How to catch data sent by Ajax using PHP - php

Question

Would like to ask for a help. I'm having an issue about getting the right data that I wanted. Maybe there's something lack in my code. I always get null.
I didn't received any error on the console tab.
Here's my ajax
$(document).ready(function() {
$(".fsubmit").click(function() {
var emailval = $("#email").val().trim();
$.ajax({
url: '/test.php',
type: 'POST',
dataType: 'json',
data: 'email='+emailval,
success: function(data) {
console.log(data);
return false;
}
});
});
});
Maybe you could help me fix the PHP. Still learning some stuff. Kindly put some details for lesson please.
Here's my PHP code
<?php
$date = new DateTime();
$datelog = $date->format('d.m.Y h:i:s');
$data = $_POST['data'];
$message = '[' . $datelog . '] - email: ' . json_encode($data);
echo($message);
?>
I appreciate your answer!

Answer 1

Not allowed to comment. Here's a little edit to Mohammad's answer. isset() returns true for blank form fields. Use empty() instead.
this runs in my system ..
ajax call
$(document).ready(function(e) {
$(".fsubmit").click(function() {
e.preventDefault();
console.log("whooowww");
var emailval = $("#email").val().trim();
$.ajax({
url: '/arrayfilter.php',
type: 'POST',
dataType: 'json',
data: 'email='+emailval,
success: function(data) {
console.log(data);
return false;
}
});
});
});
html form
<form>
<input type="text" name="user-input" id="email"/>
<button type="submit">Send</button>
</form>
php
if($_POST){
if(!empty($_POST['email']))
echo "$_POST[email]";
}
}

Answer 2

try this:
$(document).ready(function() {
$(".fsubmit").click(function() {
var emailval = $("#email").val().trim();
$.ajax({
url: '/test.php',
type: 'POST',
dataType: 'json',
data: {email:emailval},
success: function(data) {
console.log(data);
return false;
}
});
});
});
then in php:
<?php
$date = new DateTime();
$datelog = $date->format('d.m.Y h:i:s');
$data = isset($_POST['email'])?$_POST['email']:'';
$message = '[' . $datelog . '] - email: ' . json_encode($data);
echo($message);
?>
You should always check for the posted elements if its isset or not.

Answer 3

The JS
I can see - at least - one major flaw in your JS code. You are passing data: 'email='+emailval to the request. This would be fine (well - possibly not, since you are not URL-encoding your message) for a GET, but with the post I'd doubt, that jQuery can make some decent JSON out of email=Some arbitrary text. At least it will make your JS vulnerable to injections. There are answers on how to send JSON to a webservice with jQuery, please see here.
The PHP
$_POST['data'] (if this is correct) will contain a string. But according to the PHP documentation
json_encode — Returns the JSON representation of a value
I don't think that this is what you want to do. You have a string, and you'll probably want to get an object from your JSON string (actually at the moment, I doubt that it's really a JSON string what you are passing). You can achieve this with json_decode, which
json_decode — Decodes a JSON string
I'd suggest the following PHP code, which decodes the JSON into an array and then outputs the value of email
$data = json_decode($_POST['data'], true);
$message = '[' . $datelog . '] - email: ' . $data["email"];
For this to work, $_POST['data'] will have to look like this
{'email': 'Your message'}
or the like.