position the result base on row number in mysql - php

I have this function (refer below)
function showMenu(){
$result = mysqli_query($con,"SELECT * FROM user_menu");
$menu = '<nav id="main-menu"><ul class="menuwrapper extend">';
while($row = mysqli_fetch_array($result))
{
$menu .= '<li class="' . $row['status'] . '">' . $row['name'] . '</li>';
}
$menu .= '</ul></nav>';
return $menu;
}
and my database table is structured as follows (refer below)
id(BigInt 100, AUTO_INCREMENT, PRIMARY)
name(text, utf8_general_ci)
url(varchar 100, utf8_general_ci)
status(text, utf8_general_ci)
key(int 30)
The contents of that table are (refer
below)
id = "1", name = "Post", url = "http://localhost/gadgetmarket/admin/index.php", status="default", key = "0"
id = "2", name = "Pages", url = "http://localhost/gadgetmarket/admin/pages.php", status="parent", key = "1"
id = "3", name = "Menus", url = "http://localhost/gadgetmarket/admin/menus.php", status="parent", key = "2"
id = "4", name = "Theme", url = "http://localhost/gadgetmarket/admin/theme.php", status="parent", key = "3"
Now, my question is how could I output the result base on there keys number that each menu has on? my expected output base on what im trying to achieve from the datas above is (refer below)
<nav id="main-menu">
<ul class="menuwrapper extend">
<li class="default">
Post
</li>
<li class="parent">
Pages
</li>
<li class="parent">
Menus
</li>
<li class="parent">
Theme
</li>
</ul>
</nav>
As you can see the output was arrange base on keys order. Any recommendations, suggestions and ideas will be greatly appreciated. Thanks in advance.

For a start, you're not declaring the connection global in your function
function showMenu(){
global $con; // SIMPLY ADD THIS LINE
$result = mysqli_query($con,"SELECT * FROM user_menu");
$menu = '<nav id="main-menu"><ul class="menuwrapper extend">';
while($row = mysqli_fetch_array($result))
{
$menu .= '<li class="' . $row['status'] . '">' . $row['name'] . '</li>';
}
$menu .= '</ul></nav>';
return $menu;
}
and you can use ORDER BY to fetch the data the way you want it
SELECT * FROM user_menu ORDER BY key ASC

Related

dynamic multi level dropdown menu php sql

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;
}

Interrupt and return repetition within the while multiple times

I'm not new here but I also do not ask questions because whenever I have a problem I find a solution in questions related here in the forum or in blogs, but I did not have any success with this problem.
In a database I have the tables posts and categories that are related, and in a single MySQL query I am displaying the title of the posts within their categories. So far so good, everything working. However within the while loop structure the tag with the is-sub class can not be repeated. Is there any way to stop the repetition of this while and then repeat it again?
I thank you for your help
echo '
<nav class="navbar">
<ul>';
$sql = Query("SELECT * FROM posts INNER JOIN categories ON post_category = category_id ORDER BY post_category, post_title");
$is_category = NULL;
while ($row = mysqli_fetch_assoc($sql)) :
if ($row['post_category'] != $is_category) {
echo '<li class="has-sub">' . $row['category_title'] . '';
$is_category = $row['post_category'];
}
// stop loop
echo '<ul class="is-sub">';
// return loop
echo '<li>' . $row['post_title'] . '</li>';
// stop loop
echo '</ul>';
// return loop
echo '</li>';
endwhile;
echo '
</ul>
</nav>';
If I understood what you are trying to do, I think the code below would work. If the is_sub tag is really meant to show only once then this could work:
<?php
echo '
<nav class="navbar">
<ul>';
$sql = Query("SELECT * FROM posts INNER JOIN categories ON post_category = category_id ORDER BY post_category, post_title");
$is_category = NULL;
$issub = false;
while ($row = mysqli_fetch_assoc($sql)) :
if ($row['post_category'] != $is_category) {
echo '<li class="has-sub">' . $row['category_title'] . '';
$is_category = $row['post_category'];
echo '<ul class="is-sub">';
}
if(!$issub) {
echo '<ul class="is-sub">';
$issub = true;
}
echo '<li>' . $row['post_title'] . '</li>';
echo '</ul>';
echo '</li>';
endwhile;
echo '
</ul>
</nav>';
?>

How Would I Turn Mysqli Table Results into a menu with drop down

Here Is my question: What I am wanting To do is Take Results from a mysql table and turn them into a menu and a drop down menu
HERE IS A QUICK EXAMPLE:
if you see in my mysql table i have page_name and parent, So the example is:
page_name and if i have row 1 the page_name is 'Home' now it's parent is 'none' right but on id number 39 the page_name is 'Contact Us' and the Parent Is 'Far Far Away 123' so if the parent is equal to 'none' then it will show at the top of the menu not the drop down if it has a parent it will show under that parent like this:
Home | the ben page | The Brock Page | Far Far Away 123 | dsfk
Contact Us
You see Contact Us is under Far Far Away Because the parent Is Far Far Away 123
here is my table:
Here is my code That I am trying but it is not working for some reason:
<ul>
<?php
$sql = "SELECT * FROM pages ORDER by item_order";
$result = mysqli_query($db, $sql);
confirm_query($result);
while ($links = mysqli_fetch_assoc($result)) {
if($links['parent'] !== "none") {
?>
<li id = "<?php echo $links['id']; ?>"><a href="
<?php
echo "page.php?id=" . $links['id'] . "\" title=\"" . $links['page_title'] . "\"";
?>>
<?php
echo $links['page_name'];
?>
</a>
<?php
if($links['parent'] !== "none") {
$child = "";
$sql = "SELECT * FROM pages";
$result = mysqli_query($db, $sql);
while($row = mysqli_fetch_assoc($result)) {
if($row['parent'] !== "none") {
$child = $row['page_name'];
}
}
echo "<ul id=\"sub_menu\" class=\"sub_navagation" . $links['id'] . "\">";
echo "<li>";
echo $child;
echo "<li>";
echo "</ul>";
}
?>
</li>
<?php
}
}
?>
</ul>
CSS:
#sub_menu {
display: none;
}
#sub_menu:hover {
display: block;
}
Ok if as you can see i have the parent row in the MYSQL table and on id number 39 i want the 'Far Far Away123' to be the parent of Contact Us and i want to show it when i hover over 'Far Far Away123'
My suggestion is to build out an array of all the results. Then run through that array (instead of multiple database queries).
I added a function build_dropdown() that will take the page name and run through the array of pages to see if there are any items with a parent matching. If so, we make an array of those items and run through them to build the dropdown menu. If not, it does nothing and moves on to the next menu item.
<?php
function build_dropdown ($parent, $pages){
foreach($pages as $page){
if($page['parent'] == $parent){
$items = $page;
} // END if
} // END foreach
if(is_array($items)){ // If a sub
echo '<ul id="sub_menu" class="sub_navagation'. $item['id'] .'">';
foreach($items as $item){
echo '<li>'.$item['name'].'<li>';
} // END foreach
echo '</ul>';
} // END if
}
$sql = "SELECT * FROM pages ORDER by item_order";
$result = mysqli_query($db, $sql);
confirm_query($result);
while ($row = mysqli_fetch_assoc($result)) {
$pages[] = $row; // Add each row to $pages array to use later
}
foreach($pages as $key => $page){
if($page['parent'] == 'none'){ ?>
<li id = "<?php echo $page['id']; ?>">
<a href="page.php?id=<?php echo $page['id']; ?>" title="<?php echo $page['page_title']; ?>">
<?php echo $page['page_name']; ?>
</a>
<?php
build_dropdown($page['page_name'], $pages); // If there are child items then build them out
?>
</li>
<?php
} // END if
} // END foreach
?>
I suggest you will need to JOIN your table to basically query it again to get the parent value, and add that to your markup.
SELECT *
FROM Pages
LEFT JOIN Pages p2 on page_name = p2.parent
(note: the syntax above may not be right, but I wanted to give you an idea of where I would start).

Hierarchical tree menu - PHP / MySQL

In working on a dynamic menu w/ CRUD I have been having some trouble with the finishing touches. It works, but there are some extra tags getting inserted where there shouldn't be and can't figure out how to clean it up before I share it with the world. I used this as a starting point, then changed it to work with an accordion menu (http://www.phpro.org/tutorials/Simple-Mysql-PHP-Menu.html).
Below is the data from the table (I changed the names on the first 2 fields to get it to fit in the SO format from menu_item_id to id, and from menu_parent_id to pid).
id pid menu_item_name menu_url sortorder status
1 0 Settings 0 ACTIVE
2 5 Grid Demo grid.php ACTIVE
3 5 setGridOptions gridoptions.php ACTIVE
4 1 Menu Items adminmenu.php 1 ACTIVE
5 0 Grid Settings 100 ACTIVE
6 1 General Settings settings.php 100 ACTIVE
Here is the PHP that connects to the mysql database and creates the hierarchical tree array to make it work:
include 'db.php';
$sql = "SELECT * FROM menu_items WHERE status = \"ACTIVE\" ORDER BY sortorder, menu_item_name";
$query = $db->query($sql);
while($data = $query->fetch(PDO::FETCH_ASSOC))
// loop over the results
{
// Assign by reference
$thisref = &$refs[ $data['menu_item_id'] ];
// add the the menu parent
$thisref['menu_item_id'] = $data['menu_item_id'];
$thisref['menu_parent_id'] = $data['menu_parent_id'];
$thisref['menu_item_name'] = $data['menu_item_name'];
$thisref['menu_url'] = $data['menu_url'];
// if there is no parent id
if ($data['menu_parent_id'] == 0)
{
$list[ $data['menu_item_id'] ] = &$thisref;
}
else
{
$refs[ $data['menu_parent_id'] ]['children'][ $data['menu_item_id'] ] = &$thisref;
}
}
function create_list( $arr )
{
$html = "";
foreach ($arr as $key=>$v)
{
if ($v['menu_parent_id'] == '0')
{
$html .= '<a class="menuitem submenuheader" href="'. $v['menu_url'] .'">'.$v['menu_item_name']."</a>\n";
$html .= "<div class=\"submenu\">\n<ul>\n";
$html .= "<li>" . create_list($v['children']) . "</li>";
$html .= "</ul>\n";
}
else{
$html .= '<li><a id="' . $v['menu_item_id'] . '">'.$v['menu_item_name']."</a></li>\n";
}
}
$html .= "</div>\n";
return $html;
}
echo "<div class=\"glossymenu\">";
echo create_list( $list );
echo "</div>";
When I run it, it outputs the following:
<div class="glossymenu"><a class="menuitem submenuheader">Settings</a>
<div class="submenu">
<ul>
<li><li><a id="4">Menu Items</a></li>
<li><a id="6">General Settings</a></li>
</div>
</li></ul>
<a class="menuitem submenuheader" href="">Grid Settings</a>
<div class="submenu">
<ul>
<li><li><a id="2">Grid Demo</a></li>
<li><a id="3">setGridOptions</a></li>
</div>
</li></ul>
</div>
</div>
As you can see there are extra <li> tags, the </ul> is in the wrong
spot (should be after the </div>) Other than that, it is working great.
The other thing I can't figure out is if I have a root menu item with no children, I would love it to have a different output like
<a id="8">No Children Menu Item</a>
Instead of:
<a class="menuitem submenuheader">No Children Menu Item</a>
The second example would create the make it show up the little (+/-) for expanding and contracting and wouldn't allow me to click on it. For clarification on the <a> tags, I am using javascript to do a .get() based off the id which is why there is no href or url shown.
UPDATE
It is working correctly and I posted it on Github for anyone that wants it.
https://github.com/ajhalls/php-accordian-menu
Try this :
<?php
include 'db.php';
$sql = "SELECT * FROM menu_items WHERE status = 'ACTIVE' ORDER BY pid ASC, sortorder ASC, menu_item_name ASC";
$query = $db->query($sql);
$menu_items = array();
while($data = $query->fetch(PDO::FETCH_ASSOC)) {
if($data['pid'] == 0) {
$menu_items[$data['id']] = array();
$menu_items[$data['id']]['id'] = $data['id'];
$menu_items[$data['id']]['name'] = $data['menu_item_name'];
$menu_items[$data['id']]['url'] = $data['menu_url'];
$menu_items[$data['id']]['children'] = array();
} else if($data['pid'] != 0) {
$tmp = array();
$tmp['id'] = $data['id'];
$tmp['name'] = $data['menu_item_name'];
$tmp['url'] = $data['menu_url'];
array_push($menu_items[$data['pid']]['children'],$tmp);
unset($tmp);
}
}
function create_list($arr)
{
$html = "";
foreach($arr as $key => $value) {
if(count($value['children']) > 0) {
$html .= ' <a class="menuitem submenuheader" href="'. $value['url'] .'">'.$value['name'].'</a>
<div class="submenu">
<ul>';
foreach($value['children'] AS $child) {
$html .= ' <li>
<a id="'.$child['id'].'">'.$child['name'].'</a>
</li>';
}
$html .= ' </ul>
</div>';
} else{
$html .= ' <a id="'.$value['id'].'">'.$value['name'].'</a>';
}
}
return $html;
}
echo "<div class=\"glossymenu\">";
echo create_list($menu_items);
echo "</div>";
?>

How to create a nested menu from MySQL with PHP?

I need to create a menu with PHP from a MySQL database.
Table called categories has id, name, parent_id, shortdesc, etc.
The output need to have parent list and children list under the partent list as follows.
If you can show me codes or website, I will appreciate it.
<ul id="catmenu">
<li class="menulist">Cars
<ul>
<li>Ford</li>
<li>Honda</li>
<li>Toyota</li>
</ul>
</li>
<li class="menulist">Food
<ul>
<li>Pasta</li>
<li>Pizza</li>
...
</ul>
</li>
...
...
</ul>
This is specifically for two levels deep. Recommended approach should it be more is to use an optimized table structure for traversal, like http://articles.sitepoint.com/article/hierarchical-data-database/2 (pointed out elsewhere) or to pull the data you need and push it into a dictionary (associative array) and query it that way.
<?php
$query = <<<EOT
SELECT
parent.name as parent_name,
child.name as child_name,
FROM
items child
INNER JOIN
items parent
ON
child.parent_id = parent.id
ORDER BY
parent.name
EOT;
$result = mysql_query($query) or die('Failure!');
echo "<ul id=\"catmenu\">";
$last_parent = '';
while($row = mysql_fetch_array($result)){
// If this is a new category, start a new one
if($last_parent != $row['parent_name']){
// Unless this is the first item, close the last category
if($last_parent != ''){
echo "</ul></li>";
}
$last_parent = $row['parent_name'];
echo "<li class=\"menulist\">{$row['parent_name']}<ul>";
}
echo "<li>{$row['child_name']}</li>";
}
// If we actually had items, close the "category"
if($last_parent != ''){
echo "</ul></li>";
}
echo "</ul>";
?>
If you have only two levels then, you could just display them :
echo '<ul id="catmenu">';
foreach($menu as $element) {
echo '<li><ul class="menulist">';
foreach($element['submenu'] as $submenu) {
echo '<li>' . $submenu['name'] . '</li>';
}
echo '</ul></li>';
}
echo '</ul>
If you have an undefined number of submenus however you should use a Recursive Function.
function menu($item) {
$ret = '<li>' . $item['name'];
if (!empty($item['submenu'])) {
foreach($item['submenu'] as $submenu) {
$ret .= menu($submenu);
}
}
return $ret;
}
echo menu($menu);
So every of your subcategories, whatever their number is will be displayed.
You make database like this.
ID NAME PARENT
0 Cars -1
1 Foods -1
2 Ford 0
3 Honda 0
4 Toyota 0
5 Pasta 1
6 Pizza 1
...
You query them all up and put it in an array.
$Menus = array();
// In a read MySQL loop
$Menus[$i]['ID']
$Menus[$i]['NAME']
$Menus[$i]['PARENT']
// Sorry, lazy to write. I think you know what I mean.
Then you loop all menu looking for PARENT == -1. Generate all UL and IL then sub it with another nested menu.
You can simply create a function like this.
var $MenuLevelClass = array("menulist");
function CreateMenu($Menus, $Tab = 0, $Parent = -1, $Level = 0) {
global $MenuLevelClass;
$CatClass = ($Level != 0) ? '' : ' class="catmenu"';
$MenuClass = $MenuLevelClass[$Level];
if ($MenuClass != '')
$MenuClass = ' class="'.$MenuClass.'"';
$TabCount = $Level + $Tab;
$TabUL = "";
for ($t = 0; $t < $TabCount; $t++)
$TabUL = $TabUL."\t";
$TabLI = $TabUL."\t";
?>
<?=$TabUL?><ul<?=$CatClass?>>
<?php
$MenuCount = count($Menus);
for ($m = 0; $m < $MenuCount; $m++) {
$Menu = $Menu[$m];
$ID = $Menu['ID'];
if ($ID != $Parent)
continue;
?>
<?=$TabLI?><li<?=$MenuClass?>><?=$Menu['Name']?><?=CreateMenu($Menus, $Tab + 1, $ID, $Level + 1)?></li>
<?php
?>
<?=$TabUL?></ul>
<?php
}
}
And to use it just run 'CreateMenu($Menus);' or 'CreateMenu($Menus, $PrefixTabCount);'.
CreateMenu will recursively create the nested menu for you.
I have not test it so you may have to adjust it.
Hope this helps.

Categories