHP Scripts Development

PHP Scripts Development » Discuss our Web Development Tutorials

1. 

   eyolf
   Member
   

   Thanks a lot for the tutorial, which I enjoyed quite a lot: not too much basic explanation, but just enough to give a newbie a fair chance to know what to look for and where. I like that.

   Anyway, it all seems to work fine, except for one thing: I get a message:

   Notice: Undefined index: count in /home/httpd/html/test/menu.php on line 41

   I fiddled around and solved it by adding

   $parent_menu[$obj->parent_id]['count'] = 0;

   at the start of the else { } segment where 'count' first appears. I don't know if this is the right way to do it -- my php knowledge is fairly shaky -- but I just thought I'd let you know, and if you have another solution, I'd be happy to hear about it.

   Posted 3 months ago #
2. 

   Olaf
   Administrator
   

   You can use it this way too:

   (quick and dirty)

   @$parent_menu[$obj->parent_id]['count']++;

   Posted 3 months ago #
3. 

   eyolf
   Member
   

   Thanks.
   Another question: I want a two-level menu, and that's working fine. However, the directory structure goes three or four levels down, but when I visit files further down in the hierarchy, the submenu disappears. I would like all files in a directory under some of the main menu items to keep the current submenu open. So, e.g.:

   menuitem1     index.php
      subitem1   sub1.php
      subitem1   sub2.php
   menuitem2     dir1/index.php
      subitem1   dir1/subpage1.php

   where the page sub1.php may lead to a file dir2/page.php, which should keep the first submenu open. Is this possible?

   Posted 3 months ago #
4. 

   Olaf
   Administrator
   

   You need to add the QS var for the sub menu's to the URL

   Posted 3 months ago #
5. 

   eyolf
   Member
   

   Excellent! Thanks a lot.

   Posted 3 months ago #
6. 

   Olaf
   Administrator
   

   try this function

   Posted 3 months ago #

RSS feed for this topic

Reply

You must log in to post.