I want to make my website menu appear with the following conditions:
If user did not login, the menu will be like this:
Gallery | Contact Us | Login
If the user logged into the website, the menu will appear like this:
Gallery | Contact Us | Logout
This is my coding to call menu in header file:
<?php
$output = '';
$menus1 = MenuPortal::model()->findAll(array('condition' => "type='atas' AND display_status='1' AND parent_id=0 ORDER BY sort ASC"));
foreach ($menus1 as $menu) {
$submenu = MenuPortal::model()->findAll(array('condition' => "parent_id=$menu->id"));
if (sizeof($submenu) > 0) {
foreach ($submenu as $smenu) {
$output .= '<li><a href=' . $menu->url . '&id=' . PortalElement::encrypt_decrypt('encrypt', $smenu->id) . '>' . $menu->title_my . '</a></li>';
break;
}
}
else {
$output .= '<li><a href=' . $menu->url . '&id=' . PortalElement::encrypt_decrypt('encrypt', $menu->id) . '>' . $menu->title_my . '</a></li>';
}
}
echo $output;
?>
What I was thinking is to check if the session is active, then show menu #2. If there is no session, then show menu #1.
After Your foreach loop just apply following condition..
if(Yii::app()->user->isGuest()){
$output. = "<li><a href='login'>Login</a></li>";
}else{
$output. = "<li><a href='login'>Logout</a></li>";
}
Related
I have nav menu in html and i want to create a loop everything is fine but the item that have children echo twice! i know where is the problem but i can't figure out how to solve it :(
my sql table named menu
and this is my php:
$db = mysqli_connect('localhost', 'root', 'password', 'aftab');
<?php
$get = mysqli_query($db , "SELECT * from menu where parent_id is NULL");
while ($rowmenu = mysqli_fetch_assoc($get)) {
echo '<li class="menu-item menu-item-type-post_type menu-item-object-page menu-item-656"><a href="'. $rowmenu['link'] . '" >' . $rowmenu['name'] .'</a>' ;
$id = $rowmenu['id'] ;
$check = mysqli_query($db , "SELECT * from menu where parent_id = '$id'");
if ( mysqli_num_rows($check) ) {
echo '<li class="menu-item menu-item-type-post_type menu-item-object-page menu-item-has-children menu-item-656"><a href="'. $rowmenu['link'] . '" >' . $rowmenu['name'] .'</a>' ;
echo '<ul class="sub-menu">' ;
while ( $row2 = mysqli_fetch_assoc($check) ) {
echo '<li class="menu-item-302">' . $row2['name'] . '</li>' ;
}
echo '</ul>' ;
} else {
echo '</li>' ;
}
}
?>
and this the result:enter image description here
i know it happen because the father item that hold the sub menu's called in $get once and another time when it need other css class.i tried if , foreach , while and many things.
i need that item that holds submenus should have "menu-item-has-children" class otherwise its not show the sub menus.
When a menu item, have childrens then echo menu with menu-item-has-children class, otherwise echo a simple menu and move on.
<?php
$get = mysqli_query($db , "SELECT * from menu where parent_id is NULL");
while ($rowmenu = mysqli_fetch_assoc($get)) {
$id = $rowmenu['id'] ;
$check = mysqli_query($db , "SELECT * from menu where parent_id = '$id'");
$haveSubMenu = mysqli_num_rows($check);
if($haveSubMenu)
echo '<li class="menu-item menu-item-type-post_type menu-item-object-page menu-item-656 menu-item-has-children"><a href="'. $rowmenu['link'] . '" >' . $rowmenu['name'] .'</a>' ;
else
echo '<li class="menu-item menu-item-type-post_type menu-item-object-page menu-item-656"><a href="'. $rowmenu['link'] . '" >' . $rowmenu['name'] .'</a>' ;
if ($haveSubMenu)
{
echo '<ul class="sub-menu">' ;
while ( $row2 = mysqli_fetch_assoc($check) ) {
echo '<li class="menu-item-302">' . $row2['name'] . '</li>' ;
}
echo '</ul>' ;
} else {
echo '</li>' ;
}
}
?>
Here is see that Our Products from the database is printed twice.This is because you are fetching and echoing $row2['name'] twice but you can handle the first coming array by changing your SQL query.
Change your first SQL query to
$get = mysqli_query($db , "SELECT * from menu where parent_id is NULL AND name != 'OUR PRODUCTS'");
Querying this to database you will get all the NULL values in the array excluding OUR PRODUCTS and thus it will be echoed only once by the second query.
I tried the above to answer all is working only a few level dropdown menu.
If you want to really create your menu for multi-level dropdown as like tree structure just use the bellow function. for more detailing once go through
https://bootstrapfriendly.com/blog/dynamic-multi-level-dropdown-sticky-menu-in-php-mysql-using-bootstrap/
<?php
include_once("connection.php");
$query = "SELECT id, label, link, parent FROM menus ORDER BY sort ASC, label";
$result = mysqli_query($conn, $query) or die("database error:" . mysqli_error($conn));
// Create an array to conatin a list of items and parents
$menus = array(
'items' => array(),
'parents' => array()
);
// Builds the array lists with data from the SQL result
while ($items = mysqli_fetch_assoc($result)) {
// Create current menus item id into array
$menus['items'][$items['id']] = $items;
// Creates list of all items with children
$menus['parents'][$items['parent']][] = $items['id'];
//echo $items;
}
// function to create dynamic treeview menus
function createMenu($parent, $menu)
{
$html = "";
if (isset($menu['parents'][$parent])) {
// $html .= '<ul class="sina-menu sina-menu-right" data-in="fadeInLeft" data-out="fadeInOut">';
foreach ($menu['parents'][$parent] as $itemId) {
if (!isset($menu['parents'][$itemId])) {
$html .= "<li >
<a href='" . $menu['items'][$itemId]['link'] . "'>" . $menu['items'][$itemId]['label'] . "</a>
</li>";
}
if (isset($menu['parents'][$itemId])) {
$html .= "<li class='dropdown'>
<a class='dropdown-toggle' data-toggle='dropdown' href='" . $menu['items'][$itemId]['link'] . "'>" . $menu['items'][$itemId]['label'] . "</a>";
$html .= '<ul class="dropdown-menu">';
$html .= createMenu($itemId, $menu);
$html .= '</ul>';
$html .= "</li>";
}
}
// $html .= "</ul>";
}
return $html;
}
I want to make mega dropdown menu that comes from mysql database. Below is my php code. The code is working well. But the problem is I am unable to make mega dropdown menu for the code below.
I need the mega menu like the example here : https://bootsnipp.com/snippets/featured/bootstrap-mega-menu
My problem is how will I make more div with the below php code .
Plz help me the css for the mega drop down menu as shown above.
<?php
$sql = "SELECT id, product, parent_id, category_link FROM category ORDER BY parent_id, id";
$results = mysqli_query($conn,$sql) or die(mysqli_error()) ;
if($results)
{
while($result = mysqli_fetch_array($results))
{
$category['categories'][$result['id']] = $result;
$category['parent_cats'][$result['parent_id']][] = $result['id'];
}
}
function getCategories($parent, $category)
{
$html = "";
if (isset($category['parent_cats'][$parent]))
{
$html .= "<div id='wrapper'>";
$html .= "<ul class='mega-menu'>\n";
foreach ($category['parent_cats'][$parent] as $cat_id)
{
if (!isset($category['parent_cats'][$cat_id]))
{
$html .= "<li class='mega-menu-drop'>\n <a class='mega-menu-content' href='" . $category['categories'][$cat_id]['category_link'] . "'>" . $category['categories'][$cat_id]['product'] . "</a>\n</li> \n";
}
if (isset($category['parent_cats'][$cat_id]))
{
$html .= "<li class='mega-menu-drop'>\n <a class='mega-menu-content' href='" . $category['categories'][$cat_id]['category_link'] . "'>" . $category['categories'][$cat_id]['product'] . "</a> \n";
$html .= getCategories($cat_id, $category);
$html .= "</li> \n";
}
}
$html .= "</ul> \n";
$html .= "</div>";
}
return $html;
}
?>
<?php echo $data['category'] = getCategories(0, $category);?>
as of your comment. i guess you need some CSS.
this will work on mouse hover. but not on click like in your example:
.mega-menu-drop {
display:none
}
.mega-menu:hover .mega-menu-drop {
display:block
}
I need some assistance to create a foreach loop which will display each array item from an array with three items. Currently, my code correctly displays the first and last item; however, it does not display the second item.
The loop will be used to generate a navbar from the items in the array.
Here is my PHP code:
<ul>
<?php
$navOptions = array('home','services','contact');
foreach($navOptions AS $navOption) {
if ($navOption == $currentPage) {
echo '<li>' . ucfirst($navOption) . '</li>';
} else {
echo '<li><a href="' . $navOption . '.php>' . ucfirst($navOption) . '</li></a';
}
}
?>
</ul>
This code generates a list of links like this:
Home
Contact
However, it does not generate the "Services" list item.
A side note: the $currentPage variable is declared as a global variable on each individual html page ie, within index.php, services.php, contact.php.
You are missing some double quotes, and the closing > of the else statement, should be as follows:
$navOptions = array('home','services','contact');
foreach($navOptions AS $navOption) {
if ($navOption == $currentPage) {
echo '<li>' . ucfirst($navOption) . '</li>';
} else {
echo '<li>' . ucfirst($navOption) . '</li>';
}
}
I am trying to create a Dropdown Menu on my site, Ive hit a brick wall and cannot think on how to do it.
Basically i need to check 2-3 variables in a database and out put the correct data.
I have at the moment it checking if its an External Link or Not, and if it contains a submenu, but i can't get it to output the correct information.
Basically i want it to check if its a External or Non-External link, and if it has a submenu, if it has a submenu, to display the menu options underneath it. So say i have menu 1, 2 ,3, 4 and 2,4 has a submenu, i need them to list the other links under them. i have put in my database toplink_id (to represent which link this item should be under) sc_order (which will control the order the sublinks display in) also dropdown (which tells me if the menu has a submenu or not.)
Here is the start of my code
$sql = "SELECT label, url, ext, dropdown FROM content_pages WHERE top_nav='1' AND active='1' ORDER by page_order ASC";
$query = mysqli_query($dbc, $sql) or die (mysqli_error($dbc));
$menuDisplay .= '<div class="bg-2"><div class="container_12"><article class="grid_12"><nav><ul class="menu sf-js-enabled">';
while ($row = mysqli_fetch_array($query)) {
$url = $row["url"];
$nav_label = $row["label"];
$drop_down ='<ul><li>' . $nav_label . '</li></ul>';
if ($row["ext"] == 0 && $row["dropdown"] == 1){
$menuDisplay .= '<li>' . $nav_label . '' . $drop_down . '</li>';
}
elseif ($row["ext"] == 1 && $row["dropdown"] == 1){
$menuDisplay .= '<li>' . $nav_label . '' . $drop_down . '</li>';
}
elseif ($row["ext"] == 0){
$menuDisplay .= '<li>' . $nav_label . '</li>';
}
elseif ($row["ext"] == 1)
{
$menuDisplay .= '<li>' . $nav_label . '</li>';
}
}
$menuDisplay .= '</ul></nav></article></div></div></header>';
mysqli_free_result($query);
Best way I found to do this without using jQuery is use multidimensional Arrays.
// Create a multidimensional array to conatin a list of items and parents
$menu = array(
'items' => array(),
'parents' => array(),
);
// Builds the array lists with data from the menu table
while ($items = mysqli_fetch_assoc($query))
{
// Creates entry into items array with current menu item id ie. $menu['items'][1]
$menu['items'][$items['id']] = $items;
// Creates entry into parents array. Parents array contains a list of all items with children
$menu['parents'][$items['parent']][] = $items['id'];
}
// Menu builder function, parentId 0 is the root
function buildMenu($parent, $menu)
{
$html = "\n";
if ( isset($menu['parents'][$parent]) )
{
$html .= "";
foreach ($menu['parents'][$parent] as $itemId)
{
if(!isset($menu['parents'][$itemId]) && $menu['items'][$itemId]['ext'] == 0)
{
$html .= "<li>\n <a href='../pages/".$menu['items'][$itemId]['link']."'>".$menu['items'][$itemId]['label']."</a>\n</li> \n";
}
else
if(!isset($menu['parents'][$itemId]) && $menu['items'][$itemId]['ext'] == 1)
{
$html .= "<li>\n <a href='../".$menu['items'][$itemId]['link']."'>".$menu['items'][$itemId]['label']."</a>\n</li> \n";
}
if(isset($menu['parents'][$itemId]))
{
$html .= "<li>\n <a href='../pages/".$menu['items'][$itemId]['link']."'>".$menu['items'][$itemId]['label']."<span class='arrow-down'></span></a> \n";
$html .= "<ul style='border-radius: 0px 0px 6px 6px'> \n";
$html .= buildMenu($itemId, $menu);
$html .= "</ul> \n";
$html .= "</li> \n";
}
}
$html .= "\n";
}
$html .= "";
return $html;
}
I need following structure via Hook in the template.php (Drupal 7):
<ul class="first border">
<li><span class="big">Main Menu Title</span><span class="small">Main Menu Description</span><ul class="flyout"><li class="submenu">Submenu Item</li></ul></li></ul>
My current status in the template.php is following:
function mytheme_menu_link(array $variables) {
$element = $variables['element'];
$sub_menu = '';
$element['#localized_options']['html'] = TRUE;
if ($element['#below']) {
$sub_menu = drupal_render($element['#below']);
}
if ($element['#original_link']['menu_name'] == "main-menu" && isset($element['#localized_options']['attributes']['title'])){
$element['#title'] .= '<span class="description">' . $element['#localized_options']['attributes']['title'] . '</span>';
}
$output = l($element['#title'], $element['#href'], $element['#localized_options']);
return '<li' . drupal_attributes($element['#attributes']) . '>' . $output . $sub_menu . "</li>\n";
}
I tried something but without success.
If you want to set classes on your menu options, download menu_attributes.
To figure out why it's not working, did you change mytheme to your actual there name?