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POST-GET session variable (javascript-php) -javascript takes the previous variable but not last created

From the selected value (from the form) I create a variable (var parcela).
var parcela;
$(document).ready(function(){
parcela = localStorage.getItem("parcela");
if (parcela !== '') {
$('#parcela').val(parcela);
}
$("#parcela").on('change',function() {
selectBoxVal_1 = $('#parcela').val();
if (typeof(Storage) !== "undefined") {
localStorage.setItem("parcela", selectBoxVal_1);
} else {
alert('Sorry! No Web Storage support..');
}
location.reload();
});
});
From the created variable (parcela), I create a session variable in PHP.
$.post("phpscripts/session.php", {"parc_id": parcela});
PHP (session.php)
<?php
session_start();
$parcela = $_POST["parc_id"];
$parcela_int = (int)$parcela;
if($_POST){
$_SESSION['parcela_id'] = $parcela_int;
}
?>
After that, the created session variable urge to another php script
query.php
<?php
session_start();
require("common.php");
$user_id = htmlentities($_SESSION['user']['id_korisnika']);
$parc = $_SESSION['parcela_id'];
try
{
$stmt = $db->prepare("SELECT y_cent, x_cent FROM parcele WHERE id_korisnika='$user_id' AND id_parcele='$parc' ");
$stmt->execute();
}
catch(PDOException $ex)
{
die("Failed to run query: " . $ex->getMessage());
}
$rows = $stmt->fetchAll();
....
This all works perfectly!
However, when I call a php script with query (query.php) in javascript, there is a problem. JS takes the previous session variable instead of the last selected.
$.ajax({
url: 'phpscripts/query.php',
type: 'GET',
success : function(data) {
chartData = data;
//console.log(chartData);
...
Does anyone know what the problem is? I'm trying for two days to solve this ...
Note: The javascript code is contained in a single script.

I solved the problem. I had to extract part of javascript code that calls the php script into a separate script. I called the new JS script with jQuery getScript() Method.
Thank you #knets.

asynchronous commenting using ajax

I'm trying to create a comment system on my website where the user can comment & see it appear on the page without reloading the page, kind of like how you post a comment on facebook and see it appear right away. I'm having trouble with this however as my implementation shows the comment the user inputs, but then erases the previous comments that were already on the page (as any comments section, I'd want the user to comment and simply add on to the previous comments). Also, when the user comments, the page reloads, and displays the comment in the text box, rather than below the text box where the comments are supposed to be displayed. I've attached the code. Index.php runs the ajax script to perform the asynchronous commenting, and uses the form to get the user input which is dealt with in insert.php. It also prints out the comments stored in a database.
index.php
<script>
$(function() {
$('#submitButton').click(function(event) {
event.preventDefault();
$.ajax({
type: "GET",
url: "insert.php",
data : { field1_name : $('#userInput').val() },
beforeSend: function(){
}
, complete: function(){
}
, success: function(html){
$("#comment_part").html(html);
window.location.reload();
}
});
});
});
</script>
<form id="comment_form" action="insert.php" method="GET">
Comments:
<input type="text" class="text_cmt" name="field1_name" id="userInput"/>
<input type="submit" name="submit" value="submit" id = "submitButton"/>
<input type='hidden' name='parent_id' id='parent_id' value='0'/>
</form>
<div id='comment_part'>
<?php
$link = mysqli_connect('localhost', 'x', '', 'comment_schema');
$query="SELECT COMMENTS FROM csAirComment";
$results = mysqli_query($link,$query);
while ($row = mysqli_fetch_assoc($results)) {
echo '<div class="comment" >';
$output= $row["COMMENTS"];
//protects against cross site scripting
echo htmlspecialchars($output ,ENT_QUOTES,'UTF-8');
echo '</div>';
}
?>
</div>
insert.php
$userInput= $_GET["field1_name"];
if(!empty($userInput)) {
$field1_name = mysqli_real_escape_string($link, $userInput);
$field1_name_array = explode(" ",$field1_name);
foreach($field1_name_array as $element){
$query = "SELECT replaceWord FROM changeWord WHERE badWord = '" . $element . "' ";
$query_link = mysqli_query($link,$query);
if(mysqli_num_rows($query_link)>0){
$row = mysqli_fetch_assoc($query_link);
$goodWord = $row['replaceWord'];
$element= $goodWord;
}
$newComment = $newComment." ".$element;
}
//Escape user inputs for security
$sql = "INSERT INTO csAirComment (COMMENTS) VALUES ('$newComment')";
$result = mysqli_query($link, $sql);
//attempt insert query execution
//header("Location:csair.php");
die();
mysqli_close($link);
}
else{
die('comment is not set or not containing valid value');
}
The insert.php takes in the user input and then inserts it into the database (by first filtering and checking for bad words). Just not sure where I'm going wrong, been stuck on it for a while. Any help would be appreciated.

There are 3 main problems in your code:
You are not returning anything from insert.php via ajax.
You don't need to replace the whole comment_part, just add the new comment to it.
Why are you reloading the page? I thought that the whole purpose of using Ajax was to have a dynamic content.
In your ajax:
$.ajax({
type: "GET",
url: "insert.php",
data : { field1_name : $('#userInput').val() },
beforeSend: function(){
}
, complete: function(){
}
, success: function(html){
//this will add the new comment to the `comment_part` div
$("#comment_part").append(html);
}
});
Within insert.php you need to return the new comment html:
$userInput= $_GET["field1_name"];
if(!empty($userInput)) {
$field1_name = mysqli_real_escape_string($link, $userInput);
$field1_name_array = explode(" ",$field1_name);
foreach($field1_name_array as $element){
$query = "SELECT replaceWord FROM changeWord WHERE badWord = '" . $element . "' ";
$query_link = mysqli_query($link,$query);
if(mysqli_num_rows($query_link)>0){
$row = mysqli_fetch_assoc($query_link);
$goodWord = $row['replaceWord'];
$element= $goodWord;
}
$newComment = $newComment." ".$element;
}
//Escape user inputs for security
$sql = "INSERT INTO csAirComment (COMMENTS) VALUES ('$newComment')";
$result = mysqli_query($link, $sql);
//attempt insert query execution
mysqli_close($link);
//here you need to build your new comment html and return it
return "<div class='comment'>...the new comment html...</div>";
}
else{
die('comment is not set or not containing valid value');
}
Please note that you currently don't have any error handling, so when you return die('comment is not set....') it will be displayed as well as a new comment.
You can return a better structured response using json_encode() but that is outside the scope of this question.

You're using jQuery.html() which is replacing everything in your element with your "html" contents. Try using jQuery.append() instead.

How can I send javascript values to PHP script?

this is my php code that creates a table using the results of a mysql query:
echo "<table id='table' class='selectQuery'>
while($row = mysqli_fetch_array($slctQuery)) {
// ; echo $row['id']; echo
echo "<tr class='someClass' idNumber="; echo $row['id']; echo ">
<td>";
echo $row['fname'];
echo "</td>
<td>";
echo $row['lname'];
echo "</td>;
</tr>";
}
echo "</table>";
and this part is my jquery code for changing style on click on table row:
<script>
$(document).ready(function(){
$("#table tr").click(function(){
$('.someClass').removeClass('selected');
$(this).addClass('selected');
idNum = $(this).attr('idNumber');
});
$("#table tr").click(function(){
$("#DelEdtQuestion").addClass('selected1');
});
});
</script>
and this part is for style:
<style>
tr.selected {
background-color: brown !important;
color: #FFF;
}
</style>
and this is my php code for button
if(#$_POST['Search']){
/// what should I do?
}
So, now I want have my idNum value when my search button in form was clicked.
thanks for attentions

You can use ajax. If you have a form with id="myform" and (example) input fields: firstname, lastname, username and password, the following script should send data to the php:
$(document).ready(function(){
var datastring = $("#myform").serialize();
$.ajax({
type: 'POST',
url: 'ajaxfile.php',
data: datastring
}).done(function(res){
var res = $.trim(res);
alert(res);
});
});
The ajaxfile.php can be something like that:
<?php
$firstname = mysql_real_escape_string($_POST["firstname"]);
$lastname = mysql_real_escape_string($_POST["lastname"]);
$username = mysql_real_escape_string($_POST["username"]);
$password = mysql_real_escape_string($_POST["password"]);
//here you have the variables ready to do anything you want with them...
//for example insert them in mysql database:
$ins = "INSERT INTO users (firstname, lastname, username, password ) VALUES ( '$firstname', '$lastname', '$username', '$password' )";
if(mysql_query($ins)){echo "SUCCESS";}else{echo "FAILURE";}
?>
Another example, similar to yours, is to take the row id from your table, pass it to ajax, have ajax (for example) make a query to the database and return the results:
// your script, modified for ajax:
$(document).ready(function(){
$("#table tr").click(function(){
$('.someClass').removeClass('selected');
$(this).addClass('selected');
var idNum = $(this).attr('idNumber'); //use "var" to -initially- set the variable
$.ajax({
type: 'POST',
url: 'ajaxfile.php',
data: 'id='+idNum
}).done(function(res){
var res = $.trim(res);
alert(res);
});
});
$("#table tr").click(function(){
$("#DelEdtQuestion").addClass('selected1');
});
});
Modified ajaxfile.php to suit the above example:
<?php
$id = mysql_real_escape_string($_POST["id"]);
//query database to get results:
$result = "SELECT * FROM `users` WHERE `id` = '$id' LIMIT 1";
$row = mysql_fetch_assoc($result);
echo "Username: ".$row["username"]."Password: ".$row["password"]."Firstname: ".$row["firstname"]."Lastname: ".$row["lastname"].
?>
Since your question was rather ambigious, I put more effort to give you an idea about the basics of ajax so that you work out your own solution, rather than to suggest a potential solution -that at the end could not be what you were looking for...
And since we are talking about ajax basics, it is a good practice to secure your ajax files since they are accessible from any browser:
in the very beginning of any ajax file, right below the "?php" tag, you can add these lines below, to protect the file from being accessed by browser -but remain accessible to ajax calls:
//protect the file from un-authorized access
define('AJAX_REQUEST', isset($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest');
if(!AJAX_REQUEST) {die();}
Hope that helps you and others. T.
UPDATE:
It is ALWAYS a good practice to keep your php and javascript files separately... In the above examples there are ideally 3 files involved: the main php file, the scripts file and the ajax-php file.
So -preferably after the "body" tag of your "main" php file- you should include the scripts-file (after the jquery ofcourse!). Like that:
<!-- jQuery v.1.11.3-->
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<!-- include scripts file -->
<?php include("scripts.php"); ?>
(notice that for jquery I use the regular "script" tags but for the scripts file I just do a "php include").
As you see above, the javascript file has also ".php" extension (not ".js"). This is a "trick" I like to do because it gives me the ability to execute php code within the js file. Of course, all javascript code in that file is included between "script" tags.
example of a hypothetical "scripts.php":
<script>
// I create a js variable that takes value from php
var phpDate = '<?php date("Y-m-d"); ?>';
alert(phpDate);
//or pass the contents of another php variable in your app to javascript:
var myPhpVar = '<?php echo $my_php_var; ?>';
//or put a php SESSION to a js variable:
var mySess = '<?php echo $_SESSION["my_session"]; ?>';
</script>
The above comes quite handy sometimes when you want to pass to javascript php variables that already exist in your application.
It is a very long answer (more like a tutorial!)... But now should be quite clear to you how to pass values not only from js to php but also vice versa!!!

Getting a value from Javascript in PHP [duplicate]

This question already has answers here:
What is the difference between client-side and server-side programming?
(3 answers)
Closed 7 years ago.
I am not very experienced in web programming and am attempting to run a script which updates my database.
<script>
function myFunction() {
var texts = document.getElementById("content").textContent;
alert(texts)
<?php
include_once 'accounts/config.php';
$text = ...;
$tbl_name='enemies'; // Table name
$query = "UPDATE enemies SET text=('$text') WHERE id=1";
$result = mysql_query($query) or die(mysql_error());
?>
}
</script>
I have no idea what to put in the $text section as shown with $text = ...; in order to get the variable texts from above.
EDIT
I have updated my code but the function does not seem to be accessing the PHP file. I am using a button to call the function and I have also tested it so i know the function is being called. My file is called update.php and is in the same directory as this file.
<button onclick="myFunction()">Click This</button>
<script>
function myFunction() {
var texts = document.getElementById("content").textContent;
$.ajax({
url: "update.php",
type: "POST",
data: {texts:texts},
success: function(response){
}
});
}
</script>

you can post your $texts value to other php page using ajax and get the variable on php page using $_POST['texts'] and place update query there and enjoy....
function myFunction() {
var texts = document.getElementById("content").textContent;
$.ajax({
url: 'update.php',
type: "POST",
data: {texts:texts},
success: function(response)
{
}
});
And your php file will be named as update.php
<?php
include_once 'accounts/config.php';
$text =$_POST['texts'];
$tbl_name='enemies'; // Table name
$query = "UPDATE `enemies` SET `text`='".$text."' WHERE `id`=1";
$result = mysql_query($query) or die(mysql_error());
?>

PHP runs on the server and then generates output which is then returned to the client side. You can't have a JavaScript function make a call to inlined PHP since the PHP runs before the JavaScript is ever delivered to the client side.
Instead, what you'd need to do is have your function make an AJAX request to a server-side PHP script that then extracts the data from the request body and then stores it in the database.
PHP: "/yourPhpScript.php"
<?php
include_once 'accounts/config.php';
$text = $_POST['data'];
$tbl_name='enemies'; // Table name
$query = "UPDATE enemies SET text='".$text.'" WHERE id=1";
$result = mysql_query($query) or die(mysql_error());
?>
JavaScript:
function myFunction() {
var texts = document.getElementById("content").textContent;
alert(texts);
// append data as a query string
var params = 'data='+texts;
var xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
// when server responds, output any response, if applicable
if(http.readyState == 4 && http.status == 200) {
alert(http.responseText);
}
}
// replace with the filename of your PHP script that will do the update.
var url = '/yourPhpScript.php';
xmlhttp.open("POST", url, true);
xmlhttp.send(params);
}
A word of caution: This is not a safe, production-friendly way of updating data in your database. This code is open to SQL injection attacks, which is outside the scope of your question. Please see Bobby Tables: A guide to preventing SQL injection if you are writing code that will go into production.

You are wrong in approach
You should use ajax to post 'texts' value to your php script
https://api.jquery.com/jquery.post/ and create separate php file where you will get data from ajax post and update DB
javascript:
<script>
function myFunction() {
var texts = document.getElementById("content").textContent;
$.ajax({
type: "POST",
url: "update.php",
data: "texsts=" + texts,
success: success
});
}
</script>
update.php
<?php
include_once 'accounts/config.php';
$text = $_POST['texts'];
$tbl_name='enemies'; // Table name
$query = "UPDATE enemies SET text=('$text') WHERE id=1";
$result = mysql_query($query) or die(mysql_error());
?>

i will use PDO if i was you, what you do mysql_query are outdated, if you use my framework https://github.com/parisnakitakejser/PinkCowFramework you can do the following code.
<?php
include('config.php');
$text = $_POST['text'];
$query = PinkCow\Database::prepare("UPDATE enemies SET text = :text WHERE id = 1");
$bindparam = array(
array('text', $text, 'str')
);
PinkCow\Database::exec($query,$bindparam);
$jsonArray = array(
'status' => 200
);
echo json_encode($jsonArray);
?>
place this code in jsonUpdateEnemies.php file and call it width jQuery
<script>
function myFunction(yourText) {
$.post( 'jsonUpdateEnemies.php', {
'text' : yourText
}, function(data)
{
alert('Data updated');
},'json');
}
</script>
its a little more complex then you ask about, but its how i will resolved your problem, :)

Using ajax to display new database inputs without refreshing the page

I am using ajax to post comments to a certain page, I have everything working, except for when the user posts a comment I would like it to show immediately without refreshing. The php code I have to display the comments is:
<?php
require('connect.php');
$query = "select * \n"
. " from comments inner join blogposts on comments.comment_post_id = blogposts.id WHERE blogposts.id = '$s_post_id' ORDER BY comments.id DESC";
$result = mysql_query($query);
while ($row = mysql_fetch_array($result)) {
$c_comment_by = $row['comment_by'];
$c_comment_content = $row['comment_content'];
?>
<div class="comment_box">
<p><?php echo $c_comment_by;?></p>
<p><?php echo $c_comment_content;?></p>
</div>
<?php } ?>
</div>
</div>
<?php
}
}
and the code I have to post comments is:
<?php
$post_comment = $_POST['p_post_comment'];
$post_id = $_POST['p_post_id'];
$post_comment_by = "Undefined";
if ($post_comment){
if(require('connect.php')){
mysql_query("INSERT INTO comments VALUES (
'',
'$post_id',
'$post_comment_by',
'$post_comment'
)");
echo " <script>$('#post_form')[0].reset();</script>";
echo "success!";
mysql_close();
}else echo "Could no connect to the database!";
}
else echo "You cannot post empty comments!"
?>
JS:
function post(){
var post_comment = $('#comment').val();
$.post('comment_parser.php', {p_post_comment:post_comment,p_post_id:<?php echo $post_id;?>},
function(data)
{
$('#result').html(data);
});
}
This is what I have for the refresh so far:
$(document).ready(function() {
$.ajaxSetup({ cache: false });
setInterval(function() {
$('.comment_box').load('blogpost.php');
}, 3000);.
});
Now what I want to do is to use ajax to refresh the comments every time a new one is added. Without refreshing the whole page, ofcourse. What am I doing wrong?

You'll need to restructure to an endpoint structure. You'll have a file called "get_comments.php" that returns the newest comments in JSON, then call some JS like this:
function load_comments(){
$.ajax({
url: "API/get_comments.php",
data: {post_id: post_id, page: 0, limit: 0}, // If you want to do pagination eventually.
dataType: 'json',
success: function(response){
$('#all_comments').html(''); // Clears all HTML
// Insert each comment
response.forEach(function(comment){
var new_comment = "<div class="comment_box"><p>"+comment.comment_by+"</p><p>"+comment.comment_content+"</p></div>";
$('#all_comments').append(new_comment);
}
})
};
}
Make sure post_id is declared globally somewhere i.e.
<head>
<script>
var post_id = "<?= $s_post_id ; ?>";
</script>
</head>
Your new PHP file would look like this:
require('connect.php');
$query = "select * from comments inner join blogposts on comments.comment_post_id = blogposts.id WHERE blogposts.id = '".$_REQUEST['post_id']."' ORDER BY comments.id DESC";
$result = mysql_query($query);
$all_comments = array() ;
while ($row = mysql_fetch_array($result))
$all_comments[] = array("comment_by" => $result[comment_by], "comment_content" => $result[comment_content]);
echo json_encode($all_comments);
Of course you'd want to follow good practices everywhere, probably using a template for both server & client side HTML creation, never write MySQL queries like you've written (or that I wrote for you). Use MySQLi, or PDO! Think about what would happen if $s_post_id was somehow equal to 5' OR '1'='1 This would just return every comment.. but what if this was done in a DELETE_COMMENT function, and someone wiped your comment table out completely?