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JSON array accent display

I'm trying to display a json array with contents from my database but when their is accents in the database the json array doesn't want to be displayed, their is my code:
<?php
include 'connect.php';
header('Content-Type: text/plain; charset=UTF-8') ;
$serial = 1;
$statment = mysqli_prepare($conn, "SELECT * FROM lesson WHERE serial = ?");
mysqli_stmt_bind_param($statment, "i", $serial);
mysqli_stmt_execute($statment);
mysqli_stmt_store_result($statment);
mysqli_stmt_bind_result($statment, $serial, $by, $type, $description, $lesson, $nb_q);
$response = array();
$response["success"] = false;
while(mysqli_stmt_fetch($statment)){
$response["success"] = true;
$response["serial"] = $serial;
$response["by"] = $by;
$response["type"] = $type;
$response["lesson"] = $lesson;
$response["description"] = $description;
$response["nb_q"] = $nb_q;
for ($i = 1; $i <= $nb_q; $i++) {
$statment2 = mysqli_prepare($conn, "SELECT * FROM question WHERE 4_l_id = ? AND q_num = ?");
mysqli_stmt_bind_param($statment2, "ii", $serial, $i);
mysqli_stmt_execute($statment2);
mysqli_stmt_store_result($statment2);
mysqli_stmt_bind_result($statment2, $question, $nb_answer, $type, $correct_1, $for_l_id, $id, $q_num);
while(mysqli_stmt_fetch($statment2)){
$response["q".$i] = $question;
}
}
}
echo json_encode($response, JSON_UNESCAPED_UNICODE);
?>
Thank you for your answers

I found a good code but thank you for your help,
<?php
include 'connect.php';
$serial = 1;
$statment = mysqli_prepare($conn, "SELECT * FROM lesson WHERE serial = ?");
mysqli_stmt_bind_param($statment, "i", $serial);
mysqli_stmt_execute($statment);
mysqli_stmt_store_result($statment);
mysqli_stmt_bind_result($statment, $serial, $by, $type, $description, $lesson, $nb_q);
$response = array();
$response["success"] = false;
function utf8_encode_all(&$value)
{
if (is_string($value)) return utf8_encode($value);
if (!is_array($value)) return $value;
$ret = array();
foreach($dat as $i=>$d) $ret[$i] = utf8_encode_all($d);
return $ret;
}
while(mysqli_stmt_fetch($statment)){
$response["success"] = true;
$response["serial"] = $serial;
$response["by"] = utf8_encode_all($by);
$response["type"] = utf8_encode_all($type);
$response["lesson"] = utf8_encode_all($lesson);
$response["description"] = utf8_encode_all($description);
$response["nb_q"] = $nb_q;
for ($i = 1; $i <= $nb_q; $i++) {
$statment2 = mysqli_prepare($conn, "SELECT * FROM question WHERE 4_l_id = ? AND q_num = ?");
mysqli_stmt_bind_param($statment2, "ii", $serial, $i);
mysqli_stmt_execute($statment2);
mysqli_stmt_store_result($statment2);
mysqli_stmt_bind_result($statment2, $question, $nb_answer, $type, $correct_1, $for_l_id, $id, $q_num);
while(mysqli_stmt_fetch($statment2)){
$response["q".$i] = utf8_encode_all($question);
}
}
}
$array = array_map('htmlentities',$response);
$json = html_entity_decode(json_encode($array));
echo $json;
?>

Having trouble sending AND receiving information from a php file using javascript

I am sending some information to a php file that runs a query but I want to also retrieve some information from that php file at the same time. The php file executes fine but I can't get the json_encoded object.
Javascript function that sends a string and a number to a php file:
function open_close(){
var status = encodeURIComponent(SelectedTicket["Status"]);
var ticketNum = encodeURIComponent(SelectedTicket["TicketNum"]);
var info = "Status="+status+"&TicketNum="+ticketNum;
var http3 = createAjaxRequestObject();
if (http3.readyState == 4) {
if (http3.status == 200){
alert("Ticket Updated!"); //This never gets hit
getUpdatedTicket(JSON.parse(http3.responseText));
}
}
http3.open("POST", "openClose.php", true);
http3.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
http3.send(info);
}
PHP File that takes the string and number and updates a table
<?php
include("config.php");
session_start();
$status = $_POST["Status"];
$num = $_POST["TicketNum"];
$newStatus = " ";
if(strcmp($status, "Open") == 0){
$newStatus = "Closed";
}
elseif(strcmp($status, "Closed") == 0){
$newStatus = "Open";
}
$sql = "UPDATE tickets SET Status = \"$newStatus\" where TicketNum = $num ";
$r = $conn ->query($sql) or trigger_error($conn->error."[$sql]");
$sql = "SELECT * FROM tickets where TicketNum = $num";
$result = $conn->query($sql);
while($row = $result->fetch_assoc()){
$data[] = $row;
}
echo json_encode($data);
?>
How can I retrieve the json_encoded object in the same javascript function?

You'd need a readyState listener to know when the request is done, and then get the data from the responseText
function open_close() {
var status = encodeURIComponent(SelectedTicket["Status"]);
var ticketNum = encodeURIComponent(SelectedTicket["TicketNum"]);
var info = "Status=" + status + "&TicketNum=" + ticketNum;
var http3 = createAjaxRequestObject();
http3.onreadystatechange = function () {
if (http3.readyState == 4) {
if (http3.status == 200) {
console.log(http3.responseText); // <- it's there
}
}
}
http3.open("POST", "openClose.php", true);
http3.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
http3.send(info);
}

Passing 2 arrays to a javascript function with ajax

This code submits a form and an ID # to addBand.php. The php inserts the form data into a DB, and then echos an array that houses 2 arrays: 1 is an array of ids, the other is a string to update an HTML select element. I've only included the PHP that generates and echos the arrays.
i keep getting "Uncaught TypeError: Cannot read property 'selectRestults' of undefined" with the following code. what am I doing wrong?
javascript/jquery:
$(function() {
$("#addBandForm").submit(function(event) {
event.preventDefault();
var globalShowID = '&id=' + window.globalShowID;
$.ajax({
url: "womhScripts/addBand.php",
type: "POST",
data: $('#addBandForm').serialize() + globalShowID,
success: function(msg) {
var actArray = msg['actIDArray'];
var bandArray = msg['bandSelectArray'];
var result = bandArray['selectResults'];
window.globalShowID='';
$("#band1").html(result)
},
error:function(errMsg) {
console.log(errMsg);
}
});
});
});
addBand.php
$bandSelectArray = array();
$actIDArray = array();
$bandResults = "";
if (isset($_POST['id']))
{
$ShowID = $_POST['id'];
$actSQL = mysqli_query($link, "SELECT actID FROM Act WHERE showID=".$ShowID."");
while($actRow = mysqli_fetch_array($actSQL))
{
$actIDArray[] = array(
'actID' => $actRow['actID'],
);
}
}
$selectBandSQL = mysqli_query ($link, "SELECT bandID, bandName FROM Band");
while ($row = mysqli_fetch_array($selectBandSQL))
{
$bandID = $row['bandID'];
$bandName2 = $row['bandName'];
$bandResults .= '<option value="'.$bandID.'">'.$bandName2.'</option>';
}
$bandSelectArray['selectResults'] = $bandResults;
$resultArray = array();
$resultArray['actIDArray'] = $actIDArray;
$resultArray['bandSelectArray'] = $bandSelectArray;
echo json_encode($resultArray);

mysqli query not working in foreach php

My problem now when I click on the picture on my page, for the first time it will display. But for the second time it will display fail. This process will start by sending the data to ajax, then ajax(prosess.js) will send it to the php page(process1.php).
When I remove the code in blockquote ($query = "SELECT ...") it will run, but if not, it will display fail.
process1.php
<?php
include 'session.php';
include 'connection.php';
if(isset($_POST['dataS'])) {
$table = $_POST['table'];
$concat = "";
$serial = $_POST['dataS'];
$query = "SELECT * FROM product WHERE serialNum = '$serial'";
$result = mysqli_query($conn, $query);
$row = mysqli_fetch_assoc($result);
if($row) {
$prodName = $row['prodName'];
$quanProd = 1;
$priceProd = $_POST['total'] + $row['salePrice'];
if($table == "") {
$query = "SELECT * FROM product WHERE serialNum = '$serial'";
$result = mysqli_query($conn, $query);
$row = mysqli_fetch_assoc($result);
}
else{
$DOM = new DOMDocument;
$DOM->loadHTML($table);
$items = $DOM->getElementsByTagName('tr');
$check = 0;
$check_one = 0;
$y=0;
function tdrows($elements,$check,$serial,$prodName,$y) {
$quantity="";
$item = "";
$price = "";
$delete = "";
$x = 0;
foreach($elements as $element) {
if($x == 0)
$delete = $element->nodeValue;
else if($x == 1)
$item = $element->nodeValue;
else if($x == 2)
$quantity = $element->nodeValue;
else if($x == 3)
$price = $element->nodeValue;
$x++;
}
**$query = 'SELECT prodName FROM product WHERE prodName = "$item"';
$search = mysqli_query($conn, $query) or die(mysqli_error());
$row = mysqli_fetch_assoc($search);
$s = $row['prodName'];**
if($prodName == $s) {
$quantity++;
$check = 1;
}
else {
$check = 0;
}
return $check;
}
foreach ($items as $node) {
$check = tdrows($node->childNodes,$check,$serial,$prodName,$y);
$y++;
}
}
$priceProd = number_format((float)$priceProd, 2, '.', '');
echo json_encode (
array ( //this array is used to send the data back to ajax.
"success" => "1",
"concat" => $concat,
"quantity" => $quanProd,
"price" => $priceProd,
)
);
}
else {
echo json_encode (
array ( //this array is used to send the data back to ajax.
"success" => "0",
)
);
}
}
?>
process.js
$(document).ready(
function() {
$("body").on("click","#product .add",
function(e) {
var total = document.getElementById("total").value;
var table = document.getElementById('table-list').innerHTML;
table = (table.trim) ? table.trim() : table.replace(/^\s+/,'');
var serial = $(this).attr('id');
var totalQ = document.getElementById("totalQ").value;
if(total == "")
total = 0;
else
total = parseFloat(total);
if(totalQ == "")
totalQ = 0;
else
totalQ = parseInt(totalQ);
var dataS = serial;
e.preventDefault();
$.ajax({
type : "POST",
url : "process1.php",
crossDomain: true,
data : {dataS : dataS, table : table, total : total},
dataType : 'json',
})
.done(function(html) {
if(html.success == 1) {
console.log('done: %o', html);
$("#table-list").html(html.concat).show();
document.getElementById('totalQuantity').innerHTML = html.quantity;
document.getElementById("total").value = html.price;
document.getElementById("payment").value = html.price;
document.getElementById('totalQ').value = html.quantity;
document.getElementById('title').innerHTML = html.price;
document.getElementById('input').value='';
$("#input").focus();
}
else {
alert("Wrong serial number!");
document.getElementById('input').value='';
$("#input").focus();
}
})
.fail(function(html) {
console.info('fail: %o', html);
alert("fail");
});
return false;
});
});
connection.php
<?php
$conn = mysqli_connect('localhost','root','','rds');
?>

your query is wrong:try this
$query = "SELECT prodName FROM product WHERE prodName = '".$item."'";

According to your pictures, your problem is that your database connection isn't correct. When you execute the first request it won't do any database interaction (because off the blocknotes). The second request you will send table data, which will perform a query. So the first request will succeed, while the second request will give you an error on your mysqli ($conn) object.
if($table == "") {
//Database interaction
$query = "SELECT * FROM product WHERE serialNum = '$serial'";
$result = mysqli_query($conn, $query);
$row = mysqli_fetch_assoc($result);
}
else{
//No database interaction because of the blocknotes
$DOM = new DOMDocument;
$DOM->loadHTML($table);
$items = $DOM->getElementsByTagName('tr');
$check = 0;
$check_one = 0;
$y=0;
function tdrows($elements,$check,$serial,$prodName,$y) {
$quantity="";
$item = "";
$price = "";
$delete = "";
$x = 0;
foreach($elements as $element) {
if($x == 0)
$delete = $element->nodeValue;
else if($x == 1)
$item = $element->nodeValue;
else if($x == 2)
$quantity = $element->nodeValue;
else if($x == 3)
$price = $element->nodeValue;
$x++;
}
**$query = 'SELECT prodName FROM product WHERE prodName = "$item"';
$search = mysqli_query($conn, $query) or die(mysqli_error());
$row = mysqli_fetch_assoc($search);
$s = $row['prodName'];**
if($prodName == $s) {
$quantity++;
$check = 1;
}
else {
$check = 0;
}
return $check;
}
foreach ($items as $node) {
$check = tdrows($node->childNodes,$check,$serial,$prodName,$y);
$y++;
}
}
Check your the username, password and database name. I'm pretty sure you used something wrong here. As mentioned in your connection.php file you don't use a password. Are you sure the user root doens't have a password? Can you access the database with a MySQL administration tool like phpMyAdmin?

Issue on Quering Database Using jQuery Ajax and PHP

Can you please take a look at This Demo and let me know why I am getting empty array?
I have a jquery ajac request as:
$( "#appinfo" ).on( "submit", function( e ) {
var eTraget = $(".opacity").html().replace(/\D/g,'');
var senario = current;
if(qtype =="econo"){
var col = senario +"_"+eTraget;
var data='column='+col;
$.ajax({
type:"POST",
url:"assets/econo.php",
data:data,
dataType : 'json',
success:function(html) {
coords = html;
console.log(data);
st = map.set();
for (var i = 0; i < coords.length; i++) {
var circle = map.circle(coords[i][0], coords[i][1], 6);
st.push(circle);
}
e.preventDefault();
});
and econo.php as:
<?PHP
include 'conconfig.php';
$con = new mysqli(DB_HOST,DB_USER,DB_PASS,DB_NAME);
$collm = $_POST['column'];
$query = "SELECT x, y FROM econo WHERE".$collm."=1";
$results = $con->query($query);
$return = array();
if($results) {
while($row = $results->fetch_assoc()) {
$return[] = array((float)$row['x'],(float)$row['y']);
}
}
$con->close();
echo json_encode($return);
?>
As you can see I console.log(data); and the console display the result like column=ce_3000 but the $collm = $_POST['column']; is empty because whrn tried to dump it I just got an empty []. Can you please let me know why this is happening?