Get AJAX POST Using PHP - php

Question

I have a drpcategory dropdown in a form. I will just paste the dropdown code below;
<div class="form-group">
<label>Category</label>
<select class="form-control bg-dark btn-dark text-white" id="drpcategory" name="drpcategory" required>
<?php
$category = ''.$dir.'/template/post/category.txt';
$category = file($category, FILE_IGNORE_NEW_LINES);
foreach($category as $category)
{
echo "<option value='".$category."'>$category</option>";
}
?>
</select>
</div>
Then I AJAX post every time I make a selection in the above drpcategory dropdown as below;
<script>
$(function(){
$('#drpcategory').on('change',function()
{
$.ajax({
method: 'post',
data: $(this).serialize(),
success: function(result) {
console.log(result);
}
});
});
});
</script>
This seems to be currently working as I'm getting outputs like below in Chrome Browser > Inspect > Network tab every time I make a selection in drpcategory. Here is the screenshot;
The question is how can I capture this AJAX post data using PHP within the same page and echo it within the same page? So far I have tried;
<?php
if(isset($_POST['drpcategory']))
{
echo 'POST Received';
}
?>
I'm looking for a solution using only PHP, JQuery and AJAX combined.
This question was later updated and answered here:
AJAX POST & PHP POST In Same Page

Answer 1

First of all, this line -> type: $(this).attr('post') should be type: $(this).attr('method'),. So this will give the value ** type:post** and
As far as i understand, you are asking to send ajax whenever you select options from drpcategory. Why are you submitting the entire form for this. If i where you, i should have done this problem by following way
$("#drpcategory").change(function(){
e.preventDefault();
var drpcategory=$(this).val();
$.ajax({
type: 'post',
data: drpcategory,
success: function(result) {
console.log(result);
}
});
});
On you php side, you can get your data like,
echo $_POST['drpcategory'];

Answer 2

I recommend you read the documentation for the ajax function, I tried to replicate it and I had to fix this:
$.ajax({
// If you don't set the url
// the request will be a GET to the same page
url: 'YOU_URL',
method: 'POST', // I replaced type by method
data: $(this).serialize(),
success: function(result) {
console.log(result);
}
});
http://api.jquery.com/jquery.ajax/
OUTPUT:

Answer 3

First change to $value
<div class="form-group">
<label>Category</label>
<select class="form-control bg-dark btn-dark text-white" id="drpcategory" name="drpcategory" required>
<?php
$category = ''.$dir.'/template/post/category.txt';
$category2 = file($category, FILE_IGNORE_NEW_LINES);
foreach($category2 as $value)
{
echo "<option value='".$value."'>".$value."</option>";
}
?>
</select>
then add url
<script>
$(function()
{
$('#form').submit(function(e)
{
e.preventDefault();
$.ajax({
url:'folder/filename.php',
type: 'post',
data: '{ID:" . $Row[0] . "}',
success: function(result) {
console.log(result);
}
});
});
$('#drpcategory').on('change',function()
{
$("#form").submit();
});
});
try request
if(isset($_REQUEST['ID']))
The result will/should send back to the same page

Answer 4

Please try this code:
$.post('URL', $("#FORM_ID").serialize(), function (data)
{
alert('df);
}

Answer 5

I think you have an eroror syntax mistake in ajax jQuery resquest because ajax post 'http://example.com/?page=post&drpcategory=Vehicles' does not return this type url in browser Network Tab.
<?php var_dump($_POST); exit; ?> please do this statment in your php function if anything posted to php page it will dump.
Here ajax request example
$("#drpcategory").change(function(){
e.preventDefault();
var drpcategory=$(this).val();
$.ajax({
type: 'post',
data: drpcategory,
success: function(result) {
console.log(result);
}
});
});
`

Answer 6

It sounds like you're trying to troubleshoot several things at once. Before I can get to the immediate question, we need to set up some ground work so that you understand what needs to happen.
First, the confusion about the URL:
You are routing everything through index.php. Therefore, index.php needs to follow a structure something like this:
<?php
// cleanse any incoming post and get variables
// if all your POST requests are being routed to this page, you will need to have a hidden variable
// that identifies which page is submitting the post.
// For this example, assume a variable called calling_page.
// As per your naming, I'll assume it to be 'post'.
// Always check for submitted post variables and deal with them before doing anything else.
if($_POST['calling_page'] == 'post') {
// set header type as json if you want to use json as transport (recommended) otherwise, just print_r($_POST);
header('Content-Type: application/json');
print json_encode(array('message' => 'Your submission was received'));
// if this is from an ajax call, simply die.
// If from a regular form submission, do a redirect to /index.php?page=some_page
die;
}
// if you got here, there was no POST submission. show the view, however you're routing it from the GET variable.
?>
<html>
(snip)
<body>
<form id="form1" method="post">
<input type="hidden" name="calling_page" value="page" />
... rest of form ...
<button id="submit-button">Submit</button>
</form>
}
Now, confusion about JQuery and AJAX:
According to https://api.jquery.com/jquery.post/ you must provide an URL.
All properties except for url are optional
Your JQuery AJAX will send a post request to your index.php page. When your page executes as shown above, it will simply print {message: "Your submission was received"} and then die. The JQuery will be waiting for that response and then do whatever you tell it to do with it (in this example, print it to the console).
Update after discussion
<div class="form-group">
<label>Category</label>
<select class="form-control bg-dark btn-dark text-white" id="drpcategory" name="drpcategory" required>
<?php
$category = ''.$dir.'/template/post/category.txt';
$category = file($category, FILE_IGNORE_NEW_LINES);
foreach($category as $category)
{
echo "<option value='".$category."'>$category</option>";
}
?>
</select>
</div>
<!-- HTML to receive AJAX values -->
<div>
<label>Item</label>
<select class="" id="drpitem" name="drpitem"></select>
</div>
<script>
$(function(){
$('#drpcategory').on('change',function() {
$.ajax({
url: '/receive.php',
method: 'post',
data: $(this).serialize(),
success: function(result) {
workWithResponse(result);
}
});
});
});
function workWithResponse(result) {
// jquery automatically converts the json into an object.
// iterate through results and append to the target element
$("#drpitem option").remove();
$.each(result, function(key, value) {
$('#drpitem')
.append($("<option></option>")
.attr("value",key)
.text(value));
});
}
</script>
receive.php:
<?php
// there can be no output before this tag.
if(isset($_POST['drpcategory']))
{
// get your items from drpcategory. I will assume:
$items = array('val1' => 'option1','val2' => 'option2','val3' => 'option3');
// send this as json. you could send it as html, but this is more flexible.
header('Content-Type: application/json');
// convert array to json
$out = json_encode($items);
// simply print the output and die.
die($out);
}
Once you have everything working, you can take the code from receive.php, stick it in the top of index.php, and repoint the ajax call to index.php. Be sure that there is no output possible before this code snippet.