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Unable to send data from JavaScript to PHP file using ajax

I'm unable to retrieve the q parameter sent from the PHP.
When I run my code, null values get inserted in my database.
Here are the concerning parts of my code:
My JavaScript function:
function load_now(str){
//alert(str);
var id = str.split("+")[0];
var r = confirm("Start load process for scooter " + str + "?");
if (r == true) {
var xmlhttp = new XMLHttpRequest();
//console.log(str);
xmlhttp.open("GET", "load_scooter_action.php?q=" + str, true);
xmlhttp.send();
}
}
and my load_scooter_action.php file:
<?php
require "checkUserModel.php";
require "databaseController.php";
$databaseController = new DatabaseController();
$databaseController->startConnexionToDatabase();
$conn = $databaseController->getConn();
$dateObject = new DateTime();
$startTime = $dateObject->format('Y-m-d H:i:s');
$user = $_SESSION['user-id'];
// get the q parameter from URL
$q = $_REQUEST["q"];
//forme:(scooter, lat, lng, chg);
$val = explode("+",$q);
$scooter = val[0];
$lat = val[1];
$lng = val[2];
$chg = val[3];
echo "<script type='text/javascript'> console.log(".$q.")</script>";
$sql = "UPDATE `scooters` SET `disponible` = '0' WHERE `scooters`.`numero` = '$scooter';";
if ($conn->query($sql) === TRUE) {
$add = "\nScooter Taken";
} else {
$add = "Error Taking scooter" . $conn->error;
}
$sql = "
INSERT
INTO Reloads(scooter, user, initialLoad, finalLoad, sourceX, sourceY,destinationX, destinationY, startTime,endTime)
VALUES ('$scooter','$user','$chg',null,'$lat','$lng',null ,null, '$startTime', null)";
if ($conn->query($sql) === TRUE) {
//echo "\nNew record created successfully";
//echo '<script>window.location.href = "../php/scooterMapIndex.php";</script>';
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
Could you help me have a clear understanding of my mistake?
Thanks in advance.
EDIT
str is has the following form 599+50.8037+4.32782+4

Two mistakes:
The first one is a quick fix. Replace val with $val
The second one is that, although str is of the form 599+50.8037+4.32782+4 in the .js file, $q gets retrieved without the + characters in the .php file. Therefore it has to be split appropriately.

Can you show me that what is passed in str or which value is passed in str load_now function. Do you get value when you alert(str)?
If yes then try below function
$.ajax({
type: 'GET',
url: 'follow_user.php?user_id='.urlencode($user_id),
success: function(data) {
alert('done');
}
});
Thanks

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
?>

Pass array from javascript to page in wordpress

I have a page in wordpress, that uses an jquery and ajax to get information from an external api. The form sends the array generated in javascript back to the same page with another variable that the php in the page uses to determine which page to display. Outside of wordpress, the code works fine. Inside Wordpress the first portion runs, but then instead of loading the same page again it goes to a search page and says nothing found.
The url on the output is:
http://kltools.net/?s=&post_type%5B%5D=portfolio&post_type%5B%5D=post&post_type%5B%5D=page
Which seems odd considering I'm using post not get.
The javascript that generates the array and submits the form:
function submitchans(){
for (var i=0;i<chans.length;i++)
{ var newHidInp = document.createElement('input');
newHidInp.type = 'hidden';
newHidInp.name = 'chans[]';
newHidInp.value = chans[i];
form.appendChild(newHidInp);
}
}
function livearray(input){
if (input != null) {
chans.push(input);
}
if (Y === cSize){
submitchans();
document.forms[0].submit();
}
}
Previously the array was outArray[] instead of chans[], I changed it thinking that may be triggering the result, but no luck.
This is the PHP portion of the code:
<?php
$page_to_load = $_POST[view];
switch($page_to_load) {
case '':
echo "<script src=\"../scripts/jquery-3.2.0.min.js\"></script>";
echo "<script type=\"application/javascript\" src=\"../scripts/raid.js\"></script>";
echo "<font size=\"+3\" color=\"#FFFFFF\">Who should I host?<br>Please wait while channel is selected<br></font>";
echo "<font size=\"+2\">";
echo "<br><br>";
echo "<img src=\"../_images/ajax_loader_blue_350.gif\">";
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "database";
$chanarray[] = null;
$offline = 0;
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT `TwitchNames` FROM TK_Members WHERE Validated='1' AND RaidMe='1'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
array_push($chanarray, $row['TwitchNames']);
}
} else {
echo "0 results";
}
array_splice($chanarray, 0, 1);
$conn->close();
echo "<script type=\"application/javascript\">";
echo "var channels = ". json_encode($chanarray);
echo "</script>";
echo "</font>";
echo "<form id=\"form\" method=\"post\">";
echo "<input type=\"hidden\" name=\"view\" value=\"page2\">";
echo "</form>";
break;
case 'page2':
echo "<font size=\"+3\" color=\"#FFFFFF\">Who should I host?<br>";
echo "Your channel to host is:<br></font>";
echo "<font size=\"+2\">";
echo "<br><br>";
$chans[] = null;
$test = $_POST['chans'];
foreach ($test as $chan) {
$temparray = array(rand(),$chan);
array_push($chans, $temparray);
}
array_splice($chans,0,1);
sort($chans);
echo "".$chans[0][1]."";
echo "<br><br><br>";
echo "See All Live Channels";
echo "</font>";
break;
}
?>
After working with what blokeish has suggested I've modified the javascript file working out where the problem is.
The new javascript file is:
// JavaScript Document
var chans = ["test1","test2","test3"];
function submitchans(){
for (var i=0;i<chans.length;i++)
{ var newHidInp = document.createElement('input');
newHidInp.type = 'hidden';
newHidInp.name = 'chans[]';
newHidInp.value = chans[i];
document.getElementById('chansform').appendChild(newHidInp);
}
}
jQuery(function ($) {
submitchans();
document.getElementById('chansform').submit();
});
Using only the javascript clicking submit, it passes to the next page. When adding in the array pass is when it fails. This is the page log that is returning during the execution. !!--CORRECTION--!! there was a typo in the code, after correcting ID to Id the code is working as intended.

document.forms[0].submit() is likely submitting the wp search form as that could be the first form in the DOM. I see "http://kltools.net/?s=" in the URL where "s" is the search term.
Using document.getElementById('idOfForm').submit() should get you around that problem if there are multiple forms in a page and you cant be sure of its index.

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>

How to use javascript in PHP? The script tag is not recognized

I don't know why but the script tag is not working, the SELECT query is working but i am not getting the prompt from the javascript.
it is not redirecting anywhere only a blank screen is seen
$qry1="SELECT area, aadhar FROM user where username='$user'";
$result1 = $connector->query($qry1);
if($result1){
$row1=mysql_fetch_array($result1);
$userarea= $row1['area'];
$useraadhar=$row1['aadhar'];
}?>
<body>
<script type="text/javascript">
var inputarea=<?php echo $coursename; ?>;
var userarea=<?php echo $userarea; ?>;
var useraadhar=<?php echo $useraadhar;?>'
if(inputarea==userarea){
<?php/
//date
$today = date("Y-m-d");
//Create INSERT query
$qry = "INSERT INTO complain (user,category,regno,course,lecturer,room,details,address,datein) VALUES ('$userid','$category','$reg','$coursename','$lectname','$roomno','$details','$address','$today')";
//$result = #mysql_query($qry);
$result = $connector->query($qry);
//Check whether difjslk the query was successful or not
if($result) {
$errmsg_arr[] = 'Complain succesfully added, please wait for your response';
$errflag = true;
if($errflag) {
$_SESSION['ERRMSG_ARR'] = $errmsg_arr;
session_write_close();
header("location: _new_complains.php");
exit();
}
header("location: _new_complains.php");
exit();
}else {
die("Query failed, couldn't add the new record");
header("location: _new_complains.php");
exit();
}
?>
}

You are sending data (for example body tag) before header(), therefore PHP creates an error. You just don't see it. Header needs to come before anything is sent to the browser (even a space).

You have multiple JS syntax errors:
var inputarea=<?php echo $coursename; ?>;
var userarea=<?php echo $userarea; ?>;
var useraadhar=<?php echo $useraadhar;?>'
Never EVER dump out raw text from PHP into a Javascript context. You're generating code that looks like
var inputarea=foo;
var userarea=bar;
var useradhar=baz';
The data will be seen as undefined variables, and you've got a stray ' in there. All of these errors will KILL the entire <script> block.
Always use json_encode() to dump from PHP->JS:
var inputarea = <?php echo json_encode($coursename); ?>;
This will GUARANTEE that you're producing correct Javascript code. The above line would produce
var inputarea = 'foo';
and be perfectly valid and executable code.