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Jquery is not spotting that a form select has changed

I have a db table of all postcodes in the UK and want to create functionality to filter down into local locations to decrease load times of the page. The issue I am having is that I am a bit out of practice with JQuery and there seems to be an issue where selecting the country is not being picked up by the JQuery.
Here is the HTML/PHP for the select:
<label for="country_select">
Select Country
</label>
<select id="country_select" name="country_select" class="form-control search-select">
<option value=""> </option>
<?PHP
if ($result = mysqli_query($link, "SELECT DISTINCT(`country`) as countrylist FROM `postcodes`")) {
while($member = mysqli_fetch_assoc($result)) {
$country=$member['countrylist'];
?>
<option value="<?PHP echo $country; ?>"><?PHP echo $country; ?></option>
<?PHP
}
}
?>
</select>
Here is the PHP to check for the post and create a mysql statment to then create another drop down for cities. I am wondering if the issue lies here, i added an echo to see if it appears when the option has been changed, but nothing is appearing.
<?PHP
if (isset($_POST['CountryID'])) {
$country_query = $_POST['CountryID'];
$statement = " AND `country` = '".$country_query."' ";
echo $statement;
}
?>
And here is the jquery:
$(document).ready(function(){
$("#country_select").change(function () {
var country_name = $("#country_select").val();
jQuery.ajax({
type: "POST",
data: {CountryID: country_name},
success: function(data){
if(data.success == true){
alert('success');
}
}
});
});
});
I am unsure where I am going wrong as it has been a while since I have coded. I have looked around at others' issues with no luck. Any help would be greatly appreciated.

The issue was the jquery. The code should be the following:
$(document).ready(function(){
$("#country_select").change(function () {
var country_name = $("#country_select").val();
jQuery.ajax({
type: "POST",
data: {CountryID: country_name},
success: function(data){
alert(data);
}
});
});
});`

How to fetch mutiple values on select option and display in input type using ajax?

I'm trying to fetch mutiple values from database using ajax php.
I've a select option(value is fetching from database), and if i select any option then i want to display the related data which is matching with the id
of the the current option.but currently i'm able to fetch only one data column from databse.
I'm writing my current code please have a look at it and let me know how can i modify it.
My select option:-
<select data-placeholder="Choose a Vehicle..." class="chosen-select form-control" tabindex="-1" name='vno' onChange="getCity(this.value);" id="vno" required='true' >
<option value="">Select</option>
<?php
foreach($results as $vd) { ?>
<option value='<?php echo $vd['id'];?>'><?php echo $vd['vno'];?></option>";
<?php } ?>
</select>
and the js file
// Fetch city from Database
function getCity(val) {
$.ajax({
type: "POST",
url: "retrive_data.php",
data:'id='+val,
success: function(data){
$("#rate").html(data);
}
});
}
retrive_data.php
<?php
require_once ("dbController.php");
$db_handle = new DBController();
if (! empty($_POST["id"])) {
$query = "SELECT * FROM tbl_vehicle WHERE id = '" . $_POST["id"] . "' ";
$results = $db_handle->runQuery($query);
?>
<?php
foreach ($results as $city) {
?>
<option value="<?php echo $city["rate"]; ?>"><?php echo $city["rate"]; ?></option>
<?php
}
}
?>

Change your js code as below
// Fetch city from Database
function getCity(val) {
$.ajax({
type: "POST",
url: "retrive_data.php?id=" + val,
success: function(data){
$("#rate").html(data);
}
});
}

I’m making some assumptions about the desired result, and I’m not sure what the connection is between vehicles and city rates... but there are multiple issues here. Let’s work through them:
<select data-placeholder="Choose a Vehicle..." class="chosen-select form-control" tabindex="-1" name='vno' id="vno" required='true' >
<option value="">Select</option>
<?php foreach($results as $vd): ?>
<option value="<?= $vd['id']?>" ><?= $vd['vno'] ?></option>";
<?php endforeach; ?>
</select>
<!-- add a landing spot for the data coming in -->
<select id="rate"></select>
Nothing major here, just took out the onChange (typical practice is to have a listener in the JavaScript. Separation of concerns)
In your JavaScript, I don’t think you were successfully passing the id. It should be a JavaScript object. Also, send data to a function that knows how to put the data in your form:
// Fetch city from Database
function getCity(val) {
$.ajax({
type: "POST",
url: "retrive_data.php",
data:{id: val},
success: function(data){
showRate(data);
}
});
}
Monitor the select for a change. (JavaScript should be inside document ready block)
$('#vno').on('change', function (){
getCity($(this).val());
});
Function to display the results of your ajax call:
showRate(data) {
// this lets you see the data that was returned
console.log(data);
var rate = $('#rate');
// clear current content
rate.html('');
// create options, assuming this is a select
$.each(data, function() {
rate.append($("<option />").val(this.rate).text(this.rate));
});
}
retrieve.php
Need to use prepared statements, and sending data as json instead of html is recommended
<?php
// sending json (data), not html (presentation)
header('Content-Type: application/json');
require_once ("dbController.php");
$db_handle = new DBController();
if (! empty($_POST["id"])) {
// substituting variables in a query is a big no-no
// $query = "SELECT * FROM tbl_vehicle WHERE id = '" . $_POST["id"] . "' ";
// must use placeholders / prepared statement
$query = "SELECT * FROM tbl_vehicle WHERE id = ?'";
// check your database object for how to do prepared statements and row fetching. If it doesn’t do prepared statements, dump it!
$stmt = $db_handle->prepare ($query);
$stmt->execute($_POST["id"]);
$out = array();
while($row = $stmt->fetch() ) {
$rate = $row['rate'];
$out[] = array(
'rate'=>$rate
);
}
die(json_encode($out));
}
Caveat: all code is off the top of my head, and typed on a phone. Syntax errors are likely. This is intended to show concepts and ideas for further research

How do I pass MySQL rows to a specific div in html using PHP?

Good day to everyone. So I am trying to get the value of a select option and getting a set of rows in MySql depending on the column name in the database. So here is the HTML code:
<html>
<body>
<select name = "FilterDoc" onchange = "filterby(this);">
<option disabled>Filter By</option>
<option value="document_type">Document Type</option>
<option value="date">Date</option>
<option value="hei">HEI</option>
<option value="other">Other Govt.</option>
<option value="person">Person</option>
</select>
<div class="panel-body" id="container">
</div>
Here is code for Ajax:
<script type="text/javascript">
function filterby(sel){
$.ajax({ //create an ajax request to display.php
type: "POST",
data: {FilterDoc: $(sel).val()},
url: "filterdocu.php",
dataType: "html", //expect html to be returned
success: function(response)
{
$("#responsecontainer").html(response);
}
console.log(reply);
});} </script>
Now the value of the select option will pass to the PHP file. I dont know if "if statement" is the right one for this since I haven't had that much background about getting values on html and such and I'm trying to find a better way to get the rows from MySql and display them into the container.
Here's the PHP code:
<?php echo"<div class='panel panel-primary' id='container'>";
if($_POST["FilterDoc"]=="document_type")
{
echo "<script type='text/javascript'>$('container').html('""');</script>";
$result=mysqli_query($conn,"SELECT * FROM records ORDER BY document_type ASC");
while($data = json_encode(mysql_fetch_assoc($result))
{
echo json_encode($data);
}
}
else if($_POST["FilterDoc"]=="date")
{
$result=mysqli_query($conn,"SELECT * FROM records ORDER BY date_received DESC");
echo "<script type='text/javascript'>$('container').html('""');</script>";
while($data = json_encode(mysql_fetch_assoc($result))
{
echo json_encode($data);
}
}
else if($_POST["FilterDoc"]=="hei")
{
$result=mysqli_query($conn,"SELECT * FROM records ORDER BY hei ASC");
echo "<script type='text/javascript'>$('container').html('""');</script>";
while($data = json_encode(mysql_fetch_assoc($result))
($result))
{
echo json_encode($data);
}
}
else if($_POST["FilterDoc"]=="Other")
{
$result=mysqli_query($conn,"SELECT * FROM records ORDER BY other_govt ASC");
echo "<script type='text/javascript'>$('container').html('""');</script>";
while($data = json_encode(mysql_fetch_assoc($result))
{
echo json_encode($data);
}
}
else if($_POST["FilterDoc"]=="Person")
{
$result=mysqli_query($conn,"SELECT * FROM records ORDER BY contact_person ASC");
echo "<script type='text/javascript'>$('container').html('""');</script>";
while($data = json_encode(mysql_fetch_assoc($result))
{
echo json_encode($data);
}
}echo"</div>"; ?>
I'm also not entirely sure if using javascript to clear the container is the proper way before putting contents into the div container.
I would really appreciate your help. Thank you!

At first i would simplify the code like :
<?php
echo "<div class='panel panel-primary' id='container'>";
// Set Variable
$filter = $_POST["FilterDoc"]; // This needs proper escaping
$result = mysqli_query($conn,"SELECT * FROM records ORDER BY $filter ASC");
while($data = json_encode(mysql_fetch_assoc($result)) {
echo json_encode($data);
}
echo "</div>";
Thats the PHP part, now i would rewrite the javascript.
function filterby(sel) {
$.ajax({ //create an ajax request to display.php
type: "POST",
data: {
FilterDoc: $(sel).val()
},
url: "filterdocu.php",
dataType: "html", //expect html to be returned
success: function(response) {
$("#container").empty().html(response);
}
});
}
But maybe your whole filtering process may be inefficient. You could fetch all the data with one SQL call and do the filtering with JS / data-attributes.

Set initial value of dropdown output constructed with ajax

I have the following dropdown list which changes the output of <div id="item"></div> with ajax when select option is changed. I'm not using select2.
<?php
$biqsQuery = "SELECT biq.biqid, biq.name FROM biq";
$biqs = $db->query($biqsQuery);
?>
<select name="itemselector" id="itemselect">
<?php foreach ($biqs ->fetchAll() as $biq): ?>
<option value="<?php echo $biq['biqid']);?>">
<?php echo e($biq['name']);?>
</option>
<?php endforeach; ?>
</select>
<div id="item"></div>
PHP File:
if(isset($_GET['itemselector'])){
$biqQuery = "SELECT biq.biqid, biq.name, biq.img
FROM biq
WHERE biq.biqid= :biqid ";
$biq= $db ->prepare($biqQuery);
$biq->execute(['biqid' => $_GET['itemselector']]);
$selectedBiq=$biq->fetch(PDO::FETCH_ASSOC);
echo '<img src="'. $selectedBiq['img']. '">';
}
Javascript File:
$('#itemselect').on('change',function(){
var self = $(this);
$.ajax({
url: '../helpers/biq.php',
type: 'GET',
data: {itemselector : self.val()},
success: function(data){
$('#item').html(data);
}
});
});
It's currently succesfully changing the output, no problem on that part.
But when the page is first loaded, it shows the first value of the table on the dropdown menu, however it doesnt output the image of that first value into <div>.
What i need is; when the page is loaded i need the first entry in the database to be outputted into the <div id="item"></div> automatically.
Any help is appreciated, thanks in advance.

you could write on body load event to accomplish this. please correct if any type mistake will there but this will help you to get it rid
$(document).ready(function(){
var first = $("#itemselect option:first").val();
$.ajax({
url: '../helpers/biq.php',
type: 'GET',
data: {itemselector : first},
success: function(data){
$('#item').html(data);
}
});
});

Single Div refresh with jquery Ajax and PHP

Okay So I have a div on my page that has some code for display option groups in a select input. And then on the other side displaying the options in that group after the selection is made. My html/php code for this is below:
<div class="row">
<div class="col-lg-6">
<label class="control-label" for="productOptions">Select your
product options</label> <select class="form-control" id=
"productOptions">
<option>
Select an Option Group
</option><?php foreach($DefaultOptions as $option): ?>
<option value="<?php echo $option['GroupID']; ?>">
<?php echo $option['GroupName']; ?>
</option><?php endforeach; ?>
</select>
</div>
<div class="col-lg-6" id="groupOptions">
<label class="control-label">Group Options</label>
<?php if($GroupOptions): ?>
<?php foreach ($GroupOptions as $optionValue): ?>
<?php echo $optionValue['optionName']; ?> <?php endforeach; ?>
<?php endif; ?>
</div>
</div>
By default on the original page load, $GroupOptions does not exist in the form, because it is set after the user selects the Group they wish to choose from. I call the php script by using ajax to avoid page reload
$("#productOptions").change(function(){
var GroupID = $(this).val();
var dataString = 'GroupID=' + GroupID;
//alert (dataString);return false;
$.ajax({
type: "POST",
url: "#",
data: dataString,
success: function() {
$("#groupOptions").html(dataString);
}
});
return false;
});
Then the ajax goes to a php call that gets the options that match the groups id in the database.
if(isset($_POST['GroupID']))
{
$GroupID = $_POST['GroupID'];
$sql = "SELECT * from `KC_Options` WHERE GroupID=$GroupID";
$GroupOptions = $db->query($sql);
}
Now I want to refresh the div #GroupOptions to display the results from the query above, and make <?php if($GroupOptions): ?> set to true.
I managed to refresh the div with $("#groupOptions").html(dataString); in the success function of the ajax call. But that only returns well the dataString. (obviously). Is there a way to truly refresh just the div. Or a way to pass the info from the php call into the success function?

UPDATE:
You have 4 problems in your current code:
Problem #1 and Problem #2 - In your separate PHP script you are not echoing anything back to the Ajax. Anything you echo will go back as a variable to the success function. Simply the add echo statement(s) according to the format you want. Your 2nd problem is that you are trying to echo it in the HTML part, where $GroupOptions does not even exist (the Ajax simply returns an output from the PHP script, it's not an include statement so your variables are not in the same scope).
if(isset($_POST['GroupID']))
{
$GroupID = $_POST['GroupID'];
$sql = "SELECT * from `KC_Options` WHERE GroupID=$GroupID";
$GroupOptions = $db->query($sql);
//this is where you want to iterate through the result and echo it (will be sent as it to the success function as a variable)
if($GroupOptions):
foreach ($GroupOptions as $optionValue):
echo $optionValue['optionName'];
endforeach;
endif;
}
In your Ajax, add a variable named data to the success function, which will receive the output from the PHP script. Also notice that your url is incorrect, you need to post to an actual external file such as my_custom_script.php.:
$.ajax({
type: "POST",
url: "your_external_script.php",
data: dataString,
success: function(data) {
if (data && data !== '') {
//data will equal anything that you echo in the PHP script
//we're adding the label to the html so you don't override it with the new output
var output = '<label class="control-label">Group Options</label>';
output += data;
$("#groupOptions").html(output);
} else {//nothing came back from the PHP script
alert('no data received!');
}
}
});
Problem #4 - And on your HTML, no need to run any PHP. Simply change:
<div class="col-lg-6" id="groupOptions">
<label class="control-label">Group Options</label>
<?php if($GroupOptions): ?>
<?php foreach ($GroupOptions as $optionValue): ?>
<?php echo $optionValue['optionName']; ?> <?php endforeach; ?>
<?php endif; ?>
</div>
to
<div class="col-lg-6" id="groupOptions">
</div>
Hope this helps

You have to take the response in yout success callback function and actually give a response in your oho function
$("#productOptions").change(function(){
var GroupID = $(this).val();
var dataString = 'GroupID=' + GroupID;
//alert (dataString);return false;
$.ajax({
type: "POST",
url: "#",
data: dataString,
success: function(dataString) { //take the response here
// convert dataString to html...
$("#groupOptions").html(newHtml);
}
});
return false;
});
PHP:
if(isset($_POST['GroupID']))
{
$GroupID = $_POST['GroupID'];
$sql = "SELECT * from `KC_Options` WHERE GroupID=$GroupID";
$GroupOptions = $db->query($sql);
echo json_encode($GroupOptions ); //give a response here using json
}