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php file's code not executing through ajax call

I have a button in my PHP file, and when I click on that button, I want another PHP file to run and save some data in a MySQL table. For that I am using AJAX call as suggested at this link (How to call a PHP function on the click of a button) which is an answer from StackOverflow itself.
Here is my show_schedule file from which I am trying to execute code of another PHP file:
$('.edit').click(function() {
var place_type = $(this).attr("id");
console.log(place_type);
$.ajax({
type: "POST",
url: "foursquare_api_call.php",
data: { place_type: place_type }
}).done(function( data ) {
alert("foursquare api called");
$('#userModal_2').modal('show');
});
});
here 'edit' is the class of the button and that button's id is being printed in the console correctly.
here is my foursquare_api_call.php file (which should be run when the button is clicked):
<?php
session_start();
include('connection.php');
if(isset($_POST['place_type'])){
$city = $_SESSION['city'];
$s_id = $_SESSION['sid'];
$query = $_POST['place_type'];
echo "<script>console.log('inside if, before url')</script>";
$url = "https://api.foursquare.com/v2/venues/search?client_id=MY_CLIENT_ID&client_secret=MY_CLIENT_SECRET&v=20180323&limit=10&near=$city&query=$query";
$json = file_get_contents($url);
echo "<script>console.log('inside if, after url')</script>";
$obj = json_decode($json,true);
for($i=0;$i<sizeof($obj['response']['venues']);$i++){
$name = $obj['response']['venues'][$i]['name'];
$latitude = $obj['response']['venues'][$i]['location']['lat'];
$longitude = $obj['response']['venues'][$i]['location']['lng'];
$address = $obj['response']['venues'][$i]['location']['address'];
if(isset($address)){
$statement = $connection->prepare("INSERT INTO temp (name, latitude, longitude, address) VALUES ($name, $latitude, $longitude, $address)");
$result = $statement->execute();
}
else{
$statement = $connection->prepare("INSERT INTO temp (name, latitude, longitude) VALUES ($name, $latitude, $longitude)");
$result = $statement->execute();
}
}
}
?>
none of the console.log is logged in the console and also the 'temp' table is not updated. Can anyone tell me where I am making mistake? Or is it even possible to execute the code of a PHP file like this?

Your JavaScript is making an HTTP request to the URL that executes you PHP program.
When it gets a response, you do this:
.done(function( data ) {
alert("foursquare api called");
$('#userModal_2').modal('show');
}
So you:
Alert something
Show a model
At no point do you do anything with data, which is where the response has been put.
Just sending some HTML containing a script element to the browser doesn't cause it to turn that HTML into a DOM and execute all the script elements.
You'd need to do that explicitly.
That said, sending chunks of HTML with embedded JS back through Ajax is messy at best.
This is why most web services return data formatted as JSON and leave it up to the client-side JS to process that data.

to return the contents of php code you can do something like this
you can use any call to this function
function check_foursquare_api_call(place_type) {
var place_type= encodeURIComponent(place_type);
var xhttp;
//last moment to check if the value exists and is of the correct type
if (place_type== "") {
document.getElementById("example_box").innerHTML = "missing or wrong place_type";
return;
}
xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (xhttp.readyState == 4 && xhttp.status == 200) {
document.getElementById("example_box").innerHTML = xhttp.responseText;
$('#userModal_2').modal('show');
}
};
xhttp.open("GET", "foursquare_api_call.php?place_type="+place_type, true);
xhttp.send();
}
this will allow you to send and execute the code of the foursquare_api_call file and return any elements to example_box, you can return the entire modal if you want,
you can use any POST / GET method, monitor the progress, see more here
XMLHttpRequest

(populate dropdown in contact form 7 getting this error - Warning: array_keys() expects parameter 1 to be array, null given in

Ok where to start, I will try and explain as much as I can.
I am using wordpress with contact form 7 and I am trying to populate 3 dropdown items on the contact form, I found some code that I was able to use with no problem but the problem with this was that it was getting the information from a excel file, the file is now to big and will not run on my website anymore so I would like to get the information from my database now.
I have made a table in my database "vehicle_information" with 3 columns "vehicle_type", "vehicle_make", vehicle_model"
I have code in my functions.php and code in my footer to be able to use the cf7 shortcodes.
Code from funtions.php
function ajax_cf7_populate_values() {
//MySQLi information
$db_host = '***';
$db_username = '***';
$db_password = '***';
$vehicles_makes_models = array();
//connect to mysqli database (Host/Username/Password)
$connection = mysqli_connect($db_host, $db_username, $db_password) or die('Error ' . mysqli_error());
//select MySQLi dabatase table
$vehicles_makes_models = mysqli_select_db($connection, 'vehicle_information') or die('Error ' . mysqli_error());
$sql = mysqli_query($connection, 'SELECT * FROM vehicle_type');
while($row = mysqli_fetch_array($sql)) {
$vehicles_makes_models[$row[0]][$row[1]][] = $row[2]; }
}
// setup the initial array that will be returned to the the client side script as a JSON object.
$return_array = array(
'vehicles' => array_keys($vehicles_makes_models),
'makes' => array(),
'models' => array(),
'current_vehicle' => false,
'current_make' => false
);
// collect the posted values from the submitted form
$vehicle = key_exists('vehicle', $_POST) ? $_POST['vehicle'] : false;
$make = key_exists('make', $_POST) ? $_POST['make'] : false;
$model = key_exists('model', $_POST) ? $_POST['model'] : false;
// populate the $return_array with the necessary values
if ($vehicle) {
$return_array['current_vehicle'] = $vehicle;
$return_array['makes'] = array_keys($vehicles_makes_models[$vehicle]);
if ($make) {
$return_array['current_make'] = $make;
$return_array['models'] = $vehicles_makes_models[$vehicle][$make];
if ($model) {
$return_array['current_model'] = $model;
}
}
}
// encode the $return_array as a JSON object and echo it
echo json_encode($return_array);
wp_die();
// These action hooks are needed to tell WordPress that the cf7_populate_values() function needs to be called
// if a script is POSTing the action : 'cf7_populate_values'
add_action( 'wp_ajax_cf7_populate_values', 'ajax_cf7_populate_values' );
add_action( 'wp_ajax_nopriv_cf7_populate_values', 'ajax_cf7_populate_values' );
Code from my footer
<script>
(function($) {
// create references to the 3 dropdown fields for later use.
var $vehicles_dd = $('[name="vehicles"]');
var $makes_dd = $('[name="makes"]');
var $models_dd = $('[name="models"]');
// run the populate_fields function, and additionally run it every time a value changes
populate_fields();
$('select').change(function() {
populate_fields();
});
function populate_fields() {
var data = {
// action needs to match the action hook part after wp_ajax_nopriv_ and wp_ajax_ in the server side script.
'action' : 'cf7_populate_values',
// pass all the currently selected values to the server side script.
'vehicle' : $vehicles_dd.val(),
'make' : $makes_dd.val(),
'model' : $models_dd.val()
};
// call the server side script, and on completion, update all dropdown lists with the received values.
$.post('<?php echo admin_url( 'admin-ajax.php' ) ?>', data, function(response) {
all_values = response;
$vehicles_dd.html('').append($('<option>').text(' -- choose vehicle -- '));
$makes_dd.html('').append($('<option>').text(' -- choose make -- '));
$models_dd.html('').append($('<option>').text(' -- choose model -- '));
$.each(all_values.vehicles, function() {
$option = $("<option>").text(this).val(this);
if (all_values.current_vehicle == this) {
$option.attr('selected','selected');
}
$vehicles_dd.append($option);
});
$.each(all_values.makes, function() {
$option = $("<option>").text(this).val(this);
if (all_values.current_make == this) {
$option.attr('selected','selected');
}
$makes_dd.append($option);
});
$.each(all_values.models, function() {
$option = $("<option>").text(this).val(this);
if (all_values.current_model == this) {
$option.attr('selected','selected');
}
$models_dd.append($option);
});
},'json');
}
})( jQuery );
The problem is I am still learning and this is the first time I have had to use this funtion.
and I am getting an error on my website
Warning: array_keys() expects parameter 1 to be array, null given in /customers/4/0/0/motobid.co.uk/httpd.www/wp-content/themes/storevilla-child/functions.php on line 38 {"vehicles":null,"makes":[],"models":[],"current_vehicle":false,"current_make":false}
any help would be very greatful.
Just like to say code was supplied by BDMW.

Where you use the method array_keys(), instead of:
$return_array['makes'] = array_keys($vehicles_makes_models[$vehicle]);
Try this:
$return_array['makes'] = ! empty($vehicles_makes_models[$vehicle]) ? array_keys($vehicles_makes_models[$vehicle]) : [];
From what I've read, the array_keys() has been an issue depending on php versions. Hope this helps!

POST-GET session variable (javascript-php) -javascript takes the previous variable but not last created

From the selected value (from the form) I create a variable (var parcela).
var parcela;
$(document).ready(function(){
parcela = localStorage.getItem("parcela");
if (parcela !== '') {
$('#parcela').val(parcela);
}
$("#parcela").on('change',function() {
selectBoxVal_1 = $('#parcela').val();
if (typeof(Storage) !== "undefined") {
localStorage.setItem("parcela", selectBoxVal_1);
} else {
alert('Sorry! No Web Storage support..');
}
location.reload();
});
});
From the created variable (parcela), I create a session variable in PHP.
$.post("phpscripts/session.php", {"parc_id": parcela});
PHP (session.php)
<?php
session_start();
$parcela = $_POST["parc_id"];
$parcela_int = (int)$parcela;
if($_POST){
$_SESSION['parcela_id'] = $parcela_int;
}
?>
After that, the created session variable urge to another php script
query.php
<?php
session_start();
require("common.php");
$user_id = htmlentities($_SESSION['user']['id_korisnika']);
$parc = $_SESSION['parcela_id'];
try
{
$stmt = $db->prepare("SELECT y_cent, x_cent FROM parcele WHERE id_korisnika='$user_id' AND id_parcele='$parc' ");
$stmt->execute();
}
catch(PDOException $ex)
{
die("Failed to run query: " . $ex->getMessage());
}
$rows = $stmt->fetchAll();
....
This all works perfectly!
However, when I call a php script with query (query.php) in javascript, there is a problem. JS takes the previous session variable instead of the last selected.
$.ajax({
url: 'phpscripts/query.php',
type: 'GET',
success : function(data) {
chartData = data;
//console.log(chartData);
...
Does anyone know what the problem is? I'm trying for two days to solve this ...
Note: The javascript code is contained in a single script.

I solved the problem. I had to extract part of javascript code that calls the php script into a separate script. I called the new JS script with jQuery getScript() Method.
Thank you #knets.

Jquery/AJAX populated HTML form when dropdown selected

I am trying to populate a form on my page. The information required to populate the form is pulled from a MySQL database using the ID of the drop down option as the ID in the SQL statement. I was thinking that I can store the information in $_SESSION['formBookings'] and on a refresh this will populate the form (this is already happening as I am using the session variable to populate the form after a submit.
I can not have a submit button attached to the form as I already have one and the boss doesn't want another. I would like the form to eventually automatically refresh the page on the selection of an option. If the data from the SQL statement has been stored in the session array then the form will be populated.
Here is what I have so far:
The JQuery:
<script>
$(document).ready(function(){
$('select[name=recall]').on('change', function() {var recall = $(this).val()
//$(function ()
//{
//-----------------------------------------------------------------------
// 2) Send a http request with AJAX http://api.jquery.com/jQuery.ajax/
//-----------------------------------------------------------------------
$.ajax({
url: 'recall.php', //the script to call to get data
data: "recall: recall", //you can insert url argumnets here to pass to api.php
//for example "id=5&parent=6"
dataType: 'json', //data format
success: function(data) //on recieve of reply
{
var id = data[0]; //get id
var vname = data[1]; //get name
//--------------------------------------------------------------------
// 3) Update html content
//--------------------------------------------------------------------
$('div#box1').load('DFBristol.html');//html("<b>id: </b>"+id+"<b> name: </b>"+vname); //Set output element html
//recommend reading up on jquery selectors they are awesome
// http://api.jquery.com/category/selectors/
//}
});
});
});
});
</script>
The HTML:
<select name='recall' id='recall'>
<option selected></option>
<?php
session_start();
$user = 'root';
$pass = '';
$DBH = new PDO('mysql:host=localhost;dbname=nightlineDB;', $user, $pass);
$DBH->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$DBH->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
$recall = $DBH->prepare('SELECT * FROM bookings WHERE dateInputted >= now() - INTERVAL 2 DAY');
$recall->execute();
$recallResult = $recall->fetchALL(PDO::FETCH_ASSOC);
foreach ($recallResult as $key) {
echo '<option id='.$key["ID"].'>'.$key['ID'].' - '.$key['branch'].' - '.$key['title'].' '.$key['firstName'].' '.$key['lastName'].'</option>';
}
?>
</select><br />
The SQL file (recall.php):
<?php
$user = 'root';
$pass = '';
$DBH = new PDO('mysql:host=localhost;dbname=nightlineDB;', $user, $pass);
$DBH->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$DBH->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
$recall = $DBH->prepare("SELECT * FROM bookings WHERE ID = '%$recall%'");
$recall->execute();
$recallFormResult = $recall->fetchALL(PDO::FETCH_ASSOC);
echo json_encode($recallFormResult);
?>
I have tried to pass the variable 'recall' from the jquery into the SQL statement using the data argument but nothing happens.
Could someone please help me understand what I am doing wrong and how I can resolve it.

On a quick glance there seems to be two issues with the code you've posted so far:
The AJAX request
Even though $.ajax() defaults to a request type of GET by default, it's good to specify it. There is also a syntax error in your request — you have closed the success callback with a }); where it should be a } only:
$.ajax({
url: "recall.php",
data: {
recall: recall
},
type: "GET", // Declare type of request (we use GET, the default)
dataType: "json",
success: function(data)
{
var id = data[0];
var vname = data[1];
$('div#box1').load('DFBristol.html');
} // The problematic closure
});
Even better: instead of using the deprecated jqXHR.success() function, use the .done() promise object instead, i.e.:
$.ajax({
url: "recall.php",
data: {
recall: recall
},
type: "GET", // Declare type of request (we use GET, the default)
dataType: "json"
}).done(function() {
// Success
var id = data[0],
vname = data[1];
$('div#box1').load('DFBristol.html');
});
Fixing the file recall.php
When you make an AJAX GET request to recall.php, the file needs to know what variables you intend to pass, which you have not defined. You can do that using $_GET[] (see doc), i.e.:
<?php
// Assign the variable $recall with the value of the recall query string from AJAX get request
// I recommend doing a check to see if $_GET['recall'] is defined, e.g.
// if($_GET['recall']) { $recall = $_GET['recall']; }
$recall = $_GET['recall'];
// The rest of your script, unmodified
$user = 'root';
$pass = '';
$DBH = new PDO('mysql:host=localhost;dbname=nightlineDB;', $user, $pass);
$DBH->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$DBH->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
$recall = $DBH->prepare("SELECT * FROM bookings WHERE ID = '%$recall%'");
$recall->execute();
$recallFormResult = $recall->fetchALL(PDO::FETCH_ASSOC);
echo json_encode($recallFormResult);
?>
Note: However, if you choose to make a POST request, then you should use $_POST[] (see doc) instead :)

Will this huge javascript array loaded from a database crash my website?

So on this website I'm making (who knows if i'll actually finish it lol) when someone opens up the new user page, php echos into a javascript script all the usernames from the database to create an array.
<script type="text/javascript">
var allUsers = ['!' <?php
$result = mysql_query("SELECT username FROM users ") or die("error " .mysql_error());
$usersArray = array();
while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$usersArray[] = $row['username'] or die("error ". mysql_error());
}
foreach ($usersArray as $name) {
echo ',' . json_encode($name );
}
?> , ];
the point of this is to have a live checker so if you type in a username that already exists, red text shows up next to the username input. But let's say I get 1,000,000 users (completely theoretical). Fortunately, the array only gets created at the beginning of the web page load. But will the function that checks if the username already exists in the huge array and gets called everytime someone changes the text in the username input put too much stress on the script and crash the website? If so, is there a better way to do what I'm describing?
Here's the rest of the code
function contains(a, obj) {
var i = a.length;
while (i--) {
if (a[i] === obj) {
return true;
}
}
return false;
}
function onUserChange() { //gets called onkeypress, onpaste, and oninput
if(contains(allUsers, str)) {
div.innerHTML = "Username already exists";
div.style.color = "red";
userValid = false;
}
}
</script>

Something along these lines. ( with jQuery and PDO ) - note - code is not tested.
var keyTimer, request;
$('namefield').blur(function(){
onUserChange();
});
$('namefield').keyup(function(){
onUserChange();
});
function onUserChange() { //gets called onkeypress, onblur
keyTimer = setTimeout(function(){
if(request && request.readystate != 4){
//cancel a previous request if a new request is made.
request.abort();
}
request = $.post(
'http://yoursite.com/location/of/username/script.php', //post data to server
{username : $('namefield').val()},
function(data){
if(data == 0 ) { //might be a string here
alert( 'the name is ok to use.' );
}else{
alert( 'someone has this name already.' );
}
}
);
}, 500); //overwrite previous timeout if user hits key within 500 milliseconds
}
Then in the backend
$sql = 'SELECT id FROM users WHERE username = :username';
//insert from post username but we are good programers and are using PDO to prevent sql injection.
//search for the username in the db, count the number of users or rows should be 1 someone has it 0 no one has it assuming its unique.
$stmt = $Pdo->prepare($sql);
$stmt->execute(array(':username', $_POST['username']));
echo $stmt->rowCount();
exit();
etc.....

Do not do it. My counsel is to use ajax to load the php file that will make a query asking only for the user that was typed in the input and retunr only a boolean value(exists=true / notexists=false)
Code example:
HTML(yourFile.html):
<script>
jQuery(document).ready(function(){
//When the value inside the input changes fire this ajax querying the php file:
jQuery("#inputUser").change(function(){
var input = jQuery(this);
jQuery.ajax({
type:"post",
url:"path/to/file.php",
data:input.val(),
success: function(data){
//if php returns true, adds a red error message
if(data == "1"){
input.after('<small style="color:#ff0000;">This username already exists</small>');
//if php returns false, adds a green success message
} else if(data == "0"){
input.after('<small style="color:#00ff00;">You can use this username</small>');
}
}
});
});
});
</script>
<input id="inputUser" type="text" name="username" value="">
PHP(path/to/file.php):
<?php
$username = $_REQUEST['username']; // The value from the input
$res = mysqli_query("SELECT id FROM users WHERE username='".$username."'"); // asking only for the username inserted
$resArr = mysqli_fetch_array($res);
//verify if the result array from mysql query is empty.(if yes, returns false, else, returns true)
if(empty($resArr)){
echo false;
} else{
echo true;
}
?>

As I can see you need to load the PHP code when your website is loading.
First, I recommend you to separate the code. The fact that you can mix Javascript with PHP doesn't mean it is the best practice.
Second, yes, it's not efficient your code since you make Javascript load the result so you can search into it next. What I suggest you is making the search in the server side, not in client side, because as you say, if you have 100 elements maybe the best is to load all the content and execute the function, but if you have 1,000,000 elements maybe the best is to leave the server to compute so it can make the query with SQL.
Third, you can do all this using Ajax, using Javascript or using a framework like jQuery so you don't have to worry about the implementation of Ajax, but you only worry about your main tasks.