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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

Get data php with ajax without display it

Is it possible to get data php with Ajax without display them ? Simply stock data in JS variable?
I need this data to manipulate dates but no show it.
When I tried to simply return data without echo, etc. Data ajax in JS is empty
Ps : sorry my English is bad

try it this way
File *.php
<?php
$var_1 = null;
$var_2 = null;
/** ... */
$response = new stdClass;
$response->var_1 = $var_1;
$response->var_2 = $var_2;
echo json_encode($response);
?>
File *.html or *.js
<script>
var state = {};
$.ajax({
url: 'getData.php',
type: 'post',
dataType: 'json',
success: function (response) {
console.warn(response);
state = response;
}
});
</script>

Assuming you are trying to pass data from a PHP file to HTML/JS where it happens that your PHP file is also included in the HTML that's why it's displaying the echo (if I understood correctly!)
Using AJAX PHP example from w3school.
HTML sample file:
<?php include "PHP_SAMPLE_FILE.php" ?>
<header>
<meta name="temp_files" content="<?= htmlspecialchars($jsonData) ?>">
<!-- The rest of HTML content -->
JS sample file:
if (str.length == 0) {
// do something if there was nothing entered
} else {
var xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
if (this.responseText.includes('{')){
result = JSON.parse(this.responseText);
// do something if response is JSON
} else {
// do something if response is null
}
}
}
xmlhttp.open("GET", "PHP_SAMPLE_FILE.php?q="+str, true);
xmlhttp.send();
}
PHP sample file:
$q = $_REQUEST["q"] ?? $_POST["q"] ?? "";
$sql = "GET SOMETHING FROM DATABASE";
$result = mysqli_query($con, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
$json[] = $row;
}
}
$jsonData = json_encode($json ?? null);
if($q != ""){
echo $jsonData;
}
What happens exactly is that once the page loads initially it won't display the output of the PHP query as we have surrounded the echo with an if statement that requires to have query value (q) to search and it shouldn't be empty (""). Of course, assuming that once the page is loaded the data is shared with the client-side through defined PHP variables using various approaches, using a meta tag in the header for instance.
Once the data is received from the PHP file through echo, we use the JSON.parse function to parse it as in this scenario JS receives it as a string.
Hope that helped :)!

Using AJAX to get an SQL WHERE query into javascript

I am trying to pass a javascript variable into an SQL WHERE query and I keep getting null in return.
On-click of a button, the buttonClick function is ran:
<script>
var var1;
function buttonClick(elem){
var1 = elem.src //this gets the url from the element
var path = var1.slice(48); //this cuts the url to img/art/9/1.jpg
ajax = theAjax(path);
ajax.done(processData);
ajax.fail(function(){alert("Failure");});
}
function theAjax(path){
return $.ajax({
url: 'info.php',
type: 'POST',
data: {path: path},
});
}
function processData(response_in){
var response = JSON.parse(response_in);
console.log(response);
}
</script>
Here is the code stored in the info.php file:
<?php
$path = $_POST['path'];
$result3 = mysqli_query("SELECT itemName from images WHERE imgPath='$path'");
$json = json_encode($result3);
echo $json
?>
As you can see, once I click the button, the buttonClick() function is ran and a variable stores the image path or src. That path variable is send to theAjax() function where it is passed to the info.php page. In the info.php page, the SQL WHERE query is ran and returned to the processData() function to be parsed and printed in the developer console. The value printed shows null.
Below is a picture of what I am trying to get from the database:

1.Check that path is correct or not? you can check inside jquery using console.log(path); or at PHP end by using print_r($_POST['path']);
2.Your Php code missed connection object as well as record fetching code.
<?php
if(isset($_POST['path'])){
$path = $_POST['path'];
$conn = mysqli_connect ('provide hostname here','provide username here','provide password here','provide dbname here') or die(mysqli_connect_error());
$result3 = mysqli_query($conn,"SELECT itemName from images WHERE imgPath='$path'");
$result = []; //create an array
while($row = mysqli_fetch_assoc($result3)) {
$result[] = $row; //assign records to array
}
$json = json_encode($result); //encode response
echo $json; //send response to ajax
}
?>
Note:- this PHP query code is wide-open for SQL INJECTION. So try to use prepared statements of mysqli_* Or PDO.

mysqli_query() required 1st parameter as connection object.
$result3 = mysqli_query($conn,"SELECT itemName from images WHERE imgPath='$path'"); // pass your connection object here

I think your issue is that you're trying to encode a database resource.
Try adjusting your PHP to look like the following:
<?php
$path = $_POST['path'];
$result3 = mysqli_query("SELECT itemName from images WHERE imgPath='$path'");
$return_data = [];
while($row = mysqli_fetch_assoc($result3)) {
$return_data[] = $row;
}
$json = json_encode($return_data);
echo $json
?>

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 :)

Using ajax to display new database inputs without refreshing the page

I am using ajax to post comments to a certain page, I have everything working, except for when the user posts a comment I would like it to show immediately without refreshing. The php code I have to display the comments is:
<?php
require('connect.php');
$query = "select * \n"
. " from comments inner join blogposts on comments.comment_post_id = blogposts.id WHERE blogposts.id = '$s_post_id' ORDER BY comments.id DESC";
$result = mysql_query($query);
while ($row = mysql_fetch_array($result)) {
$c_comment_by = $row['comment_by'];
$c_comment_content = $row['comment_content'];
?>
<div class="comment_box">
<p><?php echo $c_comment_by;?></p>
<p><?php echo $c_comment_content;?></p>
</div>
<?php } ?>
</div>
</div>
<?php
}
}
and the code I have to post comments is:
<?php
$post_comment = $_POST['p_post_comment'];
$post_id = $_POST['p_post_id'];
$post_comment_by = "Undefined";
if ($post_comment){
if(require('connect.php')){
mysql_query("INSERT INTO comments VALUES (
'',
'$post_id',
'$post_comment_by',
'$post_comment'
)");
echo " <script>$('#post_form')[0].reset();</script>";
echo "success!";
mysql_close();
}else echo "Could no connect to the database!";
}
else echo "You cannot post empty comments!"
?>
JS:
function post(){
var post_comment = $('#comment').val();
$.post('comment_parser.php', {p_post_comment:post_comment,p_post_id:<?php echo $post_id;?>},
function(data)
{
$('#result').html(data);
});
}
This is what I have for the refresh so far:
$(document).ready(function() {
$.ajaxSetup({ cache: false });
setInterval(function() {
$('.comment_box').load('blogpost.php');
}, 3000);.
});
Now what I want to do is to use ajax to refresh the comments every time a new one is added. Without refreshing the whole page, ofcourse. What am I doing wrong?

You'll need to restructure to an endpoint structure. You'll have a file called "get_comments.php" that returns the newest comments in JSON, then call some JS like this:
function load_comments(){
$.ajax({
url: "API/get_comments.php",
data: {post_id: post_id, page: 0, limit: 0}, // If you want to do pagination eventually.
dataType: 'json',
success: function(response){
$('#all_comments').html(''); // Clears all HTML
// Insert each comment
response.forEach(function(comment){
var new_comment = "<div class="comment_box"><p>"+comment.comment_by+"</p><p>"+comment.comment_content+"</p></div>";
$('#all_comments').append(new_comment);
}
})
};
}
Make sure post_id is declared globally somewhere i.e.
<head>
<script>
var post_id = "<?= $s_post_id ; ?>";
</script>
</head>
Your new PHP file would look like this:
require('connect.php');
$query = "select * from comments inner join blogposts on comments.comment_post_id = blogposts.id WHERE blogposts.id = '".$_REQUEST['post_id']."' ORDER BY comments.id DESC";
$result = mysql_query($query);
$all_comments = array() ;
while ($row = mysql_fetch_array($result))
$all_comments[] = array("comment_by" => $result[comment_by], "comment_content" => $result[comment_content]);
echo json_encode($all_comments);
Of course you'd want to follow good practices everywhere, probably using a template for both server & client side HTML creation, never write MySQL queries like you've written (or that I wrote for you). Use MySQLi, or PDO! Think about what would happen if $s_post_id was somehow equal to 5' OR '1'='1 This would just return every comment.. but what if this was done in a DELETE_COMMENT function, and someone wiped your comment table out completely?