Wordpress plugin: Working with REST api - rest

Before I start, I want to tell you guys that I have no experience in anything Wordpress related. I do have worked in PHP and Codeigniter before.
User Case
The user enters a preferred feature in a form (Air conditioning, etc.).
When the user submits the form, the feature will be send to a REST API.
The results of the REST API ( a list of cars with AC ) will be shown on the page.
It should roughly look something like this.
What i have so far
An empty plugin that is shown in the Wordpress admin panel.
Question(s)
How do create the front-end for this plugin?
How and where do I create the form action?
How do I access the form action?
What I have/know so far
I know there are some action hooks that will place your content in the header and footer by creating something like:
<php add_action('wp_footer', 'mp_footer'); ?>

In your empty plugins php file place this:
function get_search_results(){
$args = array( 's' => $_GET['term'] );
$query = new WP_Query( $args );
if ( $query->have_posts() ) {
while ( $query->have_posts() ) {
$query->the_post();
echo '<li>'.get_the_title().'</li>';
}
} else {
echo "Nothing Found";
}
die();
}
add_action( 'wp_ajax_get_search', 'get_search_results' );
add_action( 'wp_ajax_nopriv_get_search', 'get_search_results' );
function get_searchbox( $atts ){
ob_start(); ?>
<form id="searchform">
<input type="text" id="searchbox">
<button type="submit">Search</button>
</form>
<ul id="search-result"></ul>
<?php
$output = ob_get_clean();
return $output;
}
add_shortcode( 'searchbox', 'get_searchbox' );
function add_search_script() {
wp_enqueue_script( 'search', plugins_url( '/search.js' , __FILE__ ), array( 'jquery' ) );
wp_localize_script( 'search', 'search_ajax', array( 'url'=>admin_url( 'admin-ajax.php' ), 'action'=>'get_search' ) );
}
add_action( 'wp_enqueue_scripts', 'add_search_script' );
In your plugin's directory create a new javascript file - search.js and place this:
jQuery(function($){
$('#searchform').submit(function(e){
e.preventDefault();
$.ajax({
url: search_ajax.url,
data: { 'action': search_ajax.action, 'term': $('#searchbox').val() }
}).done(function(r) {
$('#search-result').html(r);
});
});
});
Now you can use shortcode [searchbox] in your wordpress page to get your searchbox.
In php you can get same result with <? echo do_shortcode('[searchbox]') ?>
Explanation:
When we add [searchbox] shortcode, it is processed and replaced with a form via get_searchbox() function.
In jQuery code On form submit we are sending an action : get_search (defined in wp_localize_script).
Server receives that request via wp_ajax_get_search and processes get_search_results() and returns an output.
In jQuery done(function(r){ r is the response from server. Use that data to manipulate your html.
action is the essential part of REST api in wordpress. We need not have a url. Instead we define an action and for that action return a response.
Once you understand this, modify the action and response according to your need.
Helpful articles: Daniel's, wpmudev

Related

CSRF field is missing when I embed my form with a requestAction in CakePHP 3

I want to embed a contact form in multiple places on my website.
I developed a contact form in a contact() function within my MessagesController.php:
// MessagesController.php
public function contact()
{
$this->set('title', 'Contact');
$message = $this->Messages->newEntity();
... // shortened for brevity
$this->set(compact('message'));
$this->set('_serialize', ['message']);
}
I loaded the CSRF component in the initialize() function of the AppController.php:
// AppController.php
public function initialize()
{
parent::initialize();
$this->loadComponent('Csrf');
... // shortened for brevity
}
The form is rendered with a contact.ctp and it works fine.
I followed CakePHP's cookbook which suggests using requestAction() within an element, then echoing the element where I want it:
// contact_form.ctp
<?php
echo $this->requestAction(
['controller' => 'Messages', 'action' => 'contact']
);
?>
And:
// home.ctp
<?= $this->element('contact_form'); ?>
The problem is that the form is rendered fine, but the CSRF hidden field is missing. It should be automatically added to the form since the CSRF component is called in the AppController.php.
I guess either using an element with a requestAction() isn't the solution for this particular case, or I am doing something wrong.
Any ideas? Thanks in advance for the input!
Request parameters need to be passed manually
requestAction() uses a new \Cake\Network\Request instance, and it doesn't pass the _Token and _csrf parameters to it, so that's why things break.
While you could pass them yourself via the $extra argument, like
$this->requestAction(
['controller' => 'Messages', 'action' => 'contact'],
[
'_Token' => $this->request->param('_Token'),
'_csrf' => $this->request->param('_csrf')
]
);
Use a cell instead
I would suggest using a cell instead, which is way more lightweight than requesting an action, also it operates in the current request and thus will work with the CSRF component out of the box.
You'd pretty much just need to copy your controller action code (as far as the code is concerned that you are showing), and add a loadModel() call to load the Messages table, something like
src/View/Cell/ContactFormCell.php
namespace App\View\Cell;
use Cake\View\Cell;
class ContactFormCell extends Cell
{
public function display()
{
$this->loadModel('Messages');
$this->set('title', 'Contact');
$message = $this->Messages->newEntity();
// ... shortened for brevity
$this->set(compact('message'));
$this->set('_serialize', ['message']);
}
}
Create the form in the corresponding cell template
src/Template/Cell/ContactForm/display.ctp
<?php
echo $this->Form->create(
/* ... */,
// The URL needs to be set explicitly, as the form is being
// created in the context of the current request
['url' => ['controller' => 'Messages', 'action' => 'contact']]
);
// ...
And then wherever you want to place the form, just use <?= $this->cell('ContactForm') ?>.
See also
API > \Cake\Routing\RequestActionTrait::requestAction()
Cookbook > Views > Cells

laravel 4 changing url structure with parameter passing

I'm just a beginner in laravel framework. i created a simple form and controller methods . i think by default the form method is post. but now i need to make it as get method and also wants to pass the selected inputs to controller and shows that parameters in the url also. currently i did like below but failed.
index.blade.php
<?php echo Form::open(array('url' => 'home/find','method' => 'get')); ?>
<select class="location" id="location" name='location'>
<?php
$id1 = /* get user db details based on locations*/
$location_id_drop_down='';
foreach($idl as $lrow)
{
$city_name=Location::where('location_id','=',$lrow['loc_id'])->first();
if($city_name==array()) continue;
if(Input::old('location')==$lrow['loc_id'])
{
$location_id_drop_down.="<option value='" . $city_name['location_name'] . "' selected='selected'>" .$city_name['location_name'] . "</option>";
}
else
{
$location_id_drop_down.="<option value='" . $city_name['location_name'] . "'>" . $city_name['location_name']. "</option>";
}
}
echo $location_id_drop_down;
?>
</select>
<input type="submit" name="search" id="search_submit" class="search_submit" value="Search" />
{{ Form::close() }}
HomeController.php
public function anyFind($s='',$d='',$l='') {
if(Input::get('search'))
{
$location=Input::get('location');
}
else
{
if($l!='~')
$location=$l;
}
/* queries to get the user details and images based on selected location and list them*/
}
Routes.php
Route::get('/find/{location}','HomeController#anySearch');
But this shows the url as mysite.com/home/find?location=Test&search=Search
I need mysite.com/home/find/location
is there any mistake in my code?
Edit
As a part of experiment i tried this method. i gave a redirect like below at the end of my controller function anyfind()
return Redirect::to('/home/find/'.$location);
But this redirects me but did anyone knows how to load the search.search_new.blade.php with this custom url??
Having a field from the form in your url (not query string) is simply not possible with Laravel. What you can do though, is just use javascript for that.
First lets put a placeholder in the action url
<?php echo Form::open(array('url' => 'home/find/%location%','method' => 'get')); ?>
jQuery
$('form').on('submit', function(){ // you maybe need to be a bit more precise with the selector here
var location = $('#location').val();
$(this).prop('action', $(this).prop('action').replace('%location%', location));
});
Vanilla Javascript (if you can't / don't want to use jQuery)
document.getElementsByTagName('form')[0].onsubmit = function(e){
var location = document.getElementById('location').value;
e.target.setAttribute('action', e.target.getAttribute('action').replace('%location%', location));
};
By the way: the code in your question has still some weird stuff in there. I'm just assuming this has happened because of copy paste etc. So if it still doesn't work, make sure you post the correct code
Insert at the end of your method anyFind() inside the HomeController.php where you set the view for the page:
if (Input::get('location') != '') {
return Redirect::route('home.findByLocation',Input::get('location'));
}
return View::make('home', compact('location));
Also in your routes.php add the following code:
Route::get('home/find/{location?}', array('as'=>'home.findByLocation', 'uses'=>'HomeController#anyFind'));

woocommerce hook on page-new.php

I'm using woocommerce and mgates vendor software while adding hooks on the page-new.php page to make instructions for my vendors on the add product page. I'm using
add_action( 'edit_form_after_title', 'myprefix_edit_form_after_title' );
function myprefix_edit_form_after_title() {
echo 'This is my text!';
}
as well as after editor and form advanced
On my add product page I have:
Title
'edit_form_after_title'
Description
'edit_form_after_editor'
Product Short Description
How do I figure out what hook to put between these to sections?
Product Data
How or where would I put these hooks to have them only get them to show up on the Add Products post page and not ever post page?
You can fire your hook selectively using:
add_action( 'load-post-new.php', 'your_callback' );
And then check for the post type before adding the hook:
function your_callback()
{
global $typenow;
if( 'product' != $typenow )
return;
add_action( 'edit_form_after_title', 'myprefix_edit_form_after_title' );
}
Or, another option:
add_action( 'edit_form_after_title', 'myprefix_edit_form_after_title' );
function myprefix_edit_form_after_title()
{
global $pagenow, $typenow;
if( 'post-new.php' != $pagenow || 'product' != $typenow )
return;
echo 'This is my text!';
}

CakePHP: allowing database update with button click

I have a product search page with the form below. The search result is displayed on the same page with search bar at the top.
echo $this->Form->create('Searches', array('action'=>'products', 'type' => 'get', 'name' => 'textbox1'));
echo $form->input($varName1, array('label' => false));
echo $form->end('Locate');
I also have a little box next to the search result that allows (it doesn't work yet) the user to flag using checkboxes a product and accordingly update its database (table products and using model Product) with a button click. Note that I have a Searches controller for this search page.
<form method="link" action="/myapp/product/test_update_db>
<label><input type="checkbox" name="flag1" <?php echo $preCheckBox1; ?>>Flag 1</input></label>
<label><input type="checkbox" name="flag2" <?php echo $preCheckBox2; ?>>Flag 2</input></label>
<input type="submit" value="Update">
</form>
I'm having difficulty with this approach figuring out how to perform this check-box-and-DB-update routine. I'm getting to the link I'd like to go (/myapp/product/test_update_db), but I don't know how to take variables flag1 and flag2, along with row ID of this result ($results['Product']['id'])) to the new page.
Could someone guide me on how to perform this neatly? Is this general approach correct? If not, what route should I be taking? I'd prefer not to use javascript at this time, if possible.
EDIT: I think I can make this work if I use the URL for passing data.. but I'd still like to know how this could be done "under the hood" or in MVC. I feel like I'm hacking at the CakePHP platform.
UPDATE: So, I ended up using the URL parameters for retrieving information pieces like flag1 and flag2. I'm still looking for an alternative method.
To see where your is-checkbox-checked data is located, do the following in your controller:
// Cake 2.0+
debug($this->request->data);
// previous versions
debug($this->data);
If you want to pass data to your search controller from the current page, you can always add the data to your form:
$this->input
(
'Product.id',
array
(
'type' => 'hidden',
'value' => $yourProductId
)
);
I ended up using information embedded in the URL for getting submission data. Something like below..
In Products controller, when the form with flag1 and flag2 are submitted:
public function test_update_db() {
// Get variables from URL, if any, and save accordingly
$result = $this->Product->updateProduct($this->params['url'], 'url');
if ($result) {
$this->Session->setFlash('Successfully updated!', 'default', array('class' => 'success'));
$this->redirect($this->referer());
}
else {
$this->Session->setFlash('Update was unsuccessful!', 'default', array('class' => 'error'));
$this->redirect($this->referer());
}
}
This works for doing what I needed to do. I feel like there's a more proper way to do this though.
if ($result) {
$this->Session->setFlash('Successfully updated!', 'default', array('class' => 'success'));
$this->redirect($this->referer());
}

calling echo $this->action('panLogin','user') and $this->action('panRegister','user') on same script

I have a problem, i'm trying to render 2 forms (login and register) on one layout scrpt (header.phtml), every time i submit on one of the forms both actions for the controller are getting fired and i'm unsure how to fix it.
The forms are getting rendered fine within the layout, however when you click 'Login' or 'Register' on the forms the code fires in both the 'login' and 'register actions.
the header layout script snippet:-
<div class="left">
<h1>Already a member? <br>Then Login!</h1>
<?php
echo $this->action('panlogin', 'user');
?>
</div>
<div class="left right">
<h1>Not a member yet? <br>Get Registered!</h1>
<?php
echo $this->action('panregister', 'user');
?>
</div>
the action scripts (phtmls)
panregister.phtml
<div id="pan-register">
<?php
$this->registerForm->setAction($this->url);
echo $this->registerForm;
?>
</div>
panlogin.phtml
<div id="pan-login">
<?php
$this->loginForm->setAction($this->url);
?>
</div>
the user controller actions:-
class Ajfit_UserController extends Zend_Controller_Action
{
protected $_loginForm;
protected $_registerForm;
public function init()
{
$this->_loginForm = new Ajfit_Form_User_Login(array(
'action' => '/user/login',
'method' => 'post',
));
$this->_registerForm = new \Ajfit\Form\User\Registration(array(
'action' => '/user/register',
'method' => 'post'
));
}
//REGISTER ACTIONS
public function panregisterAction(){
$this->registerAction();
}
public function registerAction(){
$request = $this->_request;
if ($this->_request->isPost()){
$formData = $this->_request->getPost();
}
$this->view->registerForm = $this->_registerForm;
}
//LOGIN ACTIONS
public function panloginAction(){
$this->loginAction();
}
public function loginAction(){
$request = $this->_request;
if(!$auth->hasIdentity()){
if ($this->_request->isPost()){
$formData = $this->_request->getPost();
}
}
$this->view->loginForm = $this->_loginForm;
}
}
Please can someone with a little more knowlegde with the action('act','cont'); ?> code with in a layout script help me out with this problem.
Thanks
Andrew
While David is correct where best practices are concerned, I have on occasion just added another if() statement. Kinda like this:
if ($this->getRequest()->isPost()) {
if ($this->getRequest()->getPost('submit') == 'OK') {
just make sure your submit label is unique.
Eventually I'll get around to refactoring all those actions I built early in the learning process, for now though, they work.
Now to be nosy :)
I noticed: $formData = $this->_request->getPost(); while this works, if you put any filters on your forms retrieving the data in this manner bypasses your filters. To retrieve filtered values use $formData = $this->getValues();
from the ZF manual
The Request Object
GET and POST Data
Be cautious when accessing data from the request object as it is not filtered in any way. The router and
dispatcher validate and filter data for use with their tasks, but
leave the data untouched in the request object.
From Zend_Form Quickstart
Assuming your validations have passed, you can now fetch the filtered
values:
$values = $form->getValues();
Don't render the actions in your layout. Just render the forms:
<div class="left">
<h1>Already a member? <br>Then Login!</h1>
<?php
echo new \Ajfit\Form\User\Login(array(
'action' => '/user/login',
'method' => 'post'
));
?>
</div>
<div class="left right">
<h1>Not a member yet? <br>Get Registered!</h1>
<?php
echo new \Ajfit\Form\User\Registration(array(
'action' => '/user/register',
'method' => 'post'
));
?>
</div>
Then, whichever form gets used will post to its own action.