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How to populate one drop-down based on the selection of another drop-down from database in php and javascript/jquery

I want to populate the drop-down based on the selection of other. Here's how the database looks like. Tried the following code but it isn't giving me any results.
make
id maker
2 Apple
3 Samsung
model
id make_id model
1 3 Galaxy S III
2 3 Galaxy S IV
3 2 iphone 5s
4 2 iphone 5
Queries.php
class Queries{
public static function getMaker() {
$conn = DB::databaseConnection();
$sql = "SELECT * from make" ;
$stmt = $conn->prepare($sql);
if ($stmt->execute()) {
$result = $stmt->fetchAll(PDO::FETCH_ASSOC);
return $result;
} else {
return null;
}
}
public static function getModel($make_id) {
$conn = DB::databaseConnection();
$sql = "SELECT * from model WHERE make_id = :make_id" ;
$stmt = $conn->prepare($sql);
$stmt->bindParam(':make_id', $make_id);
if ($stmt->execute()) {
$result = $stmt->fetchAll(PDO::FETCH_ASSOC);
return $result;
} else {
return null;
}
}
}
mobilereg.php
$make_id = filter_input(INPUT_GET, "make_id");
require_once './pages/header.php';
require_once './functions/queries.php';
$maker = Queries::getMaker();
$model = Queries::getModel($make_id);
<div class="control-group form-group">
<div class="controls">
<label>Make</label>
<select class="form-control"name="maker"id="maker"onchange="get_models();">
<option selected disabled></option>
<?php
for ($i = 0; $i < count($maker); $i++) {
echo '<option value="' . $maker[$i]['id'] . '">' . $maker[$i]['maker'] . '</option>';
}
?>
</select>
</div>
</div>
<div class="control-group form-group">
<div class="controls">
<label>Model</label>
<select class="form-control" name="get_models" id="get_models">
<option selected disabled></option>
</select>
</div>
</div>
main.js
function get_models(){
var maker = $('#maker').val();
var dataString = "maker="+maker;
$.ajax({
type: "POST",
url: "functions.php",
data: dataString,
success: function(html)
{
$("#get_models").html(html);
}
});
}
functions.php
require_once './queries.php';
$make_id = filter_input(INPUT_POST, "make_id");
if($_POST){
$maker = $_POST['maker'];
if($maker!=''){
echo Queries::getModel($make_id);
echo "<select name='get_models'>";
echo "<option value='" . $row['make_id'] . "'>" . $row['model'] . "
</option>";
echo '</select>';
}
}
So maker and model are the two drop-downs. The options in the model should result on the basis of the selection we make in the maker drop-down.
I've referred the code from one the questions here but it doesn't work at all. The problem is in the functions.php I guess. I'm very new to php and js, any help would be appreciated. TIA

getting select box to display selected value in different select box

I have two select boxes in which I want to select a value for one and the second select box should get same value.
Currently I am passing id and want my designation also to pass to ajax. Can I know how this can be implemented via ajax. Any help will be highly appreciated.
<select name="designation" class="form-control" id="desig" >
<option value="">Select a Designation/Role</option>
<?php
$sql = mysql_query("SELECT id, designation FROM tbl where status =1 and designationtype_id = 1 ");
while ($rows = mysql_fetch_assoc($sql)){
echo "<option value=" . $rows['id'] . ">" . $rows['designation'] . "</option>";
}
?> <select name="dd" id="dd" class="form-control" disabled>
<option value=""></option>
</select>
My AJAX,
<script type="text/javascript">
$(document).ready(function() {
$("#desig").change(function() {
var id = $(this).val();
var dataString1 = 'id=' + id;
var des = $(this).val();
var dataString2 = 'designationname=' + des;
$.ajax({
type: "POST",
url: "escalation_ajax.php",
data: dataString,
cache: false,
success: function(html) {
var data = html.split(",");
$('#rephead').val(data[0]);
}
});
});
});
</script>
escalation_ajax.php
<?php
if ($_POST['id'])
{
if ($_POST['des'])
{
$des_id = $_POST['id'];
$designation = $_POST['des'];
$sql = mysql_query("SELECT designation_id, reporting_head FROM aafmindia_in_sbi.tbl_reporting_head WHERE status=1 and reporting_head_for='$des_id'");
if ($sql === FALSE)
{
trigger_error('Query failed returning error: ' . mysql_error() , E_USER_ERROR);
}
else
{
while ($row = mysql_fetch_array($sql))
{
$id = $row['designation_id'];
$reporting_head = $row['reporting_head'];
echo '<option value="' . $id . '">' . $reporting_head . '</option>' . ',' . '<option value="' . $des_id . '">' . $designation . '</option>';
}
}
}
}
?>

What you could do, is have the second select (the one that needs the same value as the first) in a seperate file that you load via AJAX.
AJAX function:
function selection()
{
var selectValue=$("select#dd option:selected").val();
$.ajax({
type : "POST",
url : "escalation_ajax.php",
data : { id : selectValue },
success: function (html) {
$("#secondSelectorDiv").html(html);
}
})
}
What this does, is that when the selection() function is called, it will post the selected value of the first select to "escalation_ajax.php". It will then load that page into an element (div element in my example) with the id "secondSelectorDiv".
The html for the select with the function (which I will call onchange in this example), can look like this:
<select id="dd" onchange="selection();">
<option value=""></option>
</select>
<div id="secondSelectorDiv"></div>
Now in escalation_ajax.php you can retrieve the post variable and use it to look for the id in question for the second select.
<?php
$id=$_POST['id'];
/*
If you're using the id to fetch something in your database,
which it looks like you're doing, then use the post variable
to fetch your rows and build the select from that.
*/
$sql="SELECT * FROM table_name WHERE id='$id'";
$result_set=mysql_query($sql);
$row=mysql_fetch_array($result_set);
$count=mysql_num_rows(result_set);
$counter=0;
//this is the id you will check for in order to see what's to be selected
$idToCheck=$row['id'];
?>
<select id="dd2">
while($count > $counter)
{
counter++;
echo '<option value=""'; if($idToCheck == $id){ echo 'selected="selected"'; } echo '></option>';
}
?>
If you want the second select to be displayed before the first select has a value, you can simply just call an AJAX function that loads in the second select on page load.
IMPORTANT!: You should really switch to mysqli_* or PDO instead of using the deprecated mysql_*. You should at the very least look into sanitizing your inputs.

How to get multiple data from select item in jquery

I am trying to get multiple value from multiple select textbox and use the same value in PHP for adding the SUM.
My code is:
function get_amount() {
var x = $('#work').val();;
alert(x);
$.ajax({
type: "POST",
url: "get-sum.php",
data: 'id=' + x,
success: function(data) {
alert(data);
//$('#html').load(document.URL + ' #html');
}
});
}
<select class="select2 select2-multiple" multiple="multiple" data-placeholder="Choose" name="work[]" id="work" onchange="get_amount()">
<option value="1" >option 1</option>
<option value="2"> OPtion 2 </option>
<option value="3"> OPtion 3</option>
</select>
And My php file as :
$id = $_POST['id'];
function getAmount($id){
global $mysqli;
$stmt = $mysqli->prepare("SELECT
(amount)
FROM work
WHERE id = ?
");
$stmt->bind_param("s",$id);
$stmt->execute();
$stmt->bind_result($amount);
$stmt->store_result();
$stmt->fetch();
$stmt->close();
return $amount;
}
$sql = getAmount($id);
echo $sql;
Error which i am getting is the data which i am getting from multiple select is like 5,6,7 and its always hitting the php file at id 5. And the second problem how i will be able to add all $sql data.

I have done some changes in my PHP code :
$id = $_POST['id'];
$sum = 0;
for($i = 0; $i<count($id); $i++) {
$sum = $sum + getAmount($id[$i]);
}
echo $sum;
function getAmount(&$id){
global $mysqli;
$stmt = $mysqli->prepare("SELECT SUM(amount) AS total FROM work WHERE id = (?)");
$stmt->bind_param("s",$id);
$stmt->execute();
$stmt->bind_result($amount);
$stmt->fetch();
return $amount['total'];
}

Receive post data ajax with php

I'm trying to receive post data with php from ajax in same page but seems like i have some issues that i have no idea to solve them
here is my html/php code :
<select style="width:auto; margin-left:6%;" class="form-control" name="n-omran-select" id="num_omrane">
<option value='' >Choisir...</option>
<?php
while ($row = $result->fetch_assoc())
{
echo "<option value=".$row['N_omran'].">".$row['N_omran']."</option>";
}
?>
</select><br>
<?php
if (isset($_POST["selectName"])) { // try to receive post values but it seems that's not working
$selectOption = $_POST["selectName"];
$query = "SELECT nom,prenom,tel,adress,matricule_assu FROM `personnel` WHERE N_omran ='$selectOption'";
$result = mysqli_query($db,$query);
$row=mysqli_fetch_array($result,MYSQLI_ASSOC);
if ($row) {
echo " <h4>Nom : {$row['nom']}</h4>
<h4>Prénom : {$row['prenom']}</h4>
<h4>Téléphone : {$row['tel']} </h4>
<h4>Maticule d'assurance : {$row['matricule_assu']}</h4>
<h4>Adresse : {$row['adress']}</h4>";
}
} ?>
And here is my Ajax post request :
$('#num_omrane').on('change', function () {
var n_omrane = $('#num_omrane').val();
if(n_omrane != ''){
$.ajax({
type: "POST",
url: "index.php",
data: {selectName: n_omrane},
success: function () {
alert("Post request successfully done")
}
});
}
});

the code below can replace all your data with the new ones with clean writing :)
// get data from Database
<?php
if (isset($_POST["selectName"])) { // try to receive post values but it seems that's not working
$selectOption = $_POST["selectName"];
$query = "SELECT nom,prenom,tel,adress,matricule_assu FROM `personnel` WHERE N_omran ='$selectOption'";
$result = mysqli_query($db,$query);
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
} ?>
// show rows to be selected
<select style="width:auto; margin-left:6%;" class="form-control" name="n-omran-select" id="num_omrane">
<option value='' >Choisir...</option>
<?php while ($row = $result->fetch_assoc()) { ?>
<option value="<?= $row['N_omran'] ?>"> <?= $row['N_omran'] ?></option>
<?php } ?>
</select><br>
// show recieved data
<?php if ($row) { ?>
<div id="informations">
<h4>Nom : <span id="nom"><?= $row['nom'] ?></span></h4>
<h4>Prénom : <span id="prenom"><?= $row['prenom'] ?></span></h4>
<h4>Téléphone : <span id="tel"><?= $row['tel'] ?> </span></h4>
<h4>Maticule d'assurance : <span id="matricule_assu"><?= $row['matricule_assu'] ?></span></h4>
<h4>Adresse : <span id="adress"><?= $row['adress'] ?></span></h4>
</div>
<?php } ?>
// script for making ajax call
<script>
$('#num_omrane').on('change', function () {
var n_omrane = $('#num_omrane').val();
if(n_omrane != ''){
$.ajax({
type: "POST",
url: "index.php",
data: {selectName: n_omrane},
success: function (response) {
$("#nom").text(response.nom);
$("#prenom").text(response.prenom);
$("#tel").text(response.tel);
$("#matricule_assu").text(response.matricule_assu);
$("#adress").text(response.adress);
}
});
}
});
</script>

$json_data=file_get_contents('php://input');
$json=json_decode($json_data,true);
if(array_key_exists("selectName",$json)){
$selectOption =$json["selectName"];
}

Values Are Combining

`I have a script that is returning the right values except for one little problem. Lid is returning 892 which is right, However; Cid which should return 16 its returning 89216 a combination of the two. How can I get Cid to just return 16?
$(document).ready(function()
{
$(".Doggie").change(function()
{
var LocationString ='Lid='+ $(this).val();
$.ajax({
type: "POST",
url: "ajax_city.php",
data: LocationString,
cache: false,
success: function (html) {
$(".Kitty").html(html);
}
});
});
$('.Kitty').live("change",function(){
var Lid = $('#Doggie').val(), // This is the value of the id="Doggie" selected option
Cid = $(this).val(); // This is the value of the id="Kitty" selected option
$.ajax({
type: "POST",
url: "ajax_area.php",
data: {"Lid":Lid,"Cid":Cid},
cache: false,
success: function (html) {
$(".Pig").html(html);
}
});
});
});
</script>
</head>
<body>
<div id="frame1">
<label>Place :</label>
<select name="Doggie" class="Doggie" id="Doggie">
<option selected="selected">--Select Place--</option>
<?php
$sql = mysql_query("SELECT tblLocations.RestID as Lid, tblRestaurants.RestName as name
FROM tblRestaurants INNER JOIN tblLocations ON tblRestaurants.RestID = tblLocations.RestID
GROUP BY tblLocations.RestID, tblRestaurants.RestName
ORDER BY tblRestaurants.RestName ASC");
while($row=mysql_fetch_array($sql))
{
echo '<option value="'.$row['Lid'].'">'.$row['name'].'</option>';
} ?>
</select>
<label>City :</label>
<select name="Kitty" class="Kitty" id="Kitty">
<option selected="selected">--Select City--</option>
</select>
<label>Area: :</label>
<select name="Pig" class="Pig" id="Pig">
<option selected="selected">--Select Area--</option>
</select>
</div>
</body>
</html>
ajax_city.php
<?php
require('config.php');
if($_POST['Lid'])
{
$Lid=$_POST['Lid'];
//$Cid=$_POST['Cid'];
$sql=mysql_query("SELECT tblLocations.RestId as Lid, tblLocations.CityID as Cid, tblCities.CityName as name
FROM tblLocations INNER JOIN tblCities ON tblLocations.CityID = tblCities.CityID
WHERE tblLocations.RestID = $Lid
GROUP BY tblLocations.RestID, tblCities.CityName
ORDER BY tblCities.CityName ASC");
echo '<option selected="selected">--Select City--</option>';
while($row=mysql_fetch_array($sql))
{
echo '<option value="'.$row['Lid'].''.$row['Cid'].'">'.$row['name'].'</option>';
}
}
?>
ajax_area.php
<?php
require('config.php');
if($_POST['Lid'])
{
//$Lid=$_POST['Lid'];
$Cid=$_POST['Cid'];
$sql=mysql_query("SELECT tblLocations.RestId as Lid, tblLocations.CityID as Cid, tblAreas.AreaName as name
FROM tblLocations INNER JOIN tblAreas ON tblLocations.AreaID = tblAreas.AreaID
WHERE tblLocations.RestID = $Lid AND tblLocations.CityID = $Cid
GROUP BY tblLocations.RestID, tblAreas.AreaName
ORDER BY tblAreas.AreaName ASC");
echo '<option selected="selected">--Select Area--</option>';
while($row=mysql_fetch_array($sql))
{
echo '<option value="'.$row['Lid'].''.$row['Cid'].'">'.$row['name'].'</option>';
}
}
?>

It was these two lines in the ajax-city.php file.
"SELECT tblLocations.CityId as Cid, tblCities.CityName as name
And
echo '<option value="'.$row['Cid'].'">'.$row['name'].'</option>';