WordPress and External API - php

I'm attempting to use an AJAX call to post to an external API. I primarily want to do this to protect the API Key from prying eyes, though I'm a little lost on where to go with this next.
I've defined the API Key in my WP_config.php file and am able to echo that back in a PHP file. I've created a PHP file that creates a shortcode and added it to my custom post page template. I'd like to call a PHP function to handle the AJAX request to obfuscate the API Key and return the result so I can then modify the interface to notify the user that the feedback has been received. The shortcode works well and the form is posting correctly based on the console.log.
This is my page-feedback.php plugin code at the moment but I don't know where to go next. I was investigating the WP Rest API but I don't need to leverage any the WP functionality for this. Can I reference this existing PHP file for handling posting to the API? How would I approach that?
// Initialization
defined('ABSPATH') or die('Unauthorized access.');
add_shortcode('page-feedback', 'page_feedback_shortcode');
add_action('wp_footer', 'load_scripts');
function page_feedback_shortcode()
{ ?>
<div class="page-feedback">
<form id="page-feedback-form__form">
<input name="page" type="hidden" value="<?php echo urlencode(get_the_permalink()) ?>" />
<input name="username" type="hidden" value="<?php echo wp_get_current_user()->user_login ?>" />
<input id="page-feedback-form-helpful" name="helpful" type="hidden" />
<p id="page-feedback-form-helpfulQuestion">Was this page helpful? <button type="submit">Yes</button> or <button type="submit">No</button></p>
</form>
</div>
<?php }
function load_scripts()
{ ?>
<script>
jQuery('#page-feedback-form__form').submit(function(event) {
event.preventDefault();
var form = jQuery(this).serialize();
console.log(form);
jQuery.ajax({
method: 'post',
url: '',
data: form,
success: function(response) {
console.log(response);
}
})
});
</script>
<?php }
?>

Related

Ajax - page reloading on submit with jquery

I have a simple webstore and I'm trying to add multiple shipping options. There are two options and I want to store the option selected by the customer in a session (or alternatively a cookie).
The php script seems to work on its own but results in page reloading so I have been trying to implement ajax using jquery to send the data but either the code doesn't run or the page reloads.
I have tried to follow several answers, including substituting button for type="submit" but that results in the code not seeming to execute at all. Adding 'click' to .on triggers the code but also results in a reload. I've checked the console and can't see any issues. Thanks for any help.
jQuery
$(function(){
$('#shipping_form').on('submit', function(e){
e.preventDefault();
var ship = $('input[name="shipping_val"]').val();
$.ajax({
type: 'GET',
data: { discount: ship },
success: function(data) {
//go to next section
}
});
return false;
});
});
HTML/PHP
<?php
Session_start();
if(isset($_GET['shipping_submit'])){
$shipping_get = $_GET['shipping_val'];
}else{
$shipping_get = '3.99';
}
$_SESSION['shipping'] = $shipping_get ;
?>
<html>
<main class="container">
<form method="GET" name="shipping_form" id="shipping_form" action="">
<p>Please choose your prefered shipping method.</p>
<input type="radio" name="shipping_val" value="3.99" checked>
<label for="shipping_val">
Standard delivery
</label>
<input type="radio" name="shipping_val" value="6.99" >
<label for="shipping_val">
Express delivery
</label>
<button type="submit" id="shipping_submit" name="shipping_submit" >Update</button>
<?php
echo '<h1>' . $_SESSION['shipping'] . '</h1>';
?>
</form>```
Your problem is likely that you are using AJAX to submit data to the same file you are calling it from. Your ajax end point (PHP-side) needs to be a different PHP file.
See this other question:
values not updated after an ajax response

Hiding a form upon click of the submission button

<?php
'<form method="post" action="postnotice.php");>
<p> <label for="idCode">ID Code (required): </label>
<input type="text" name="idCode" id="idCode"></p>
<p> <input type="submit" value="Post Notice"></p>
</form>'
?>
Alright, so that's part of my php form - very simple. For my second form (postnotice.php):
<?php
//Some other code containing the password..etc for the connection to the database.
$conn = #mysqli_connect($sql_host,$sql_user,$sql_pass,$sql_db);
if (!$conn) {
echo "<font color='red'>Database connection failure</font><br>";
}else{
//Code to verify my form data/add it to the database.
}
?>
I was wondering if you guys know of a simple way - I'm still quite new to php - that I could use to perhaps hide the form and replace it with a simple text "Attempting to connect to database" until the form hears back from the database and proceeds to the next page where I have other code to show the result of the query and verification of "idCode" validity. Or even database connection failure. I feel it wrong to leave a user sitting there unsure if his/her button click was successful while it tries to connect to the database, or waits for time out.
Thanks for any ideas in advance,
Luke.
Edit: To clarify what I'm after here was a php solution - without the use of ajax or javascript (I've seen methods using these already online, so I'm trying to look for additional routes)
what you need to do is give form a div and then simply submit the form through ajax and then hide the div and show the message after you get the data from server.
<div id = "form_div">
'<form method="post" id = "form" action="postnotice.php";>
<p> <label for="idCode">ID Code (required): </label>
<input type="text" name="idCode" id="idCode"></p>
<p> <input type="submit" value="Post Notice"></p>
</form>'
?>
</div>
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
<script>
$(function () {
$('form').on('submit', function (e) {
$.ajax({
type: 'post',
url: 'postnotice.php',
data: $('form').serialize(),
success: function (data) {
if(data == 'success'){
echo "success message";
//hide the div
$('#form_div').hide(); //or $('#form').hide();
}
}
});
e.preventDefault();
});
});
</script>
postnotice.php
$idCode = $_POST['idCode'];
// then do whatever you want to do, for example if you want to insert it into db
if(saveSuccessfulIntoDb){
echo 'success';
}
try using AJAX. this will allow you to wait for a response from the server and you can choose what you want to do based on the reponse you got.
http://www.w3schools.com/ajax/default.ASP
AJAX with jQuery:
https://api.jquery.com/jQuery.ajax/

sending form data to php using ajax

I Have an requirement to pass form data to php using ajax and implement it in php to calculate the sum , division and other arithmetic methods I am a new to ajax calls trying to learn but getting many doubts....
It would be great help if some one helps me out with this
index.html
<script type="text/javascript">
$(document).ready(function(){
$("#submit_btn").click(function() {
$.ajax({
url: 'count.php',
data: data,
type: 'POST',
processData: false,
contentType: false,
success: function (data) {
alert('data');
}
})
});
</script>
</head>
<form name="contact" id="form" method="post" action="">
<label for="FNO">Enter First no:</label>
<input type="text" name="FNO" id="FNO" value="" />
label for="SNO">SNO:</label>
<input type="text" name="SNO" id="SNO" value="" />
<input type="submit" name="submit" class="button" id="submit_btn" value="Send" />
</form>
In count.php i want to implement
<?php
$FNO = ($_POST['FNO']);
$SNO=($_post['SNO']);
$output=$FNO+$SNO;
echo $output;
?>
(i want to display output in count.php page not in the first page index.html)
Thanks for your help in advance.
You can use a simple .post with AJAX. Take a look at the following code to be able to acheive this:
$('#form').submit(function() {
alert($(this).serialize()); // check to show that all form data is being submitted
$.post("count.php",$(this).serialize(),function(data){
alert(data); //check to show that the calculation was successful
});
return false; // return false to stop the page submitting. You could have the form action set to the same PHP page so if people dont have JS on they can still use the form
});
This sends all of your form variables to count.php in a serialized array. This code works if you want to display your results on the index.html.
I saw at the very bottom of your question that you want to show the count on count.php. Well you probably know that you can simply put count.php into your form action page and this wouldn't require AJAX. If you really want to use jQuery to submit your form you can do the following but you'll need to specify a value in the action field of your form:
$("#submit_btn").click(function() {
$("#form").submit();
});
I have modified your PHP code as you made some mistakes there. For the javscript code, i have written completely new code for you.
Index.html
<html>
<head>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js"></script>
</head>
<body>
<form name="contact" id="contactForm" method="post" action="count.php">
<label for="FNO">Enter First no:</label>
<input type="text" name="FNO" id="FNO" value="" />
<label for="SNO">SNO:</label>
<input type="text" name="SNO" id="SNO" value="" />
<input type="submit" name="submit" class="button" id="submit_btn" value="Send" />
</form>
<!-- The following div will use to display data from server -->
<div id="result"></div>
</body>
<script>
/* attach a submit handler to the form */
$("#contactForm").submit(function(event) {
/* stop form from submitting normally */
event.preventDefault();
/* get some values from elements on the page: */
var $form = $( this ),
//Get the first value
value1 = $form.find( 'input[name="SNO"]' ).val(),
//get second value
value2 = $form.find( 'input[name="FNO"]' ).val(),
//get the url. action="count.php"
url = $form.attr( 'action' );
/* Send the data using post */
var posting = $.post( url, { SNO: value1, FNO: value2 } );
/* Put the results in a div */
posting.done(function( data ) {
$( "#result" ).empty().append( data );
});
});
</script>
</html>
count.php
<?php
$FNO = $_POST['FNO'];
$SNO= $_POST['SNO'];
$output = $FNO + $SNO;
echo $output;
?>
There are a few things wrong with your code; from details to actual errors.
If we take a look at the Javascript then it just does not work. You use the variable data without ever setting it. You need to open the browser's Javascript console to see errors. Google it.
Also, the javascript is more complicated than is necessary. Ajax requests are kind-of special, whereas in this example you just need to set two POST variables. The jQuery.post() method will do that for you with less code:
<script type="text/javascript">
$(document).ready(function(){
$("#form").on("submit", function () {
$.post("/count.php", $(this).serialize(), function (data) {
alert(data);
}, "text");
return false;
});
});
</script>
As for the HTML, it is okay, but I would suggest that naming (i.e. name="") the input fields using actual and simple words, as opposed to abbreviations, will serve you better in the long run.
<form method="post" action="/count.php" id="form">
<label for="number1">Enter First no:</label>
<input type="number" name="number1" id="number1">
<label for="number2">Enter Second no:</label>
<input type="number" name="number2" id="number2">
<input type="submit" value="Calculate">
</form>
The PHP, as with the Javascript, just does not work. PHP, like most programming languages, are very picky about variables names. In other words, $_POST and $_post are not the same variable! In PHP you need to use $_POST to access POST variables.
Also, you should never trust data that you have no control over, which basically means anything that comes from the outside. Your PHP code, while it probably would not do much harm (aside from showing where the file is located on the file system, if errors are enabled), should sanitize and validate the POST variables. This can be done using the filter_input function.
<?php
$number1 = filter_input(INPUT_POST, 'number1', FILTER_SANITIZE_NUMBER_INT);
$number2 = filter_input(INPUT_POST, 'number2', FILTER_SANITIZE_NUMBER_INT);
if ( ! ctype_digit($number1) || ! ctype_digit($number2)) {
echo 'Error';
} else {
echo ($number1 + $number2);
}
Overall, I would say that you need to be more careful about how you write your code. Small errors, such as in your code, can cause everything to collapse. Figure out how to detect errors (in jQuery you need to use a console, in PHP you need to turn on error messages, and in HTML you need to use a validator).
You can do like below to pass form data in ajax call.
var formData = $('#client-form').serialize();
$.ajax({
url: 'www.xyz.com/index.php?' + formData,
type: 'POST',
data:{
},
success: function(data){},
error: function(data){},
})

Codeigniter, getting ajax to run on button press

My original goal was to get a php file to execute on a button press. I used ajax. When the javascript was in the view, it worked.
However, I tried to switch the javascript to its own .js file and include it in the header. It doesn't work anymore. I am confused.
the model code:
public function insert_build($user_id)
{
$query = "INSERT INTO user_structure (str_id, user_id) VALUES ('7', '$user_id')";
mysql_query($query) or die ('Error updating database');
}
Something interesting to note here is that when I include $user_id as a value, it completely negates my headertemplate. As in, it simply doesnt load. When I replace $user_id with a static value (i.e. '7') it works no problem.
This is my view code :
<div id="structures">
<h1>Build</h1>
<form name="buildForm" id="buildForm" method="POST">
<select name="buildID" class="buildClass">
<option value="0" selected="selected" data-skip="1">Build a Structure</option>
<?php foreach ($structures as $structure_info): ?>
<option name='<?php echo $structure_info['str_name'] ?>' value='<?php echo $structure_info['str_id'] ?>' data-icon='<?php echo $structure_info['str_imageloc'] ?>' data-html-text='<?php echo $structure_info['str_name'] ?><i>
<?php echo $structure_info['timebuildmins'] ?> minutes<br><?php echo $structure_info['buy_gold'] ?> gold</i>'><?php echo $structure_info['str_name'] ?></option>
<?php endforeach ?>
</select>
<div id="buildSubmit">
<input id ="btnSubmit" class="button" type="submit" value="Submit"/>
</div>
</form>
</div>
Heres my .js file :
$(".button").click(function(e){
e.preventDefault();
$.ajax({
type: "POST",
url: "<?php $this->structure_model->insert_build($user_id) ?>", //the script to call to get data
data: "", //you can insert url arguments here to pass to api.php
//for example "id=5&parent=6"
dataType: 'json', //data format
success: function(data) //on receive of reply
{
alert("success!");
}
});
});
I am almost sure I know the problem: That structure_model->insert_build($user_id) ?> doesn't work when its outside the view. Though, I dont know the alternative.
I excluded the header file. I confirmed that the .js file is indeed being directed to the correct path.
Could someone please explain the correct way to do this? Thank you!
Did you move your javascript to a .js that is being directly accessed by the browser? I.E: If you view source, so you see the <?php ... ?> in the javascript code?
To me, it sounds as though the PHP is not getting parsed. If this is not the case, then can you please clarify.
If you need to include PHP variables in your javascript, you should use CI to generate the JS page for inclusion. You can even create a View that is purely JS and call it like a normal page.
Otherwise, if you want to seperate the JS from CI, you should reference JS variables instead of PHP. Then in your CI page somewhere, define them with a <script>var jsVar = <?php echo phpvar(); ?></script> tag.
When you move the js file to it's own file, php variables will not be accessible anymore. You can either move the js code back to your view file, or fetch the url through javascript. See below for example.
HTML:
<div id="structures">
<h1>Build</h1>
<form name="buildForm" id="buildForm" method="POST">
<input type="hidden" name="url" value="<?php $this->structure_model->insert_build($user_id) ?>" />
<!-- Rest of your code -->
</form>
</div>
Javascript:
$(".button").click(function(e){
var form_url = $(this).closest('form').find('input[name=url]').val();
e.preventDefault();
$.ajax({
type: "POST",
url: form_url, //the script to call to get data
data: "", //you can insert url arguments here to pass to api.php
//for example "id=5&parent=6"
dataType: 'json', //data format
success: function(data) //on receive of reply
{
alert("success!");
}
});
});

JQuery to Reload DIV Layer with PHP GET from Text Box

Greetings from some noob trying to learn JQuery,
I am attempting to make it when you type something in a box below a div layer it reloads that layer upon submission of the form with a php get of the text box in the form. Expected behavior is it would reload that box, actual behavior is it don't do anything. Can someone help me out here.... Below is the code.
<div id="currentwxdiv">This is where the new stuff happens
</div>
<form name="changewx" action="/">
<input type="text" id="city">
<input type="submit" name="submit" class="button" id="submit_btn" value="New City" />
</form>
<script>
/* attach a submit handler to the form */
$('form[name="changewx"]').submit(function(event) {
/* get some values from elements on the page: */
var $form = $( this ),
city = $('#city').val()
/* Send the data using post and put the results in a div */
$('#currentwxdiv').load('http://api.mesodiscussion.com/?location=' + city);
return false;
});
</script>
Its giving the Javascript Console Error Error....
"XMLHttpRequest cannot load http://api.mesodiscussion.com/?location=goodjob. Origin http://weatherofoss.com is not allowed by Access-Control-Allow-Origin."
You are using POST method? is impossible to post to an external url because with ajax, the url fails the "Same Origin POlice".
If you use GET method, is possible to do that.
Another solution is to make a proxy. A little script that recive the params and then... using CURL or another thing you have to post to the external URL... finally, you jquery have to do the post thing to the proxy:
For example:
$.ajax({
url: '/proxy.php?location=' + city,
success: function(data) {
$('#currentwxdiv').html(data);
}
});
I do it so:
<div id="currentwxdiv">This is where the new stuff happens
</div>
<form name="changewx" action="/">
<input type="text" id="city">
</form>
<script>
$('#city').keyup(function() {
var city = $('#city').val()
$.ajax({
url: 'http://api.mesodiscussion.com/?location=' + city,
success: function(data) {
$('#currentwxdiv').html(data);
}
});
});
</script>
To help you out, i need to test this.
What's the url address of your html code working ?
http://api.mesodiscussion.com/?location= doesn't work... only list the directory content... maybe that's de problem?
Greatings.

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