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What is the best way to enter data obtained from a database in PHP into seperate html <div>'s?

I'm currently working on a Facebook like chat, with 3 different chat boxes that should work simultaneously. I can send and read messages from my database, but I'm having difficulty displaying this information in the right place. In chat.php I have this snippet of code:
$.ajax({
url: "fetch_user_chat_history.php",
method: "POST",
data: jQuery.param({receiver_id:receiver_id, num:num}),
success: function(data) {
$('$chat_history_'+receiver_id).html(data);
}
});
Now I am able to read the data from my database correctly in fetch_user_chat_history.php, but when I iterate over my messages I'm unable to output them correctly back to chat.php. Here is my fetch_user_chat_history.php:
<?php
include "opendb.php";
session_start();
$output1 = '';
$output2 = '';
$increment = 0;
$sender = $_SESSION['user_id'];
$receiver_id = $_POST['receiver_id'];
$chatboxnum = $_POST['num'];
$query = 'SELECT content, timestamp_chat, sender_id FROM messages WHERE (sender_id = '.$sender.' AND receiver_id = '.$receiver_id.' OR (sender_id = '.$receiver_id.' AND receiver_id = '.$sender.'))';
$statement = $db->prepare($query);
$statement->execute();
$result = $statement->fetchAll(PDO::FETCH_ASSOC);
switch ($chatboxnum) {
case 1:
foreach ($result as $row) {
if ($row['sender_id']==$sender) {
echo '<script>var para = document.createElement("div");';
echo 'para.innerHTML = '.$row["content"].';';
echo 'var att = document.createAttribute("class");';
echo 'att.value = "msg-send";';
echo 'para.setAttributeNode(att);';
echo 'document.getElementById("sendbox-1").appendChild(para);';
echo 'document.body.appendChild(para); </script>';
}
else {
echo "var element = document.getElementById('receivebox-1'); element.classList.add('msg-receive');";
}
$increment += 1;
}
break;
}
?>
None of my echo statements add anything to chat.php. I'm sure there's a very easy fix as I have done this before but I can't seem to get it working. I apologise for the long post.

ajax not work and not show php data

when one record then show data when multiple record come then not show data other site.
ajaxx.php
<?php
include 'database.php';
session_start();
$post = $_POST;
$search = $post['search'];
$searchType = $post['searchType'];
if ($searchType == 'all')
{$sql = "SELECT DISTINCT title FROM hadees WHERE title LIKE '$search%' AND (type='Bukhari' OR type='Muslim') ";}
else
{$sql = "SELECT DISTINCT title FROM hadees WHERE title LIKE '$search%' AND type='$searchType' ";}
$result = mysqli_query($db,$sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$row['title'];
echo json_encode($row);
}
} else
{ echo "Not Found Result" ; }
?>
when data record is one then append the data successfully when multiple record come then not show data and append not work
javascript code
function searchh()
{
var type = $("input[name='type']:checked").val();
var searchhh = $( ".myButton option:selected" ).text();
debugger;
$.ajax({
url: 'ajaxx.php',
type: "POST",
data: {'searchType':type, 'search':searchhh},
success: function (data) {
var duce = jQuery.parseJSON(data);
alert(duce.title);
}
});
}

I think your issue is in the while loop. You don't want to encode each row one-by-one, but as a whole like this.
$myResults = [];
while($row = $result->fetch_assoc()) {
$row['title'];
$myResults[] = $row;
}
echo json_encode($myResults);

You are producing invalid JSON by using echo json_encode($row); within a loop.
Try to craft an array of rows, and then display it.
if($result->num_rows > 0)
{
$output = array();
while($row = $result->fetch_assoc())
{
output[] = $row;
}
if($searchType == 'all')
{
echo json_encode($output);
}
else
{
echo json_encode(current($output)); // print just one
}
}

move some variable from one php page to onother and move between to page

I'm new at php. I get the last id from database. For each id I want the state and the link. I'll check if state == 1, then get the content of the link (there's JavaScript variable that I need that is in the content of link). I'll send that variable with location.href.
Then I get that variable with $_GET in the second page. I want to store that var into database, then come back to first page and get the second link from the database and again do the same works.
How can I send the $j into second page for saving the x_cor and y_cor, and how to increase the $j when it comes to first page again?
<html>
<head>
<title>firstpage</title>
</head>
<body>
<?php
include_once ('simple_html_dom.php');
include_once ('split_unicode_function.php');
// getting the last id from db
$connection = #mysql_connect("localhost", "root", "kkuser");
$select_db = mysql_select_db("kk", $connection);
$sql = "SELECT id FROM netbarg ORDER BY id DESC LIMIT 1";
$result = mysql_query($sql, $connection);
$rows = mysql_fetch_array($result);
$last_id = $rows['id'];
// getting id and link for each column
for ($j = 1; $j <= 2; $j++)
{
$select_db = mysql_select_db("kk", $connection);
$id = "SELECT state FROM `table` WHERE id='$j' ";
$result = mysql_query($id, $connection);
$row = mysql_fetch_array($result);
echo $state = $row[0] . '<br />';
// getting link
$link = "SELECT link FROM `table` WHERE id='$j' ";
$result = mysql_query($link, $connection);
$rows = mysql_fetch_array($result);
$link = $rows[0];
// (state is just 1 or 2)check if the state is 2 or not...
if ($state == 1)
{
$f = file_get_contents($link);
echo "<div>$f</div>";
}
}
?>
<script>
$("body").hide();
location.href ='secondpage.php?val='+point0+;
</script>
</body>
</html>
second page
<html>
<head>
<title>second page</title>
</head>
<body>
<?php
if (isset($_GET['val']))
{
$hat = $_GET['val'];
echo $hat;
$coords = trim($hat, '()');
// echo $coords.'<br />';
$a = array();
$a = explode(",", $coords);
var_dump($a);
echo $long = $a[0];
echo '<br />';
if ($long == "undefined") $lat = "undefined";
else echo $lat = $a[1];
if ($_SERVER['REQUEST_METHOD'] == "GET")
{
$connection = #mysql_connect("localhost", "root", "kkuser");
$select_db = mysql_select_db("kk", $connection);
$update = "UPDATE `table` SET `x_cor`='$long',`y_cor`='$lat' , `state`='2' WHERE `id`='$j' ";
$insert_todb = mysql_query($update, $connection);
if ($insert_todb) echo "coordinates has been updated", '<br />';
}
}
?>
<script>
location.href ='firstpage.php';
</script>
</body>
</html>

if get the last insert id use mysqli_insert_id($con);
it return last insert id from database using this geted id you can

Assigning PHP Variables To Javascript Variables Not Working

I am currently having trouble with this. I would like to make one of my variables in Javascript have a PHP value. Here is what I mean:
<script>
JSvariable = <?php echo $PHPvariable; ?>;
</script>
For some reason that is not working. Here is my full (snippet) of code:
<script>
currentreplyid = <?php echo $allpostcomments[$key]['replyid']; ?>;
$('#parentcommentholder').val(currentreplyid);
</script>
I am sure it is some stupid mistake, but I can not seem to find it! What is the problem? Thank you!
PS #parentcommentholder is an input field, and it just had the value 0 after the field is supposed to of been changed.
Here is some source:
<?php
$postcommentsquery = "SELECT * FROM comments WHERE parent = :parent AND postid = :postid ORDER BY datecreated DESC";
$postcommentsparams = array(':parent' => $allreplies[$key]["postid"],
':postid' => $postid);
try{
$postcommentsstmt = $connection->prepare($postcommentsquery);
$postcommentsresult = $postcommentsstmt->execute($postcommentsparams);
}
catch(PDOException $ex){
echo ("Failed to run query: " . $ex->getMessage());
}
$allpostcomments = $postcommentsstmt->fetchAll();
foreach ($allpostcomments as $key => $value) {
?>
<script>
var currentreplyid = <?php echo $allpostcomments[$key]['replyid']; ?>;
$('#parentcommentholder').val(currentreplyid);
</script>
<input id="parentcommentholder"></div>

Don't forgot for give quotes ' or ". Use following:
<script>
var JSvariable = '<?php echo $PHPvariable; ?>';
//or
var JSvariable = "<?php echo $PHPvariable; ?>";
</script>
Reason: If php variable contains string and if while assigning it to javascript variable we shall not give quote like:
<?php $PHPvariable = 'String';?>
var JSvariable = <?php echo $PHPvariable; ?>;
Will transform into :
var JSvariable = String;//which will give error in javascript
But this will work fine if PHP variable contains a numeric value like:
<?php $PHPvariable = 2;?>
var JSvariable = <?php echo $PHPvariable; ?>;
Will transform into :
var JSvariable = 2;//which will work perfect
Complete code should be:
<script>
var currentreplyid = "<?php echo $allpostcomments[$key]['replyid']; ?>";
//or if you are sure your variable contains int value
var currentreplyid = parseInt("<?php echo $allpostcomments[$key]['replyid']; ?>");
$('#parentcommentholder').val(currentreplyid);
</script>

Try the below instead of using javascript (as I don't think you need it):
<?php
$postcommentsquery = "SELECT * FROM comments WHERE parent = :parent AND postid = :postid ORDER BY datecreated DESC";
$postcommentsparams = array(':parent' => $allreplies[$key]["postid"],
':postid' => $postid);
try{
$postcommentsstmt = $connection->prepare($postcommentsquery);
$postcommentsresult = $postcommentsstmt->execute($postcommentsparams);
}
catch(PDOException $ex){
echo ("Failed to run query: " . $ex->getMessage());
}
$allpostcomments = $postcommentsstmt->fetchAll();
foreach ($allpostcomments as $key => $value) {
?>
<input id="parentcommentholder" value="<?php echo ((int)$allpostcomments[$key]['replyid']>0) ? $allpostcomments[$key]['replyid'] : 0; ?>" />
<?php
}
?>
If your defiantly sure $allpostcomments[$key]['replyid'] is bringing back a value, this should work without any issues.

Echo alert if date is today

I am new to PHP, I am trying to echo a javascript alert box if taskDueDate is Today.
Any suggestions:
$varTaskAlert = mysql_query("SELECT * FROM tasks");
while ( $rows = mysql_fetch_array($varTaskAlert)) {
$varTaskID = $rows['taskid'];
$varTaskTitle = $rows['tasktitle'];
$varTaskDetail = $rows['taskdetail'];
$varTaskResource = $rows['taskresource'];
$varTaskDue = $rows['taskduedate'];
$varTaskStatus = $rows['taskstatus'];
$varTaskType = $rows['tasktype'];
$active = Active;
$curdate = date('y/m/d');
if ($varTaskStatus == $active && $varTaskDue == $curdate) {
echo '<script type="text/javascript">alert("The Task <?php echo $varTaskTitle ; ?> is Today.")</script>';
}
}

Please check date format of
$varTaskDue = $rows['taskduedate'];
Is $varTaskDue formate similar to 'y/m/d' format?
Hope this will help.

On first look you should change
echo '<script type="text/javascript">alert("The Task <?php echo $varTaskTitle ; ?> is Today.")</script>';
to
echo "<script type='text/javascript'>alert('The Task {$varTaskTitle} is Today.')</script>";

Your Echo is wrong. You are already in php you just include js alert so you can directly use that variable. Try this
echo "<script type='text/javascript'>alert('The Task {$curdate} is Today.')</script>";