We Keep Coding

External PHP file not get variable from PHP link

Building a messaging system for my site and i have been stuck for days. I have this PHP link
<a href='user_msg.php?hash=$hash'>$name</a>
When you click on the link, it takes you to a page where you can send message to a user you've connected to (this connection is binded by the $hash)
in the page for sending the message, i hid the $hash in and hidden input value="$hash" and it sends the message to row in database with the $hash with the following scripts (They have no issue and work fine)
var msg_area = $('.msg_area');
msg_area.scrollTop(msg_area.prop("scrollHeight"));
$('#send_rep').submit(function (e) {
e.preventDefault();
var $form = $(this), url = $form.attr('action');
var posting = $.post(url, {rep_msg: $('#rep_msg').val(), hash: $('#hash').val()});
posting.done(function (data) {
alert('success');
});
});
PHP script to send
require_once ("db.php");
$db = new MyDB();
session_start();
if (isset($_POST['rep_msg']) && !empty($_POST['rep_msg']) || isset($_POST['hash']) && !empty($_POST['hash']))
{
$hash = (int)$_GET['hash'];
$my_id = $_SESSION['log_id'];
$rep_msg = $_POST['rep_msg'];
$hash = $_POST['hash'];
$rsql = <<<EOF
INSERT INTO messager (message, group_hash, from_id) VALUES('$rep_msg', '$hash', '$my_id');
EOF;
$rret = $db->exec($rsql);
$ursql = <<<EOF
SELECT * FROM User WHERE ID = '$my_id';
EOF;
$urret = $db->query($ursql);
while ($urrow = $urret->fetchArray(SQLITE3_ASSOC)) {
$from_fname = $urrow['fname'];
$from_img = $urrow['image'];
header('Location: user_msg.php?hash=' . $hash);
}
}
The above Ajax Request and php script work fone to sedn the messages to database.
The issue not is getting the messages from database
This is the script i am currently using (not working)
PHP script to get message
require_once ("db.php");
$db = new MyDB();
session_start();
if (isset($_GET['hash']) && !empty($_GET['hash']))
{
$hash = (int)$_GET['hash'];
$us_id = $_SESSION['log_id'];
$mesql =<<<EOF
SELECT from_id, message FROM messager WHERE group_hash = '$hash';
EOF;
$meret = $db->query($mesql);
while ($merow = $meret->fetchArray(SQLITE3_ASSOC))
{
$from_id = $merow['from_id'];
$messages = $merow['message'];
$usql =<<<EOF
SELECT * FROM User WHERE ID = '$from_id';
EOF;
$uret = $db->query($usql);
while ($urow = $uret->fetchArray(SQLITE3_ASSOC)) {
$from_fname = $urow['fname'];
$from_img = $urow['image'];
if ($from_id != $_SESSION['log_id']) {
echo "
<div class='from_bubble'><div class='from_img'><img src='$from_img'></div><div class='from_txt'><p>$messages</p></div></div>";
} else {
echo "
<div class='rep_bubble'><div class='rep_img'><img src='$from_img'></div><div class='rep_txt'><p>$messages</p></div></div>";
}
}
echo "<input style='display: none' type='text' class='hash' name='hash' value='$hash' id='hash'>";
}
}
Ajax Request
setInterval(function() {
$('.msg_area').load("get_msg.php");
}, 2000);
But the get is not working. I suspect fro some reason, its not getting the $hash. Please is there a solution to this or am i trying something impossible.
Any help would be appreciated. If more info is needed please ask. Thanks in advance

Maybe $hash = (int)$_GET['hash']; would be $hash = (int)$_POST['hash'];
And you have more than just one $_GET...

In your ajax request you are using POST so you will need to get the hash with the POST as well: if (isset($_POST['hash']) && !empty($_POST['hash']))
{
$hash = (int)$_POST['hash'];

Using '' in PHP indicates the actual string you are writing.
If you want to use a PHP variable in a string you should use "", so try changing $name
Check this for more info:
http://php.net/manual/en/language.types.string.php
Then, how do you hide your hash in the input?
You could use <input value="$_GET['hash']" ... />
In the script to send , does it enter the if? Try adding statements to see if SQL returns any error.
Hope this helps.

Answer to the question was to create a PHP SESSION for the $hash($_SESSION['hash'] = $hash) and to use this session all over the site with session_start().

AJAX request to fill a popover working only once to a php file

I have made a simple search bar in which on every .keyup() event,an asynchronous request goes to a php file which then fills the data in the bootstrap popover.
The problem is that in the popover,the data is filled only once,i.e.,when I type the first character,after that the same data is shown even after multiple .keyup() events.
Here is the code:
HTML:
<input type="text" data-placement="bottom" id="search" name="search1" class="search-box" placeholder="Search..." title="Results"/>
AJAX:
$("#search").keyup(function(){
console.log('erer');
//var searchString = $("#search").val();
var data = $("#search").val();
console.log(data);
var e=$(this);
//if(searchString) {
$.ajax({
type: "POST",
url: "do_search.php",
data: {search:data},
success: function(data, status, jqXHR){
console.log('html->'+data+'status->'+status+'jqXHR->'+jqXHR);
e.popover({
html: true,
content: data,
}).popover('show');
},
error: function() {
alert('Error occured');
}
});
//}
});``
PHP:
$word = $_POST['search'];
//echo $word;
//$word=htmlentities($word)
$sql = "SELECT FName FROM user WHERE FName LIKE '%$word%' ";
//echo $sql;
// get results
//$sql = 'SELECT * FROM Profiles';
$end_result = '';
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
#$end_result.='<li>'.$row["FName"].'</li>';
#$_SESSION['fnamer'] = $row['Fname'];
#$end_result.='<li>'.'<a href =view_search.php>'.$row["Fname"].'</a>'.'</li>';
echo '<a href =view_search.php>'.$row["FName"].'</a>';
}
#echo $end_result;
}
Even though in the success parameter of the $.ajax,the data is being printed fine,i.e.,it changes as I enter different alphabets,but the popover content does not change.
Please provide some suggestions to resolve this problem.

The popover is already shown. This is not the correct way of changing the popover content dynamically.
Your code: https://jsfiddle.net/gsffhLbn/
Instead, address the content of the popover directly:
var popover = e.attr('data-content',data);
popover.setContent();
Working solution
Fiddle: https://jsfiddle.net/gsffhLbn/1/

Ajax live search results not shown on my page

I got a simple ajax live search script that works fine when I type the keyword in my url and visit the php file. But for some reason when I type the keyword in my input field, nothing happens.
What am I doing wrong?
My input field on products.php:
<input type="search" name="keyword" class="producten-icon divider" placeholder="Zoeken..." id="s_search">
And further down the page I got my result div:
<div id="results"></div>
My ajax script:
<script>
$(document).ready(function () {
$("#s_search").on('keyup',function () {
var key = $(this).val();
$.ajax({
url:'includes/fetch_results.php',
type:'GET',
data:'keyword='+key,
beforeSend:function () {
$("#results").slideUp('fast');
},
success:function (data) {
$("#results").html(data);
$("#results").slideDown('fast');
}
});
});
});
</script>
My fetch_results.php:
<?php
include 'connection.php';
$conn = new Connection;
if($_GET['keyword'] && !empty($_GET['keyword']))
{
// Results names
$results = "SELECT `naam` FROM `producten` WHERE `naam` LIKE '%".$_GET['keyword']."%'";
$resultscon = $conn->query($results);
$resultscr = array();
while ($resultscr[] = $resultscon->fetch_assoc());
$eend = #array_map('current', $resultscr);
// echo '<pre>';
// print_r($eend);
// echo '</pre>';
$resultsoverzicht .= '<div style="height:100%;border:10px solid red;">';
foreach($eend as $result){
$resultsoverzicht .= '
<p>'.$result.'</p>';
}
$resultsoverzicht .= '</div>';
echo $resultsoverzicht;
};
?>
When I use the network inspector I don't see anything posted when typing in the input field. Which should be the case with keyup right?

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.

Check/Set button status on page load

I am using a custom mvc framework and have added a favourite button. Once pressed this displays a 'successfully added to favourites' div, when clicked again it displays a 'successfully removed from favourites' div.
My query works fine, adding and deleting from my favourite table as it should.
What I would like to do now is change the state of the button depending on the selection. For example, if the user has the book in their favourites add btn-success class, if the user hasn't, use the btn-default class.
I'm not sure the best way to approach this. I'm new to php and js so any advice or direction is appreciated. I have tried adding toggleClass to my JS but it's not working. Do I need to perform a query/check on pageLoad?
I have included my code below for reference.
itemView.php
echo
'<td>
<button id="fav" value="'.$book->id.'" type="button" class="btn btn-default"></button>
</td>';
JS (in itemView.php)
$(document).ready(function(){
$( "#fav" ).click(function(){
book_id = $(fav).val();
$.ajax({
type: 'POST',
url: '<?php echo URL; ?>books/checkFav',
data: {book_id:book_id},
success: function () {
window.location.reload(true);
$("#fav").addClass( "btn-success" );
}//end success
});//end ajax
});
});
my checkFav function
public function checkFav($bookid,$userid)
{
$bookid=$_REQUEST['book_id'];
$userid=$_SESSION['user_id'];
$sql = "SELECT * FROM favourite WHERE book_id = :book_id AND user_id = :user_id";
$query = $this->db->prepare($sql);
$query->bindParam(':user_id', $userid);
$query->bindParam(':book_id', $bookid);
$query->execute();
$rows_found = $query->fetchColumn();
if(empty($rows_found)) {
$sql = "INSERT INTO favourite (book_id, user_id) VALUES (:book_id, :user_id)";
$query = $this->db->prepare($sql);
$query->bindParam(':user_id', $userid);
$query->bindParam(':book_id', $bookid);
$query->execute();
if ($query->rowCount() == 1) {
// successful add to favs
$_SESSION["feedback_positive"][] = FEEDBACK_ADDED_TO_FAVS;
return true;
}
} else {
$sql = "DELETE FROM favourite WHERE book_id = :book_id AND user_id = :user_id";
$query = $this->db->prepare($sql);
$query->bindParam(':user_id', $userid);
$query->bindParam(':book_id', $bookid);
$query->execute();
if ($query->rowCount() > 0) {
// successful remove from favs
$_SESSION["feedback_negative"][] = FEEDBACK_REMOVED_FROM_FAVS;
return true;
}
}
}

Use session variable in the ajax request script and using that session variable in page where button exist you can play with button css. for example:
Put this code where Button exists.
$css = "btn_default";
if($_SESSION['btnClicked'] == "success") {
$css = "btn_success";
}
Use $css variable in the button class like--
<button id="fav" value="'.$book->id.'" type="button" class="btn <?php echo $css?>"></button>
This session will manage in the ajax script where in you are adding and deleting favourite.
set session value
$_SESSION['btnClicked'] = 'success'
below the line
$_SESSION["feedback_positive"][] = FEEDBACK_ADDED_TO_FAVS;
and unset the session
unset($_SESSION['btnClicked']);
after the line.
$_SESSION["feedback_negative"][] = FEEDBACK_REMOVED_FROM_FAVS;