» PHP Forum Archive » PHP snippets and example code » Discuss our PHP snippets here
Hi Guys,
I have to create a dynamic menu using php and mysql. I have tried to used script given at this site at the following url: http://www.finalwebsites.com/tutorials/dynamic-navigation-list.php
I am able to get the first level of category menu but not able to get the second and any other level of menu. i think i am not able to get the CSS for this code, can somebody tell me what exactly the thing i am missing.
Hi,
do you have an live example somewhere or paste the php code here.
or better check the arrays, use this code before you call the function
print_r($sub_menu);
print_r($parent_menu);Hi
here is the code i am using:
$link = mysql_connect('localhost','root','');
mysql_select_db('globalbiz',$link);
echo $sql = "SELECT id, cat_name, page_url, parent_id FROM test_cat ORDER BY parent_id, id ASC";
$items = mysql_query($sql);
while ($obj = mysql_fetch_object($items)) {
if ($obj->parent_id == 0) {
$parent_menu[$obj->id]['label'] = $obj->cat_name;
$parent_menu[$obj->id]['link'] = $obj->page_url;
} else {
$sub_menu[$obj->id]['parent'] = $obj->parent_id;
$sub_menu[$obj->id]['label'] = $obj->cat_name;
$sub_menu[$obj->id]['link'] = $obj->page_url;
$parent_menu[$obj->parent_id]['count']++;
}
}
mysql_free_result($items);
/////end first block
//echo"
"; print_r($sub_menu);exit;
///second block
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 .= "
Hi,
do you tried the original code first?
Dear Olaf,
I am having the same issue with the sub-menus. I have been trying to use the code that you have provided. In tracing through it, I believe it is in how I am marking the rows in the database. The arrays trace out based on feedback you have provided(see below), but I was wondering if you have a sample table of what the rows look like in MYSQL? Or is it maybe the CSS? Or does another column need an index? Any help would be great.
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
<link rel="nofollow" href="style/structure.css" rel="stylesheet" type="text/css" />
</head>
<body>
<ul id="nav">
<li><a class="foldout" rel="nofollow" href="test?menu=1">Design</a></li>
<li><a class="foldout" rel="nofollow" href="where?menu=2">New Category</a></li>
</ul>
Array
(
[0] => Array
(
[parent] => 1
[label] => Design
[link] => design_link
)
[3] => Array
(
[parent] => 2
[label] => New Category
[link] => new
)
[4] => Array
(
[parent] => 2
[label] => New Category
[link] => new2
)
)
Array
(
[1] => Array
(
[label] => Design
[link] => test
[count] => 1
)
[2] => Array
(
[label] => New Category
[link] => where
[count] => 2
)
)
</body>
</html>\Hi I'm confused, where is the css cod for this? I can get the parent menu but not the sub menu. What does $extra_style refer to?
Cheers
do you read the tutorial behind the link above?
there is no stylesheet (not part of the tutorial)
$extrastyle is just a variable that can hold a css pseudoclass
"Note: sources at finalwebsites.com are for developer, it's (mostly) not ready-to-use code, but code which is easy to use if you know a little bit about PHP" ;)
I know a little php but not loads - I did read the tutorial cheers :)
Is it ok to ask for help on here?
I can get the parent menu and the sub menu to appear but not mixed. And I am not sure why. The first line only shows the parent menu:
echo dyn_menu($parent_menu, $sub_menu, "menu", "nav", "subnav");
echo dyn_menu($parent_menu, "menu", "nav", "subnav");
echo dyn_menu($sub_menu, "menu", "nav", "subnav");
The others are parent_menu and sub_menu. I created a record sub of 1...
Please advise -
Cheers,
Colin
you need the function only ones, how does you array looks like ?(show it with print_r())
Array ( [1] => Array ( [label] => Start [link] => # [count] => 1 ) [2] => Array ( [label] => Info [link] => http://www.yahoo.co.uk [count] => 2 ) ) Array ( [5] => Array ( [parent] => 1 [label] => About Us [link] => # ) [3] => Array ( [parent] => 2 [label] => fef [link] => http://www.google.co.uk ) [4] => Array ( [parent] => 2 [label] => efe [link] => # ) ) ArrayArray
Cheers for looking :)
OK not sure what is going on but I can see this post on this forum if I am logged in - but not if I'm not.
I posted this - see below:
Array ( [1] => Array ( [label] => Start [link] => # [count] => 1 ) [2] => Array ( [label] => Info [link] => http://www.yahoo.co.uk [count] => 2 ) ) Array ( [5] => Array ( [parent] => 1 [label] => About Us [link] => # ) [3] => Array ( [parent] => 2 [label] => fef [link] => http://www.google.co.uk ) [4] => Array ( [parent] => 2 [label] => efe [link] => # ) ) ArrayArray
The post was marked as spam, could you post the array info between the code tags like
echo 'hello World';This way it hard to read
Ok I was wondering what was happening. Cheers for your help.
Array (
[1] => Array ( [label] => Start [link] => # [count] => 1 )
[2] => Array ( [label] => Info [link] => http://www.yahoo.co.uk [count] => 2 ) )
Array (
[5] => Array ( [parent] => 1 [label] => About Us [link] => # )
[3] => Array ( [parent] => 2 [label] => fef [link] => http://www.google.co.uk )
[4] => Array ( [parent] => 2 [label] => efe [link] => # ) ) ArrayArrayI guess this will not work, use the full code from the tutorial on your website, dump the array and send me the URL
This is my code - adapted from above:
<?php
// Dynamic menu from http://www.finalwebsites.com/tutorials/dynamic-navigation-list.php
// Connect to the db.
require_once 'dyn_mysql_connect.php';
$sql = "SELECT id, label, link_url, parent_id FROM dyn_menu ORDER BY parent_id, id ASC";
$items = mysql_query($sql);
while ($obj = mysql_fetch_object($items)) {
if ($obj->parent_id == 0) {
$parent_menu[$obj->id]['label'] = $obj->label;
$parent_menu[$obj->id]['link'] = $obj->link_url;
} else {
$sub_menu[$obj->id]['parent'] = $obj->parent_id;
$sub_menu[$obj->id]['label'] = $obj->label;
$sub_menu[$obj->id]['link'] = $obj->link_url;
$parent_menu[$obj->parent_id]['count']++;
}
}
mysql_free_result($items);
print_r($parent_menu);
print_r($sub_menu);
function dyn_menu($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."\" rel="nofollow" href=\"".$pval['link']."?".$qs_val."=".$pkey."\">".$pval['label']."</a></li>\n";
} else {
$menu .= " <li><a rel="nofollow" href=\"".$pval['link']."\">".$pval['label']."</a></li>\n";
}
if (!empty($_REQUEST[$qs_val])) {
$menu .= "<ul id=\"".$sub_id."\">\n";
$menu .= "</ul>\n";
}
}
$menu .= "</ul>\n";
return $menu;
}
function rebuild_link($link, $parent_var, $parent_val) {
$link_parts = explode("?", $link);
$base_var = "?".$parent_var."=".$parent_val;
if (!empty($link_parts[1])) {
$link_parts[1] = str_replace("&", "##", $link_parts[1]);
$parts = explode("##", $link_parts[1]);
$newParts = array();
foreach ($parts as $val) {
$val_parts = explode("=", $val);
if ($val_parts[0] != $parent_var) {
array_push($newParts, $val);
}
}
if (count($newParts) != 0) {
$qs = "&".implode("&", $newParts);
}
return $link_parts[0].$base_var.$qs;
} else {
return $link_parts[0].$base_var;
}
}
echo dyn_menu($parent_menu, $sub_menu, "menu", "nav", "subnav");
echo dyn_menu($parent_menu, "menu", "nav", "subnav");
echo dyn_menu($sub_menu, "menu", "nav", "subnav");
?>and my table sql
CREATE TABLE IF NOT EXISTSdyn_menu` (
id int(11) NOT NULL auto_increment,
label varchar(50) NOT NULL default '',
link_url varchar(100) NOT NULL default '#',
parent_id int(11) NOT NULL default '0',
PRIMARY KEY (id)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=6 ;
INSERT INTO dyn_menu (id, label, link_url, parent_id) VALUES
(1, 'Start', '#', 0),
(2, 'Info', 'http://www.yahoo.co.uk', 0),
(3, 'fef', 'http://www.google.co.uk', 2),
(4, 'efe', '#', 2),
(5, 'About Us', '#', 1);`
Cheers Olaf,
Colin
How come my posts don't show up unless I'm logged in? Am I still considered a spammer?
Cheers
looks like that the forum soft has some issues (need to upgrade very soon)
It seems that you did everything alright, where is the link to your example page?
as said before you need to call the function only ONES:
echo dyn_menu($parent_menu, $sub_menu, "menu", "nav", "subnav");:)
PS. maybe you can use a different email-address for your account (that should help akismet that you're not a spammer
Yeah I remebered and did do that. I wanted to see that the submenu was returning as well - also just playing with the code. Thing is I wanted to get the parent and sub menus displaying together like:
Parent
- Child
- Child
Parent
- Child
- Child
Parent
- Child
which echo dyn_menu($parent_menu, $sub_menu, "menu", "nav", "subnav"); looks like it should be doing.
Baffled.
Cheers,
Colin
Changed my email to my googlemail account. I think I am marked for life...
With echo dyn_menu($parent_menu, $sub_menu, "menu", "nav", "subnav"); all I get is the parent_menu.
Cheers.