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I cannot figure out why my ajax is not sending the data to my php file

I'm having a problem with my Ajax. It seems to not be sending the data to my php file even though it worked properly 2 days ago. HTML:
<form id='comment' action='process.php' method="POST">
<textarea></textarea>
<button type='submit'>Comment</button>
</form>
My ajax code:
$('#comment').submit(function(event) {
var form = $(this);
var method = form.attr('method');
var url = form.attr('action');
info = {
comment: $('textarea').val()
};
console.log(method);
console.log(url);
console.log(info);
$.ajax({
type: method,
url: url,
data: info,
success: function(data){
alert(data);
}
});
event.preventDefault();
});
I'm doing this for a friend and I'm using this exact same Ajax code (slightly modified) on my website and it's working flawlessly.
I think the biggest red flag here is that in my php file I have an if-else that should send an alert in case the textarea is empty but for some reason it's not doing that here even though nothing is getting through. I used console.log on all the variables to see if their values are correct and they are. The alert(data) just returns an empty alert box.
EDIT: As requested, PHP code from process.php
<?php
session_start();
include_once 'db_connection.php';
date_default_timezone_set('Europe/Zagreb');
if(isset($_POST['comment'])){
function SQLInsert($id, $date, $komentar, $conn){
$sql = "INSERT INTO comments (user, date, comment) VALUES ('$id', '$date',
'$comment')";
$conn -> query($sql);
$conn -> close();
}
$id = $_SESSION['username'];
$date = date('Y-m-d H:i:s');
$comment = htmlspecialchars($_POST['comment']);
SQLInsert($id, $date, $komentar, $conn);
} else {
echo '<script>';
echo 'alert("Comment box is empty.");';
echo '</script>';
}
?>
EDIT: Problem solved, thanks for the help everyone.

You are no getting alert because you are no displaying anything as response in php file. Add the insert function out side the if condition too
function SQLInsert($id, $date, $komentar, $conn){
$sql = "INSERT INTO comments (user, date, comment) VALUES ('$id', '$date',
'$comment')";
if($conn -> query($sql)){
return true;
}else{
return false;
}
$conn -> close();
}
if(isset($_POST['comment'])){
$id = $_SESSION['username'];
$date = date('Y-m-d H:i:s');
$comment = htmlspecialchars($_POST['comment']);
$insert = SQLInsert($id, $date, $komentar, $conn);
//On based on insert display the response. After that you will get alert message in ajax
if($insert){
echo 'insert sucess';
die;
}else{
echo 'Error Message';
die;
}
}

<form id='comment' action='process.php' method="POST">
<textarea></textarea>
<button id="submit_button">Comment</button>
</form>
starting from this html you have to trigger your function as:
$("#submit_button").click(function(e){
I have added an id to your button for simplicity and removed the type because it is useless in this case.
If you want to catch the submit event of the form you have to change your html as:
<form id='comment' action='process.php' method="POST">
<textarea></textarea>
<input type='submit'>Comment</button>
</form>
and then you can keep the same javascript

This here is the issue. Have you tried providing a "method" ?
$.ajax({
**type: method,**
method : method,
url: url,
data: info,
success: function(data){
alert(data);
}
});
Also if this doesn't solve it. show me the console output

<form name="fileInfoForm" id='comment' method="post" enctype="multipart/form-data">
<textarea id="textarea"></textarea>
<button type="submit"></button>
</form>
<script type="text/javascript">
$('#comment').submit(function (e) {
e.preventDefault();
var textarea=$('#textarea').val();
var form=document.getElementById('comment');
var fd=new FormData(form);
fd.append('textarea',textarea);
$.ajax({
type: "POST",
enctype: 'multipart/form-data',
url: 'action.php',
data: fd,
dataType: "json",
processData: false,
contentType: false,
cache: false,
success: function (data) {
alert(data);
}
})
});
</script>
in action.php
$textarea= $_POST['textarea'];
echo $textarea;

Accessing JSON returned by php script using jquery ajax

Basically my program is a web page with 5 radio buttons to select from. I want my web app to be able to change the picture below the buttons every time a different button is selected.
My problem is coming in the JSON decoding stage after receiving the JSON back from my php scrip that accesses the data in mysql.
Here is my code for my ajax.js file:
$('#selection').change(function() {
var selected_value = $("input[name='kobegreat']:checked").val();
$.ajax( {
url: "kobegreat.php",
data: {"name": selected_value},
type: "GET",
dataType: "json",
success: function(json) {
var $imgEl = $("img");
if( $imgEl.length === 0) {
$imgEl = $(document.createElement("img"));
$imgEl.insertAfter('h3');
$imgEl.attr("width", "300px");
$imgEl.attr("alt", "kobepic");
}
var link = json.link + ".jpg";
$imgEl.attr('src', link);
alert("AJAX was a success");
},
cache: false
});
});
And my php file:
<?php
$db_user = 'test';
$db_pass = 'test1';
if($_SERVER['REQUEST_METHOD'] == "GET") {
$value = filter_input(INPUT_GET, "name");
}
try {
$conn = new PDO('mysql: host=localhost; dbname=kobe', $db_user, $db_pass);
$conn->setAttribute(PDO:: ATTR_ERRMODE, PDO:: ERRMODE_EXCEPTION);
$stmt = $conn->prepare('SELECT * FROM greatshots WHERE name = :name');
do_search($stmt, $value);
} catch (PDOException $e) {
echo 'ERROR', $e->getMessage();
}
function do_search ($stmt, $name) {
$stmt->execute(['name'=>$name]);
if($row = $stmt->fetch()) {
$return = $row;
echo json_encode($return);
} else {
echo '<p>No match found</p>;
}
}
?>
Here's my HTML code where I am trying to post the image to.
<h2>Select a Great Kobe Moment.</h2>
<form id="selection" method="get">
<input type="radio" name="kobegreat" value="kobe1" checked/>Kobe1
<input type="radio" name="kobegreat" value="kobe2"/>Kobe2
<input type="radio" name="kobegreat" value="kobe3"/>Kobe3
</form>
<div id="target">
<h3>Great Kobe Moment!</h3>
</div>
And here's is what my database looks like:
greatshots(name, link)
name link
------ --------
kobe1 images/kobe1
kobe2 images/kobe2
kobe3 images/kobe3
Whenever I run the web app right now, the rest of the images on the page disappear and the image I am trying to display won't show up. I get the alert that "AJAX was a success" though, but nothing comes of it other than the alert. Not sure where I am going wrong with this and any help would be awesome.

As mentioned you should parse the JSON response using JSON.parse(json);.
Also, you should specifically target the div element with a simpler setup:
$("#target").append('<img width="300px" src="' + link + '.png"/>');

Using AJAX to send form information to another page using a button

Hello I have two files that are supposed to be connected to one another. I want to send an AJAX request to another page that uses a sql query to send form information.
The application that I'm trying to create is a questionnaire with eight questions, each questions has four answers paired together with the same id (qid) and each answer has a value from the database. After you answer eight questions you will see a button that sends an AJAX request to the page test.php, (named submitAJAX).
The problem is that although my connection with AJAX is working, the values from the form are not being sent to my database. Previously I thought that the problem may lie with the form page, but now I I think the problem lies in this file:
test.php (file with json)
<?php
$localhost = "localhost";
$username = "root";
$password = "";
$connect = mysqli_connect($localhost, $username, $password) or die ("Kunde inte koppla");
mysqli_select_db($connect, 'wildfire');
if(count($_GET) > 0){
$answerPoint = intval($_GET['radiobtn']);
$qid = intval($_GET['qid']);
$tid = intval($_GET['tid']);
$sql2 = "INSERT INTO result (qid, points, tid) VALUES ($qid, $answerPoint, $tid)";
$connect->query($sql2);
$lastid = $connect->insert_id;
if($lastid>0) {
echo json_encode(array('status'=>1));
}
else{
echo json_encode(array('status'=>0));
}
}
?>
I think that the problem may lie in the row where: if($lastid>0) {
$lastid should always be more than 0, but whenever I check test.php I get this message: {"status":0} What's intended is that I get this message: {"status":1}
<html>
<head>
<meta charset="utf-8">
</head>
<body>
<?php
$localhost = "localhost";
$username = "root";
$password = "";
$connect = mysqli_connect($localhost, $username, $password) or die ("Kunde inte koppla");
mysqli_select_db($connect, 'wildfire');
$qid = 1;
if(count($_POST) > 0){
$qid = intval($_POST['qid'])+1;
}
?>
<form method="post" action="">
<input type="hidden" name="qid" id="qid" value="<?=$qid?>">
<?php
$sql1 = mysqli_query($connect,"SELECT * FROM question where answer != '' && qid =".intval($qid));
while($row1=mysqli_fetch_assoc($sql1)){
?>
<input type='radio' name='answer1' class="radiobtn" value="<?php echo $row1['Point'];?>">
<input type='hidden' name='tid' class="tid" value="<?php echo $row1['tid'];?>">
<?php echo $row1['answer'];?><br>
<?php
}
?>
<?php if ($qid <= 8) { ?>
<button type="button" onclick="history.back();">Tillbaka</button>
<button type="submit">Nästa</button>
<?php } else { ?>
<button id="submitAjax" type="submit">Avsluta provet</button>
<?php } ?>
</form>
<script src="https://code.jquery.com/jquery-1.11.3.js"></script>
<script type="text/javascript">
function goBack() {
window.history.go(-1);
}
$(document).ready(function(){
$("#submitAjax").click(function(){
if($('.radiobtn').is(':checked')) {
var radiobtn = $('.radiobtn:checked').val();
var qid = $('#qid').val();
var answer = $('input[name=answer1]:radio').val();
$.ajax(
{
type: "GET",
url: 'test.php',
dataType: "json",
data: "radiobtn="+radiobtn+"&qid="+qid,
success: function (response) {
if(response.status == true){
alert('points added');
}
else{
alert('points not added');
}
}
});
return false;
}
});
});
</script>
</body>
The values that I want to send to my database from test.php are:
qid(int), tid(int), Point(int)
There is a database connection, and my test.php file's sql query should work, but its not sending form information. Is there something that I need to rewrite or fix to make it work?

First, your data parameter in the AJAX call is not using the correct syntax. You're missing brackets. It should look like:
data: JSON.stringify({ radiobtn: radiobtn, qid: qid }),
Second, I'd suggest using POST instead of GET:
type: "POST",
which means that you need to look for your data in $_POST['radiobtn'] and $_POST['qid'] on test.php. NOTE: you should check for the key you expect using isset() before assigning the value to a variable, like so:
$myBtn = isset($_POST['radiobtn']) ? $_POST['radiobtn'] : null;
Third, for testing, use a console.log() inside your condition that checks for the checkbox being checked in order to verify that condition is working as expected.
if($('.radiobtn').is(':checked')) {
console.log('here');
UPDATE:
Fourth: You should specify the content type in your AJAX call, like so:
contentType: "application/json; charset=utf-8",

After you execute your query that inserts the result you can use a sql statement to select the last insert id. Try something like
$sql2 = "INSERT INTO result (qid, points, tid) VALUES ($qid, $answerPoint, $tid)";
$connect->query($sql2);
$result = $connect->query("SELECT LAST_INSERT_ID()");
$row = $result->fetch_row();
$lastid = $row[0];
That should return the correct last insert id, if that was where your error was occurring.

mysqli_insert_id() returns the ID generated by a query on a table with a column having the AUTO_INCREMENT attribute.
In your SQL, you are providing the ID yourself, there is no auto-increment. So you should get 0 from $connect->insert_id, because the function returns zero if there was no previous query on the connection or if the query did not update an AUTO_INCREMENT value.
For your purpose, you can use the return value of mysqli_query() instead, which returns TRUE on success and FALSE on failure.
if($connect->query($sql2)) {
echo json_encode(array('status'=>1));
}
else{
echo json_encode(array('status'=>0));
}

How to define an element with a a sql row id usng JSON encoded data

I'm using jQuery AJAX to process form data, the PHP side of it should delete two files on the server and then the SQL row in the database (for the id that was sent to it). The element containing the SQL row should then change color, move up, delete and the next SQL rows move into its place. The animation stuff occurs in the beforeSend and success functions of the ajax callback.
This script is not working, when user clicks button, the page url changes to that of the php script but the item and files do not get deleted either on the server or in the database. Nor does any of the animation occur.
This is my first time using jQuery ajax, I think there is a problem with how I define the element during the call back. Any help would be great:
js
$("document").ready(function(){
$(".delform").submit(function(){
data = $(this).serialize() + "&" + $.param(data);
if (confirm("Are you sure you want to delete this listing?")) {
$.ajax({
type: "POST",
dataType: "json",
url: "delete_list.php",
data: data,
beforeSend: function() {
$( "#" + data["idc"] ).animate({'backgroundColor':'#fb6c6c'},600);
},
success: function() {
$( "#" + data["idc"] ).slideUp(600,function() {
$( "#" + data["idc"] ).remove();
});
}
});
return false;
}
});
});
php
if (isset($_POST["id"]))
{
$idc = $_POST["id"];
if (isset($_POST["ad_link"]) && !empty($_POST["ad_link"]))
{
$ad_linkd=$_POST["ad_link"];
unlink($ad_linkd);
}
if (isset($_POST["listing_img"]) && !empty($_POST["listing_img"]))
{
$listing_imgd=$_POST["listing_img"];
unlink($listing_imgd);
}
try {
require('../dbcon2.php');
$conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sql = "DELETE FROM listings WHERE id = $idc";
$conn->exec($sql);
}
catch (PDOException $e) {
echo $sql . "<br>" . $e->getMessage();
}
echo json_encode($idc);
}
html
<div id="record-<?php echo $id; ?>">
*bunch of stuff*
<form method="post" class="delform">
<input name="id" type="hidden" id="id" value="<?php echo $id; ?>" />
<input name="ad_link" type="hidden" id="ad_link" value="<?php echo $ad_link; ?>" />
<input name="listing_img" type="hidden" id="listing_img" value="<?php echo $listing_img; ?>" />
<button type="submit">Delete</button>
</form>
</div>

You should fix your php code like this
try {
require('../dbcon2.php');
// It's better, if you will going to use MySQL DB, use the class designed to connect with it.
$conn = mysqli_connect("Servername", "usernameDB", "PasswordDB", "NameDB");
$sql = "DELETE FROM listings WHERE id = $idc";
mysqli_query($conn, $sql);
// you have to create a asociative array for a better control
$data = array("success" => true, "idc" => $idc);
// and you have to encode the data and also exit the code.
exit(json_encode($data));
} catch (Exception $e) {
// you have to create a asociative array for a better control
$data = array("success" => false, "sentence" => $sql, "error" => $e.getMessage());
// and you have to encode the data and also exit the code.
exit(json_encode($data));
}
Now in you JS code Ajax change to this.
$.ajax({
type: "POST",
dataType: "json",
url: "delete_list.php",
data: data,
beforeSend: function() {
$( "#" + data["idc"] ).animate({'backgroundColor':'#fb6c6c'},600);
},
success: function(response) {
// the variable response is the data returned from 'delete_list.php' the JSON
// now validate if the data returned run well
if (response.success) {
$( "#" + response.idc ).slideUp(600,function() {
$( "#" + response.idc ).remove();
});
} else {
console.log("An error has ocurred: sentence: " + response.sentence + "error: " + response.error);
}
},
// add a handler to error cases.
error: function() {
alert("An Error has ocurred contacting with the server. Sorry");
}
});

POST session variables to a relational database with ajax

I am trying to add some data to a relational database, and would like the session_user_id to be the foreign key for that database. When a user clicks a button, I want to make a database entry with the session_user_id and some other information I have POSTed to the page. My ajax posts to the php webpage page which it is run on (meaning all my scripts are on the same page)
I am currently getting a Uncaught ReferenceError: $sess_user_id1 is not defined. The jquery is firing. While I would love to get the undefined variable fixed, overall this does not seem like a very direct way to to this, and has added a bunch of confusing variables, when all the variables I need were already in my PHP statement. Is there any way to trigger the PHP entry without going through ajax and having to define the variables again?
Here is my php, which is at the header which is on the same page as my JS and HTML:
<?php
$markerid = $_POST["id"];
$name = $_POST["name"];
$type = $_POST["type"];
$point = $_POST["point"];
$lat2 = $_POST["lat"];
$lng2 = $_POST["lng"];
$locationdescription = $_POST["locationdescription"];
$locationsdirections = $_POST["locationdirections"];
session_start();
if (!isset($_SESSION['sess_user_id']) || empty($_SESSION['sess_user_id'])) {
// redirect to your login page
exit();
}
$sess_user_id1 = $_SESSION['sess_user_id'];
if ((isset($_POST['usid'])) && (isset($_POST['usid']))) {
$user_id_follow = strip_tags($_POST['usid']);
echo $user_id_follow;
$query = "INSERT INTO markerfollowing ( userID, markerID, type )
VALUES ('$user_id_follow', '$markerid', '$type');";
$result = mysql_query($query);
if (!$result) {
die('Invalid query: ' . mysql_error());
}
mysql_close();
}
?>
Here is the HTML button:
<div class="btn pull-right">
<button class="btn btn-large btn-followmarker" type="submit"id="followmarker">Add me to the list</button>
</div>
Here is the jquery/ajax post:
<script/javascript>
$(document).ready(function () {
$("#followmarker").click(function(){
$.ajax({
type: "POST",
url: "", //
data: { usid: <?php echo '$sess_user_id1'; ?>},
success: function(msg){
alert("success");
$("#thanks").html(msg);
},
error: function(){
alert("failure");
}
});
});
});
</script
A sincere thanks for any and all help. I haven't worked with relational databases before.

<?php echo '$sess_user_id1'; ?>
is wrong. If you wont to get
data: { usid: 123} at $sess_user_id1 is 123, you should write
data: { usid: <?php echo "$sess_user_id1"; ?>}
See your html source code in your brawser. I think there is data: { usid: $sess_user_id1}, and javascript is not understand what is the $sess_user_id1
This is the only one problem that I can see now, but I don't understand your current task whole to say more.