So I'm utilizing a for each to try to generate a menu dynamically from a MySQL database. Because there's more than one it will always return an array. I use multiple files to generate the menu.
I use a class to create the menu
class menu extends db{
public function LoadMainMenu() {
global $db;
$query = <<<SQL
SELECT id,name
FROM menu
WHERE enabled = :active
AND location = :mainmenu
SQL;
$resource = $db->sitedb->prepare( $query );
$resource->execute( array(
':active' => '1',
':mainmenu' => '1',
));
foreach($resource as $row){
echo '<li>'.$row['name'].'</li>';
}
}
$menu = new menu();
My next file is my base.class.php file
function LoadMainMenu() {
global $menu;
$menu->LoadMainMenu();
}
Then I have my index.php file within my theme settings where I have it called
<ul class='topmenu'>
<?php LoadMainMenu(); ?>
</ul>
It then gives me the error that it's a call to a member function on an array on base.class.php; If anymore code would help please let me know.
Not Really an end all cure all, but it works. I added it within the db.class.php and then just added global $db and then called $db->GetMainMenu() within the main index.php Template and it seems to have worked. Other than that I'm not sure what else to do.
Related
I'm having a problem with some PHP / MySQL code.
I need a view called gameview for a Star Wars game I'm writing.
If I created the view in MySQL then the code runs perfectly. However, I need this view to be dropped every time the game starts. So if I start without the view "gameview" present in the DB, the page cannot be displayed due to the view not existing. However, the moment I manually add the view into MySQL, it works. I can't see why.
Class code
<?php
class gameView
{
protected $Conn;
public function __construct($Conn)
{
$this->Conn = $Conn;
}
public function dropGameView()
{
$drop = "DROP VIEW if EXISTS gameview;";
$stmt = $this->Conn->prepare($drop);
$stmt->execute(array());
}
public function createGameView()
{
$view = "CREATE VIEW gameview AS SELECT id, name, image, quote FROM person;";
$stmt = $this->Conn->prepare($view);
$stmt->execute(array());
}
public function useGameView()
{
$query = "SELECT * from gameview";
$stmt = $this->Conn->prepare($query);
$stmt->execute(array());
$gameView = $stmt->fetchAll(PDO::FETCH_ASSOC);
return $gameView;
}
}
?>
PHP code
<?php
$gameView = new gameView($Conn);
$finalCharacter = $gameView->useGameView();
$smarty->assign('game_view', $finalCharacter);
?>
Well.... stone the crows. I thought this would be too simple to work, but it did!
<?php
$gameView = new gameView($Conn);
$dropGameView = $gameView->dropGameView();
$smarty->assign('drop_gameview', $dropGameView);
$createGameView = $gameView->createGameView();
$smarty->assign('create_gameview', $createGameView);
$finalCharacter = $gameView->useGameView();
$smarty->assign('game_view', $finalCharacter);
?>
Now to crack on and use the view.
i'm working on a simple plugin to count FB shares of post and
i want to create instance of the plugin class using a cron job.
the main part of the plugin is this:
class PostShareCount
{
public function __construct()
{
global $wpdb;
$this->db = $wpdb;
add_action('hourly_event', 'updateAllPostsShares');
add_shortcode('social-count', array($this, 'social_shares'));
}
public function updateAllPostsShares()
{
$this->db->query('CREATE TABLE IF NOT EXISTS wp_facebook_shares(
post_id INT UNSIGNED AUTO_INCREMENT PRIMARY KEY,
post_shares INT NOT NULL,
post_url VARCHAR(255) NOT NULL
)');
$posts = $this->db->get_results('SELECT ID, post_date, post_name FROM '.$this->db->posts.' WHERE ID < 3');
$fb_graph = 'http://graph.facebook.com/?id=';
$site = 'http://www.example.com/';
$posts_shares = [];
foreach ($posts as $post) {
$post_url = $site.$post->post_name;
$posts_shares[$post->ID] = array();
$posts_shares[$post->ID]['post_id'] = $post->ID;
$posts_shares[$post->ID]['post_url'] = $post_url;
$api_call = $fb_graph.$site.$post->post_name;
if (isset($this->get_response_body($api_call)->shares)) {
$posts_shares[$post->ID]['post_shares'] = $this->get_response_body($api_call)->shares;
} else {
$posts_shares[$post->ID]['post_shares'] = rand(80, 1200);
}
$this->db->replace('wp_facebook_shares', $posts_shares[$post->ID], array('%d', '%s', '%d'));
}
return $posts_shares;
}
}
and for a test to my cron i created this simple file:
<?php
require_once ('post-share-count.php');
$obj = new PostShareCount();
$obj->updateAllPostsShares();
?>
but whenever i try to run it i get this error:
Call to undefined function add_action() in ..
any idea what is causing this and how can i fix it? thx
There is simpler way, how to test your cronjobs:
Install WP Crontrol plugin.
Now, you can run your crons from administration.
Go to WordPress Administration: Tools -> Cron Events
If your plugin is working properly, you should see your scheduled hook name here.
Click on Run Now
Calling function directly from PHP file
If you need for some reason to call plugin functions directly from testing PHP file, don't forget to include wp-load.php.
Example (file is in same directory as wp-load.php):
<?php
include ('wp-load.php');
$obj = new PostShareCount();
$obj->updateAllPostsShares();
?>
I have my main (user visible) file which displays posts, and I need to set-up pagination.
It would be easy if I fetch DB in the same file (but I want to avoid that), that is why I created a seperate (user hidden) file which contains class' which are then called from main file(blog.php):
BLOG.php(simplified):
<?php
require 'core.php';
$posts_b = new Posts_b();
$posts_bx = $posts_b->fetchPosts_b();
foreach($posts_hx as $posts_hy){
echo $posts_hy['title'];
}
?>
core.php(simplified);
class Posts_b extends Core {
public function fetchPosts_b(){
$this->query ("SELECT posts_id, title FROM posts");
//return
return $this->rows();
}
}
This works like a charm, but now I need to do the count within query, which works fine, and which gives me a variable $pages=5 (handled inside class posts_b - in file core.php),
core.php(simplified-with variable);
class Posts_b extends Core {
public function fetchPosts_b(){
$this->query ("SELECT posts_id, title FROM posts");
$pages=5;
//return
return $this->rows();
}
}
Now I need a way to return this variable value to blog.php (the way I return rows())
Please help, anyone,
Thank you...
A function can only have a single return value.
There are ways to get around this though. You can make your return value be an array that contains all of the values you want. For example:
return array("pages"=>$pages, "rows"=>$this->rows());
Then in your code
require 'core.php';
$posts_b = new Posts_b();
$posts_bx = $posts_b->fetchPosts_b();
$pages = $posts_bx["pages"];
foreach($posts_hx["rows"] as $posts_hy){
echo $posts_hy['title'];
}
?>
Or you can adjust a input parameter provided it was supplied as a reference
public function fetchPosts_b(&$numRows){
$this->query ("SELECT posts_id, title FROM posts");
//return
return $this->rows();
}
In your code
require 'core.php';
$posts_b = new Posts_b();
$pages = 0;
$posts_bx = $posts_b->fetchPosts_b(&$pages);
foreach($posts_hx["rows"] as $posts_hy){
echo $posts_hy['title'];
}
?>
Or you can opt to figure out your pagination outside of the fetchPosts_b method.
$posts_bx = $posts_b->fetchPosts_b();
$pages = floor(count($posts_bx)/50);
I have the following class:
<?php
class photos_profile {
// Display UnApproved Profile Photos
public $unapprovedProfilePhotosArray = array();
public function displayUnapprovedProfilePhotos() {
$users = new database('users');
$sql='SELECT userid,profile_domainname,photo_name FROM login WHERE photo_verified=0 AND photo_name IS NOT NULL LIMIT 100;';
$pds=$users->pdo->prepare($sql); $pds->execute(array()); $rows=$pds->fetchAll();
$unapprovedProfilePhotosArray = $rows;
echo 'inside the class now....';
foreach($rows as $row) {
echo $row['userid'];
}
}
}
I can display the data successfully from the foreach loop.
This is a class that is called as follows and want to be able to use the array in the display/view code. This why I added the "$unapprovedProfilePhotosArray = $rows;" but it doesn't work.
$photos_profile = new photos_profile;
$photos_profile->displayUnapprovedProfilePhotos();
<?php
foreach($photos_profile->unapprovedProfilePhotosArray as $row) {
//print_r($photos_profile->unapprovedProfilePhotosArray);
echo $row['userid'];
}
?>
What is the best way for me to take the PHP PDO return array and use it in a view (return from class object). I could loop through all the values and populate a new array but this seems excessive.
Let me know if I should explain this better.
thx
I think you're missing the $this-> part. So basically you're creating a local variable inside the method named unapprovedProfilePhotosArray which disappears when the method finishes. If you want that array to stay in the property, then you should use $this->, which is the proper way to access that property.
...
$pds=$users->pdo->prepare($sql); $pds->execute(array()); $rows=$pds->fetchAll();
$this->unapprovedProfilePhotosArray = $rows;
...
I am using Kohana 3.2 and I am having problems calling the ouput of a controller in another controller.
What I want...
In some pages I have got a menu, and in others I don't. I want to use make use of the flexability of the HMVC request system. In the controller of a page I want to call another controller which is responsible for the creation of the menu.
What I have a the moment:
file menu.php:
<?php defined('SYSPATH') or die('No direct script access.');
class Controller_Menu extends Controller
{
private $_model = null;
public function __construct(Request $request, Response $response)
{
parent::__construct($request, $response);
$this->_model = Model::factory('menu');
}
public function action_getMenu()
{
$content = array();
$content['menuItems'] = $this->_model->getMenuItems();
// Render and output.
$this->request->response = View::factory('blocks/menu', $content);
//echo '<pre>'; print_r($this->request->response->render()); echo '</pre>'; die();
}
}
somepage.php
public function action_index()
{
$this->template->title = 'someTitle';;
$contentData['pageTitle'] = 'someTitle';
$contentData['contentData'] = 'someData';
#include the menu
$menuBlock = Request::factory('menu/getMenu')->execute();
$menuData = array('menu' => $menuBlock);
$this->template->menu = View::factory('pages/menu')->set('menu',$menuData);
$this->template->content = View::factory('pages/somePage', $contentData);
$view = $this->response->body($this->template);
$this->response->body($view);
}
If I uncomment the following line in menu.php, I see the menu rendered:
//echo '<pre>'; print_r($this->request->response->render()); echo '</pre>'; die();
So I guess that part is alright. The problem is in the following line in somepage.php:
$menuBlock = Request::factory('menu/getMenu')->execute();
This gives me back a response object. Whatever I do, I do not get the output in $this->template->menu.
$this->template->menu = View::factory('pages/menu')->set('menu',$menuData);
What must I do to have $this->template->menu contain the view, so I can use it correctly?
I hope this all makes sense. This is the way I would like to do it, but maybe I am completely on the wrong track.
I would do it this way:
class Controller_Menu extends Controller
{
public function action_build()
{
// Load the menu view.
$view = View::factory('navigation/menu');
// Return view as response-
$this->response->body($view->render());
}
}
In your controller get the menu as follows:
// Make request and get response body.
$menu = Request::factory('menu/build')->execute()->body();
// e.g. assign menu to template sidebar.
$this->template->sidebar = Request:.factory('menu/build')->execute()->body();
I would not use the __construct method in your controllers. Use before() instead, this is sufficient for most of the problems (for example auth):
public function before()
{
// Call aprent before, must be done here.
parent::before();
// e.g. heck whether user is logged in.
if ( !Auth::instance()->logged_in() )
{
//Redirect if not logged in or something like this.
}
}
I found the answer to my problem in less than an hour after asking.
I just forgot to put it here.
In somePage.php change :
$menuBlock = Request::factory('menu/getMenu')->execute();
$menuData = array('menu' => $menuBlock);
$this->template->menu = View::factory('pages/menu')->set('menu',$menuData);
To:
$this->template->menu = Request::factory('menu/getMenuBlock')->execute()->body();
And in menu.php change:
$this->request->response = View::factory('blocks/menu', $content);
To:
$request = View::factory('blocks/menu', $content);
$this->response->body($request);
I hope this will help someone else.