Wordpress Magic Members: Checking permissions for an individual page - php

I'm just using the Magic Members plugin, and wondering if anyone has any experience checking permissions for an individual page. I'm looping through all of the child pages of one of the main pages, but different user levels will be able to view different child pages, so I need to be able to check and display that information based on whether the user has access to that particular child.
<?php
foreach($pages as $page):
if ($has_access):
?>
content for this child page prints out to the screen!
<?php
endif;
$i++;
endforeach;
?>
How can I programmatically find out whether the user has the proper permissions to view that page ($has_access either true or false)?
Thanks!

It's not too difficult, actually. I got the answer from their custom support. Here it is, in case you are looking:
//current user
$current_user = wp_get_current_user();
//get membership type
$mgm_member = mgm_get_member($current_user->ID);
//mgm_array_dump($mgm_member); //you can uncomment this to see all of the available data for the member
$membership_type=$mgm_member->membership_type;
//then you can check for a certain user type
if ($membership_type=='client'){
//do something incredible here!
}

//current user
$current_user = wp_get_current_user();
//get membership type
$mgm_member = mgm_get_member($current_user->ID);
//mgm_array_dump($mgm_member); //you can uncomment this to see all of the available data for the member
$membership_type=$mgm_member->membership_type;
//then you can check for a certain user type
if ($membership_type=='client'){
//do something incredible here!
}
This code will only see if they were a member at one time in the life of your site. If they are payed up active member is another story.
just add this code:
//get member status
$membership_status=$mgm_member->status;
//then you can check for a certain user type
if ($membership_status=='Active'){
//do something incredible here!
}

Related

wordpress php can't return value from database

I am building a plugin that determines if a user is signed up to an affiliates program and if not, it shows a dashboard widget.
Here is the function I am using to determine if the current user is signed up to the affiliates program:
function is_affiliate() {
global $wpdb;
$user_id = get_current_user_id();
$is_affiliate = "SELECT enable_affiliate FROM {$wpdb->base_prefix}usermeta WHERE user_id = '$user_id'";
return;
}
The output should either be "yes or "no", based on the output I create a dashboard widget as follows:
if(is_affiliate() != 'yes') {
// Dashboard widget goes here
}
However, my initial function to determine if the current user is signed up is not returning anything. Is there something blatently obviously wrong with my code?
You should be using get_user_meta to retrieve that value. The WordPress API already has what you need.
To answer your question, though, you need to look at the global wpdb object which can prepare & execute queries for you.

PHP Cart with Object Oriented PHP

I am developing a simple system of sample products with Object Oriented PHP, very simple thing. So far, no problem, but I have to create a button that adds a product code recorded in the database to a form in the sidebar. I do not know to develop a shopping cart with OO PHP, and codes that I find always give error because of the call of the database, or when the data list. I've been thinking of doing for JS, any help?
sorry my bad english
I got it, first I did when I click a link step by GET and the ID Code I need after seto it in a cookie, each cookie with an ID. Then I check if cookie has registered with the IDs. Not the best and most correct, but it works (more or less). Now another problem, I need to click two times to get the result as it passes by id and need to be caught refresh = /
I think it was a bit confusing but it is the maximum that dyslexia allows me to do hehehe
Here I set up the structure with the data I have in my product page:
add
<?php
$cod_get = $_GET['cod'];
setcookie("SITENAME_cod_".$id_get."", $cod_get, time()+3600, "/","", 0);
?>
And here I have a loop checking if cookie with ids, I think it will give problems, but for now I think it works ...
Thank you all.
$produto = new produtos();
$i = 0;
$produto->selecionaTudo($produto);
$produto->selecionaCampos($produto);
while($res = $produto->retornaDados()):
$res->id;
$i++;
$get_cookie = $_COOKIE['SITENAME_cod_'.$res->id.''];
if (isset($get_cookie)) {
echo $get_cookie.', ';
}else{
echo "";
}
endwhile;

Display data if usertype is admin otherwise display error message

I currently have a list of users in my mysql database. One of the columns is "type". I am trying to display certain data if type is equal to admin. If type is equal to anything else, it should just echo an error message.
Unfortunately, I have tried multiple methods but it just does not seem to be working out for me. Can anyone help me get this to work properly?
This is what I have, but obviously I am doing something wrong....
<?php
$usertype = $_SESSION['type'];
if ($usertype == "admin" ){
?>
admin stuff only goes here
<?
}
else
{
echo "not priveleged usertype";
}
?>
EDIT:
The following code works when displaying via username, however, I need content displayed by usertype, not the username.
<?php
if($_SESSION['user']['username'] == "oneoftheadminusernames" )
{
?>
Each page has to start with
<?php
#session_start();
?>
otherwise, php does not "see" the sessions contents. So that's probably it.
The # prevents the php error: A session has already been started... by the way.
Now, every page that uses the session must have this directive at the top.
At least, in a quick example, that reproduces your error perfectly.
If you are saving each logged in users type field in $_SESSION['type'] variable than the code you are writing is correct. Or if you are storing type in another variable than you that variable to check.
i have an idea like add a field EnableFlag in the table. if enablee flag is set to 1 consider it as a admin else as a User;

Update table value on page view?

I am trying to create a PHP trigger for when a user views certain pages on my website it will update the user table in the points section.
I understand the process would work something like this
on page view > update user > where user id is (**get username from session**) > add 5 to points row
Anyone have any idea how to set up something simple like this for giving users simple points for viewing pages?
My site is using PHP and mySQL for the database.
Use cookies or session variables to keep track of the user details like the username or ID. So making a pageview trigger would be as easy as adding a mysql query at the top of every page which would update the database table for views. Kinda the same way that forums operate.
E.g
<?php
session_start();
$db_connection = mysqli_connect('host','username','password','db');
$user_id = $_SESSION['userid']; //That is asssuming that you had gotten the user id on login
mysqli_query($db_connection, 'UPDATE page_views SET views_column=views_column+1 WHERE userid=$user_id');
?>
Yes, you could do something like (if you own the page the user has to visit):
<?php
$pointsForThisSite = 5;
include "points_adder.php";
?>
While Points_adder looks whether $pointsForThisSite is defined and > 0, then adds the Points to the database as you descripbed.
Is that what you are looking for?
Create a php function and call it everytime the user enter the page.
You don't need a mysql trigger because, the action is at the webpage.
function add_points($user, $page){
//If users visits too many maybe you don't want to gave him some points.
//add points
}
and invoke the function in that pages you want to score
The most unobtrusive way to do this is with an AJAX call after the page has loaded. The call should be to an include file that performs the database update operation and returns a 204 response so that the visitor's browser doesn't wait for response content.
For an Apache server;
header('HTTP/1.0 204 No Content');
header('Content-Length: 0', true);
header('Content-Type: text/html', true);
flush();
// do the table update here

How to show a users last visited pages in wp / buddypress?

im trying to figure out how i can show the last 3-5 or so pages within my site a person has visited. I did some searching, and I couldn't find a WP plugin that does so, if anyone knows of one, please point me in that direction :) if not, I'll have to write it from scratch, and thats where i'll need the help.
I've been trying to understand the DB and how it works. I'm assuming that this is where the magic will happen, with PHP, unless there is a javascript option using cookies to do it.
Im open to all ideas :P & Thank you
If i were to code such a plugin, i'd use the session cookies to populate an array via array_unshift() and array_pop(). it'd be as simple as :
$server_url = "http://mydomain.com";
$current_url = $server_url.$_SERVER['PHP_SELF'];
$history_max_url = 5; // change to the number of urls in the history array
//Assign _SESSION array to variable, create one if empty ::: Thanks to Sold Out Activist for the explanation!
$history = (array) $_SESSION['history'];
//Add current url as the latest visit
array_unshift($history, $current_url);
//If history array is full, remove oldest entry
if (count($history) > $history_max_url) {
array_pop($history);
}
//update session variable
$_SESSION['history']=$history;
Now i've coded this on the fly. There might be syntax errors or typos. If such a mistake appears, just put a notice and i'll modify it. The purpose of this answer is mostly to make a proof of concept. You can adapt this to your liking. Please note that i assume that session_start() is already in your code.
Hope it helps.
===============
Hey! Sorry about the late answer, i was out of town for a couple of days! :)
This addon is to answer your request for a print out solution with LI tags
Here's what i'd do :
print "<ol>";
foreach($_SESSION['history'] as $line) {
print "<li>".$line.</li>";
}
print "</ol>";
Simple as that. you should read the foreach loop here : http://www.php.net/manual/en/control-structures.foreach.php
As for the session_start();, put it before you use any $_SESSION variables.
Hope it helped! :)
I'm going to update and translate the code above for WordPress 5+ because the original question has the wordpress tag. Note that you don't need session_start() anywhere.
Here it goes, add the code below to your singular.php template (or single.php + page.php templates, depending on what you need):
/**
* Store last visited ID (WordPress ID)
*/
function so7035465_store_last_id() {
global $post;
$postId = $post->ID; // or get the post ID from your template
$historyMaxUrl = 3; // number of URLs in the history array
$history = (array) $_SESSION['history'];
array_unshift($history, $postId);
if (count($history) > $historyMaxUrl) {
array_pop($history);
}
$_SESSION['history'] = $history;
}
// Display latest viewed posts (or pages) wherever you want
echo '<ul>';
foreach ($_SESSION['history'] as $lastViewedId) {
echo '<li>' . get_permalink($lastViewedId) . '</li>';
}
echo '</ul>';
You can also store latest viewed custom post types (CPT) by placing the so7035465_store_last_id() function in your single-cpt.php template.
You can also add it to a hook or inject it in your template as an action, but that is beyond the scope of this question.

Categories