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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

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.

Populating an array through jquery AJAX in php

I have a function in a compare.php that takes a parameter $data and uses that data to find certain things from web and extracts data and returns an array.
function populateTableA($data);
So to fill array I do this
$arrayTableA = populateTableA($name);
now this array is then used to iterate tables..
<table id="tableA">
<input type="text" name="search"/><input type="submit"/>
<?php foreach($arrayTableA as $row) { ?>
<tr>
<td><?php echo $row['name']?></td>
<td><?php echo $row['place']?></td>
</tr>
</table>
Now what I want to do is to enter some data on input and then through jquery ajax
function populateTableA($data);
should be called and $array should be refilled with new contents and then populated on tableA without refreshing the page.
I wrote this jquery but no results.
$(document).on('submit',function(e) {
e.preventDefault(); // Add it here
$.ajax({ url: 'compare.php',
var name = ('search').val();
data: {action: 'populateTableA(name)'},
type: 'post',
success: function(output) {
$array = output;
}
});
});
I have been doing web scraping and the above was to understand how to implement that strategy... original function in my php file is below
function homeshoppingExtractor($homeshoppingSearch)
{
$homeshoppinghtml = file_get_contents('https://homeshopping.pk/search.php?category%5B%5D=&search_query='.$homeshoppingSearch);
$homeshoppingDoc = new DOMDocument();
libxml_use_internal_errors(TRUE);
if(!empty($homeshoppinghtml)){
$homeshoppingDoc->loadHTML($homeshoppinghtml);
libxml_clear_errors();
$homeshoppingXPath = new DOMXPath($homeshoppingDoc);
//HomeShopping
$hsrow = $homeshoppingXPath->query('//a[#class=""]');
$hsrow2 = $homeshoppingXPath->query('//a[#class="price"]');
$hsrow3 = $homeshoppingXPath->query('(//a[#class="price"])//#href');
$hsrow4 = $homeshoppingXPath->query('(//img[#class="img-responsive imgcent"])//#src');
//HomeShopping
if($hsrow->length > 0){
$rowarray = array();
foreach($hsrow as $row){
$rowarray[]= $row->nodeValue;
// echo $row->nodeValue . "<br/>";
}
}
if($hsrow2->length > 0){
$row2array = array();
foreach($hsrow2 as $row2){
$row2array[]=$row2->nodeValue;
// echo $row2->nodeValue . "<br/>";
}
}
if($hsrow3->length > 0){
$row3array = array();
foreach($hsrow3 as $row3){
$row3array[]=$row3->nodeValue;
//echo $row3->nodeValue . "<br/>";
}
}
if($hsrow4->length > 0){
$row4array = array();
foreach($hsrow4 as $row4){
$row4array[]=$row4->nodeValue;
//echo $row3->nodeValue . "<br/>";
}
}
$hschecker = count($rowarray);
if($hschecker != 0) {
$homeshopping = array();
for($i=0; $i < count($rowarray); $i++){
$homeshopping[$i] = [
'name'=>$rowarray[$i],
'price'=>$row2array[$i],
'link'=>$row3array[$i],
'image'=>$row4array[$i]
];
}
}
else{
echo "no result found at homeshopping";
}
}
return $homeshopping;
}

As mentioned in the comments PHP is a server side language so you will be unable to run your PHP function from javascript.
However if you want to update tableA (without refreshing the whole page) you could create a new PHP page that will only create tableA and nothing else. Then you could use this ajax call (or something similar) -
$(document).on('submit','#formReviews',function(e) {
e.preventDefault();
$.ajax({
url: 'getTableA.php', //or whatever you choose to call your new page
data: {
name: $('search').val()
},
type: 'post',
success: function(output) {
$('#tableA').replaceWith(output); //replace "tableA" with the id of the table
},
error: function() {
//report that an error occurred
}
});
});

Hi You are doing it in wrong way.You must change your response to html table and overwrite older one.
success: function(output) {
$("#tableA").html(output);
}
});
In your ajax page create a table with your result array

You are in a very wrong direction my friend.
First of all there are some syntax error in your JS code.
So use JavaScript Debugging
to find where you went wrong.
After that Basic PHP with AJAX
to get a reference how ajax and PHP work together
Then at your code
Create a PHP file where you have to print the table part which you want to refresh.
Write an AJAX which will hit that PHP file and get the table structure from the server. So all the processing of data will be done by server AJAX is only used for request for the data and get the response from the server.
Put the result in your html code using JS.
Hope this will help

Getting a value from Javascript in PHP [duplicate]

This question already has answers here:
What is the difference between client-side and server-side programming?
(3 answers)
Closed 7 years ago.
I am not very experienced in web programming and am attempting to run a script which updates my database.
<script>
function myFunction() {
var texts = document.getElementById("content").textContent;
alert(texts)
<?php
include_once 'accounts/config.php';
$text = ...;
$tbl_name='enemies'; // Table name
$query = "UPDATE enemies SET text=('$text') WHERE id=1";
$result = mysql_query($query) or die(mysql_error());
?>
}
</script>
I have no idea what to put in the $text section as shown with $text = ...; in order to get the variable texts from above.
EDIT
I have updated my code but the function does not seem to be accessing the PHP file. I am using a button to call the function and I have also tested it so i know the function is being called. My file is called update.php and is in the same directory as this file.
<button onclick="myFunction()">Click This</button>
<script>
function myFunction() {
var texts = document.getElementById("content").textContent;
$.ajax({
url: "update.php",
type: "POST",
data: {texts:texts},
success: function(response){
}
});
}
</script>

you can post your $texts value to other php page using ajax and get the variable on php page using $_POST['texts'] and place update query there and enjoy....
function myFunction() {
var texts = document.getElementById("content").textContent;
$.ajax({
url: 'update.php',
type: "POST",
data: {texts:texts},
success: function(response)
{
}
});
And your php file will be named as update.php
<?php
include_once 'accounts/config.php';
$text =$_POST['texts'];
$tbl_name='enemies'; // Table name
$query = "UPDATE `enemies` SET `text`='".$text."' WHERE `id`=1";
$result = mysql_query($query) or die(mysql_error());
?>

PHP runs on the server and then generates output which is then returned to the client side. You can't have a JavaScript function make a call to inlined PHP since the PHP runs before the JavaScript is ever delivered to the client side.
Instead, what you'd need to do is have your function make an AJAX request to a server-side PHP script that then extracts the data from the request body and then stores it in the database.
PHP: "/yourPhpScript.php"
<?php
include_once 'accounts/config.php';
$text = $_POST['data'];
$tbl_name='enemies'; // Table name
$query = "UPDATE enemies SET text='".$text.'" WHERE id=1";
$result = mysql_query($query) or die(mysql_error());
?>
JavaScript:
function myFunction() {
var texts = document.getElementById("content").textContent;
alert(texts);
// append data as a query string
var params = 'data='+texts;
var xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
// when server responds, output any response, if applicable
if(http.readyState == 4 && http.status == 200) {
alert(http.responseText);
}
}
// replace with the filename of your PHP script that will do the update.
var url = '/yourPhpScript.php';
xmlhttp.open("POST", url, true);
xmlhttp.send(params);
}
A word of caution: This is not a safe, production-friendly way of updating data in your database. This code is open to SQL injection attacks, which is outside the scope of your question. Please see Bobby Tables: A guide to preventing SQL injection if you are writing code that will go into production.

You are wrong in approach
You should use ajax to post 'texts' value to your php script
https://api.jquery.com/jquery.post/ and create separate php file where you will get data from ajax post and update DB
javascript:
<script>
function myFunction() {
var texts = document.getElementById("content").textContent;
$.ajax({
type: "POST",
url: "update.php",
data: "texsts=" + texts,
success: success
});
}
</script>
update.php
<?php
include_once 'accounts/config.php';
$text = $_POST['texts'];
$tbl_name='enemies'; // Table name
$query = "UPDATE enemies SET text=('$text') WHERE id=1";
$result = mysql_query($query) or die(mysql_error());
?>

i will use PDO if i was you, what you do mysql_query are outdated, if you use my framework https://github.com/parisnakitakejser/PinkCowFramework you can do the following code.
<?php
include('config.php');
$text = $_POST['text'];
$query = PinkCow\Database::prepare("UPDATE enemies SET text = :text WHERE id = 1");
$bindparam = array(
array('text', $text, 'str')
);
PinkCow\Database::exec($query,$bindparam);
$jsonArray = array(
'status' => 200
);
echo json_encode($jsonArray);
?>
place this code in jsonUpdateEnemies.php file and call it width jQuery
<script>
function myFunction(yourText) {
$.post( 'jsonUpdateEnemies.php', {
'text' : yourText
}, function(data)
{
alert('Data updated');
},'json');
}
</script>
its a little more complex then you ask about, but its how i will resolved your problem, :)

Trouble with php variables and ajax javascript

ok I have edited this to another couple of questions I've asked on a similar issue, but I really am in a rush so thought I'd start a new one, sorry if it bothers anyone.
first I have a php script on test.php on the apache server
<?php
//create connection
$con = mysqli_connect("localhost", "user", "password", "dbname");
//check connection
if (mysqli_connect_errno()){
echo "failed to connect to MySQL: " . mysqli_connect_error();
}
$grab = mysqli_query($con, "SELECT * FROM table");
$row = mysqli_fetch_array($grab);
$name = $row["name"];
$color = $row["color"];
$price = $row["price"];
$n1 = $name[0];
$c1 = $color[0];
$p1 = $price[0];
?>
Then I've got this ajax script set to fire onload of page a webpage written in html. so the load() function is onload of the page in the body tag. This script is in the head.
function load(){
var xmlhttp = new XMLHttpRequest();
xmlhttp.open("GET", "test.php", true);
xmlhttp.send();
xmlhttp.onreadystatecahnge = function(){
if(xmlhttp.readyState == 4 && xmlhttp.status == 200){
document.getElementById("itemNameLink1").innerHTML = "<?php echo $n1;?>;
}
}
}
ok so what I want is the $n1 variable in the php script to be used in the javascript ajax code. Where the script is, but I'm not sure where or how to make use of the variable, I've tried a few things. All that happens right now is the innerHTML of itemNameLink1 just disappears.
I'm quite new so any advise would be appreciated, thanks.

The response (this is what you echo in php) returned from request you can get by responseText attribute of XMLHttpRequest object.
So first your JS code should be:
function load(){
var xmlhttp = new XMLHttpRequest();
xmlhttp.open("GET", "test.php", true);
xmlhttp.send();
xmlhttp.onreadystatecahnge = function(){
if(xmlhttp.readyState == 4 && xmlhttp.status == 200){
document.getElementById("itemNameLink1").innerHTML = xmlhttp.responseText;
}
}
}
now in php echo $n1 variable:
....
$grab = mysqli_query($con, "SELECT * FROM table");
$row = mysqli_fetch_array($grab);
$name = $row["name"];
$color = $row["color"];
$price = $row["price"];
$n1 = $name[0];
$c1 = $color[0];
$p1 = $price[0];
// echo it to be returned to the request
echo $n1;
Update to use JSON for multiple variables
so if we do this:
$name = $row["name"];
$color = $row["color"];
$price = $row["price"];
$response = array
(
'name' => $name,
'color' => $color,
'price' => $price
);
echo json_encode($response);
Then in javascript we can parse it again to have data object containing 3 variables.
var data = JSON.parse(xmlhttp.responseText);
//for debugging you can log it to console to see the result
console.log(data);
document.getElementById("itemNameLink1").innerHTML = data.name; // or xmlhttp.responseText to see the response as text
Fetching all the rows:
$row = mysqli_fetch_array($grab); // this will fetch the data only once
you need to cycle through the result-set got from database: also better for performance to use assoc instead of array
$names = $color = $price = array();
while($row = mysqli_fetch_assoc($grab))
{
$names[] = $row['name'];
$color[] = $row['color'];
$price[] = $row['price'];
}
$response = array
(
'names' => $names,
'color' => $color,
'price' => $price
);

You can dynamically generate a javascript document with php that contains server side variables declared as javascript variables, and then link this in the head of your document, and then include this into your document head whenever server side variables are needed. This will also allow you to dynamically update the variable values upon page generation, so for example if you had a nonce or something that needs to change on each page load, the correct value can be passed upon each page load. to do this, you need to do a few things. First, create a php script and declare the correct headers for it to be interpreted as a script:
jsVars.php:
<?php
//declare javascript doc type
header("Content-type: text/javascript; charset=utf-8");
//tell the request not to cache this file so updated variables will not be incorrect if they change
header('Cache-Control: no-cache, no-store, must-revalidate'); // HTTP 1.1.
header('Pragma: no-cache'); // HTTP 1.0.
header('Expires: 0'); // Proxies.
//create the javascript object
?>
var account = {
email: <?= $n1; ?>,
//if you need other account information, you can also add those into the object here
username: <?= /*some username variable here for example */ ?>
}
You can repeat this for any other information you need to pass to javascript on page load, and then reference your data using the namespaced javascript object (using object namespacing will prevent collisions with other script variables that may not have been anticipated.) wherever it is needed as follows:
<script type="text/javascript>
//put this wherever you need to reference the email in your javascript, or reference it directly with account.email
var email = account.email;
</script>
You can also put a conditional statement into the head of your document so it will only load on pages where it is needed (or if any permission checks or other criteria pass as well). If you load this before your other scripting files, it will be available in all of them, provided you are using it in a higher scope than your request.
<head>
<?php
//set the $require_user_info to true before page render when you require this info in your javascript so it only loads on pages where it is needed.
if($require_user_info == TRUE): ?>
<script type="text/javascript" href="http://example.com/path-to-your-script/jsVars.php" />
<?php endif; ?>
<script type="text/javascript" href="your-other-script-files-that-normally-load" />
</head>
You can also do this for any other scripts that have to load under specific criteria from the server.

You should define the PHP variable. And use that variable in your javascript:
<?php
$n1 = "asd";
?>
<html>
<head></head>
<body>
<div id="itemNameLink1"></div>
<script>
function load()
{
var xmlhttp = new XMLHttpRequest();
xmlhttp.open('GET', '/test.php', true);
xmlhttp.send(null);
//Note you used `onreadystatecahnge` instead of `onreadystatechange`
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById("itemNameLink1").innerHTML = '<?=$n1?>';
}
}
}
load();
</script>
</body>
</html>