We Keep Coding

Wordpress cannot get_current_user_id() from ajax request

I try to do an ajax request with wordpress. So I've created a simple js request:
$.ajax({
url: '?',
type: 'POST',
data: {
'pr_post': post,
'pr_rating': rating
},
success: function (response) {
console.log(response);
}
});
Here is my function to handle the request.
function pr_request()
{
if (isset($_REQUEST['pr_post']) && isset($_REQUEST['pr_rating']) && isset($_REQUEST['pr_user'])) {
$post = $_REQUEST['pr_post'];
$rating = ($_REQUEST['pr_rating'] > 5 ? 5 : $_REQUEST['pr_rating']);
$user = get_current_user_id();
if (!pr_has_user_already_voted($user, $post)) {
global $wpdb;
$table = $wpdb->prefix . 'mitmach_ratings';
$wpdb->query($wpdb->prepare("insert into $table values (null, $post, $rating, '$user');"));
wp_send_json(['message' => 'success']);
} else {
wp_send_json(['message' => 'duplicate'], 403);
}
}
}
As you see I call the get_current_user_id() function. This function always returns true even if the user logged in. How can I get the user id in my handler without sending it via ajax?

For a start check docs - WP Ajax.
You need to send action key
Notice how the 'action' key's value 'my_action', defined in our JavaScript above, matches the latter half of the action 'wp_ajax_my_action' in our AJAX handler below. This is because it is used to call the server side PHP function through admin-ajax.php. If an action is not specified, admin-ajax.php will exit, and return 0 in the process.
$.ajax({
url: '?',
type: 'POST',
data: {
action : 'pr_post',
pr_rating : rating
},
success: function (response) {
console.log(response);
}
});
and call like this:
add_action( 'wp_ajax_pr_request', 'pr_request' );
add_action( 'wp_ajax_nopriv_pr_request', 'pr_request' );
function pr_request() {
// Code
}

AJAX request is not sending with certain Database

I have a page where I do different requests via AJAX, also I have the developer DB and the production DB but the JS is the same. In certain module, if I open it using the developer DB it shows me the response but if I use the production database it shows nothing but not error is displayed. If I use the production DB in any other module it works perfectly. I already checked in the network console and it says that in both cases the request is sent. Here is the JS code:
function consultaAdministradores(id, tipo, btn){
var datos = {
'id': id,
'tipo': tipo
}
console.log("DATOSSSS");
console.log(datos);
$.ajax({
url: '../send/get_AdminEmpresas.php',
type: 'POST',
data: datos,
dataType: 'JSON',
success: function(res) {
console.log(" res --------- ");
console.log(res);
imprimeAdmins(res);
}
});
console.log("Se pasó el AJAX");
And in the PHP file I have this:
$res = $con->consulta($sql);
if ($res->num_rows > 0 ) {
$i = 0;
while ($dato = $res->fetch_assoc()) {
$administradores[$i] = $dato;
$i++;
}
if ($tipo_page == 'prev') {
array_multisort($administradores);
}
$array = array('status' => "Success", 'administradores' => $administradores, 'id' => $id_page, 'tipo' => $tipo_page);
}
else
{
$array = array('status' => "Fail", 'message' => 'Sin resultados');
}
$json = json_encode($array);
echo $json;

The problem wasn't the JS or the PHP files, the problem was the codification of the DB, I inserted some data in UTF-8 codification and later I converted to UTF-8 codification giving me as a result some weird characters.

Ajax Alert Response from PHP

Hopefully an easy question here. I actually used an example I found on SO but can't figure out why its not working. No errors in console or anything.
I have an ajax Post function I am using to pass data to a php script.
Its passing the data correct, but the response each time is coming back as an error alert. I can confirm that server side is getting the data and processing it correctly, just can't figure out why its never returning a success response.
Here is the Ajax:
$(function () {
$('#pseudoForm').on('click', '#submit', function (e) {
e.preventDefault();
$.ajax({
type: "POST",
url: "psu_output.php",
data: $('#pseudoForm').serialize(),
datatype: 'json',
success: function (response) {
if(response.type == 'success') {
$('#messages').addClass('alert alert-success').text(response.message);
} else {
$('#messages').addClass('alert alert-danger').text(response.message);
}
}
});
return false;
});
});
</script>
And in my php script I used this:
<?php
$success = true;
if($success == true) {
$output = json_encode(array('type'=>'success', 'message' => 'YAY'));
} else {
$output = json_encode(array('type'=>'error', 'message' => 'WHOOPS'));
}
die($output);
?>

The problem is that datatype: 'json' should be dataType: 'json'. Javascript is case-sensitive.

The error is because you received the returned data as json but the content type is a simple string (text/html) so you need to JSON.parse() the received data first like so:
$(function () {
$('#pseudoForm').on('click', '#submit', function (e) {
e.preventDefault();
$.ajax({
type: "POST",
url: "psu_output.php",
data: $('#pseudoForm').serialize(),
datatype: 'json',
success: function (response) {
response = JSON.parse(response);
if(response.type == 'success') {
$('#messages').addClass('alert alert-success').text(response.message);
} else {
$('#messages').addClass('alert alert-danger').text(response.message);
}
}
});
return false;
});
});
The second option is to send json headers from php itself thus removing the need of parsing JSON in javascript. You can do that by using the following line of code BEFORE ECHOING OR PRINTING ANYTHING ELSE FROM THE PHP SCRIPT:
header('Content-Type: application/json');
and then
echo $output;

If you are working with JSON responses, you need to set the header so your browser and your JavaScript could interpret it correctly:
<?php
$success = true;
if ($success == true) {
$output = json_encode(array(
'type' => 'success',
'message' => 'YAY'
));
} else {
$output = json_encode(array(
'type' => 'error',
'message' => 'WHOOPS'
));
}
header('Content-Type: application/json');
echo $output;

JQuery confirmation dialog after AJAX request

I need to validate, on server side, if a person with a given registration number is already on the database. If this person is already registered, then I proceed with the program flow normally. But, if the number is not already registered, then I'd like to show a confirmation dialog asking if the operator wants to register a new person with the number entered and, if the operator answers yes, then the person will be registered with the number informed on the form on it's submission.
I've tried
Server side(PHP):
if (!$exists_person) {
$resp['success'] = false;
$resp['msg'] = 'Do you want to register a new person?';
echo json_encode($resp);
}
Client side:
function submit(){
var data = $('#myForm').serialize();
$.ajax({
type: 'POST'
,dataType: 'json'
,url: 'myPHP.php'
,async: 'true'
,data: data
,error: function(response){
alert('response');
}
});
return false;
}
I can't even see the alert, that's where I wanted to put my confirmation dialog, with the message written on server side. Other problem, how do I resubmit the entire form appended with the operator's answer, so the server can check if the answer was yes to register this new person?
EDIT
I was able to solve the problem this way:
Server side(PHP):
$person = find($_POST['regNo']);
if ($_POST['register_new'] === 'false' && !$person) {
$resp['exists'] = false;
$resp['msg'] = 'Do you want to register a new person?';
die(json_encode($resp)); //send response to AJAX request on the client side
} else if ($_POST['register_new'] === 'true' && !$person) {
//register new person
$person = find($_POST['regNo']);
}
if($person){
//proceed normal program flow
}
Client side:
function submit(e) {
e.preventDefault();
var data = $('#myForm').serialize();
var ajax1 = $.ajax({
type: 'POST'
, dataType: 'json'
, async: 'true'
, url: 'myPHP.php'
, data: data
, success: function (response) {
if (!response.exists && confirm(response.msg)) {
document.getElementById('register_new').value = 'true'; //hidden input
dados = $('#myForm').serialize(); //reserialize with new data
var ajax2 = $.ajax({
type: 'POST'
, dataType: 'json'
, async: 'true'
, url: 'myPHP.php'
, data: data
, success: function () {
document.getElementById('register_new').value = 'false';
$('#myForm').unbind('submit').submit();
}
});
} else if (response.success) {
alert(response.msg);
$('#myForm').unbind('submit').submit();
}
}
});
}

There doesn't appear to be anything wrong with your PHP.
The problem is (1) You are doing the alert inside of an error callback, and your request isn't failing, so you don't see the alert. (2) You are alerting the string 'response' instead of the variable response.
It is also worth noting that you should be using the .done() and .fail() promise methods (http://api.jquery.com/jquery.ajax/#jqXHR).
Here is the fixed JS:
function submit() {
var data = $('#myForm').serialize();
// Same as before, with the error callback removed
var myAjaxRequest = $.ajax({
type: 'POST',
dataType: 'json',
url: 'myPHP.php',
async: 'true',
data: data
});
// The request was successful (200)
myAjaxRequest.done(function(data, textStatus, jqXHR) {
// The data variable will contain your JSON from the server
console.log(data);
// Use a confirmation dialog to ask the user your question
// sent from the server
if (confirm(data.msg)) {
// Perform another AJAX request
}
});
// The request failed (40X)
myAjaxRequest.fail(function(jqXHR, textStatus, errorThrown) {
console.log(errorThrown);
});
return false;
}
Also, you are setting a 'status' in PHP and checking that in the JS (I presume). What you want to be doing is setting a HTTP status code from the server, as below:
if (!$exists_person)
{
$resp['msg'] = 'Do you want to register a new person?';
// 400 - Bad Request
http_response_code(400);
echo json_enconde($resp);
}
Then, jQuery will determine whether the request failed based on the status code you respond with. 200 is a successful request, and 400 numbers are fail.
Check out this page for a full list: https://httpstatuses.com/

Okay so this is a two part question; I'll try my best to answer both parts:
Part 1: How to detect if success is false and trigger the confirmation popup?
In jQuery.ajax the error handler is triggered based on response code. This is probably not what you want. You can use your success handler and test the value res.success to see if it's true or false. It would be something along the lines of:
function submit(e) {
e.preventDefault();
var data = $('#myForm').serialize();
$.ajax({
type: 'POST',
dataType: 'json',
url: 'myPHP.php',
async: 'true',
data: data
}).done(function(res) {
if (!res.success) {
alert(res.msg);
}
});
}
Part 2: How do I resubmit with a confirmation?
Working off of our previous code we will make some changes that allow for submit() to be passed an argument registerNew. If registerNew is true we will pass it as a param to the ajax handler in the PHP so it knows we want to register a new person. The Javascript will look something like this:
function submit(e, registerNew) {
if (e) e.preventDefault();
var data = $('#myForm').serialize();
var ajax_options = {
type: 'POST',
dataType: 'json',
url: 'myPHP.php',
async: 'true',
data: data
};
ajax_options.data.register_new = !!registerNew;
$.ajax(ajax_options).done(function(res) {
if (!res.success && confirm(res.msg)) {
submit(null, true);
}
});
}
As you can see here, we are passing a new register_new param in the data in our ajax options. Now we need to detect this on the PHP side, which is easy enough and looks like this (this goes in your php ajax handler):
if ($_POST["register_new"]) {
// new user registration code goes here
} else {
// your existing ajax handler code
}

Add confirm inside submit function
function submit(){
var data = $('#myForm').serialize();
if (confirm('Are you ready?')) {
$.ajax({
type: 'POST'
,dataType: 'json'
,url: 'myPHP.php'
,async: 'true'
,data: data
,error: function(response){
alert('response');
}
});
}
return false;
}

Symfony2 Save Form Data via Ajax

I have a form in Symfony that needed to be submitted via ajax call. I worte the code but it's not saving any data in db but also doesn't give/show any error.
send_dict = {
type: 'POST',
url: $(this).attr('action'),
processData: true,
data: $('#Form').serialize(),
beforeSend: function(request) {alert('before send');},
success: function (data) {alert("success")},
error: function(xhr, textStatus, thrownError) {
alert('Some Thing Went Wrong, Please Refresh and Try Again...');
}
}
$.ajax(send_dict);
public function createAction(Request $request)
{
$user = $this->getUser();
$address = new Addresses();
if($request->isXmlHttpRequest()) {
// Do something...
if ($request->isMethod('POST')) {
$request = $this->get('request');
$permanent_is_present = $request->get('permanent_is_present');
$present_address = $request->get('present_address');
$present_address_country = $request->get('present_address_country');
// Persisting Objects to the Database
if($permanent_is_present==true){
$address->isIsPresent(true);
$address->isIsPermanent(true);
}else{
$address->isIsPresent(true);
}
$address->setUser($user);
$address->setStreet1($present_address);
$address->setCountry($present_address_country);
$address->setState($present_address_state);
$address->setCity($present_address_city);
//exit(\Doctrine\Common\Util\Debug::dump($address));
// Entity Manager To Get Connected with Doctrine
$em = $this->getDoctrine()->getManager();
// Persists the entire objects....
$em->persist($address);
// Flush queries into database
$em->flush();
$output = array();
$response = new Response();
$output[] = array('success' => true);
$response->headers->set('Content-Type', 'application/json');
$response->setContent(json_encode($output));
return $response;
}else{
return $this->render('AddressBundle:Addresses:new.html.twig');
}
} else {
return $this->redirect($this->generateUrl('address_new'));
}
}
both alerts in before send and success on ajax function are showing. but the data is not saved in the database? New to symfony, also don't know how to trace/debug this?
UPDATE:
One more thing i notice that there is no user login in the system, and em trying to made ajax calls. and it is sending to controller and because user id is not found so it is not saving into database

You need to serialize form data. Give your form an Id attribute then:
data: $('#Form').serialize(),