We Keep Coding

Refresh page if there is change in database

I am trying to refresh my a page if there is a change in orderStatus from database using Ajax and PHP. I set the current orderStatus as predefined data and then use Ajax to get the current orderStatus from database and finally compare if they are not the same. I want to refresh the page if they are not the same.
PHP (autorefresh.php)
<?php
$orderId = $_POST["orderId"];
$query = "SELECT * FROM orderinhomeonlinecall WHERE orderId='$orderId'";
$result = mysqli_query($db, $query);
while($row = mysqli_fetch_array($result))
{
$orderStatus = $row['orderStatus'];
$data = array(
'orderStatus' => $orderStatus
);
echo json_encode($data);
}
?>
Javascript
<script type="text/javascript" >
var predefined_val = '<?php echo $orderStatus; ?>';// your predefined value.
$.document(ready(function(){
setInterval(function(){
$.ajax({
type:"POST",
url:"autorefresh.php", //put relative url here, script which will return php
data:{orderId: <?php echo $orderId; ?>}, // if any you would like to post any data
success:function(response){
var data = response; // response data from your php script
if(predefined_val !== data){
window.location.href=window.location.href;
}
}
});
},5000);// function will run every 5 seconds
}));

The below code should work, Need to mention dataType:"json" else use JSON.stringify(data) to parse response
<script type="text/javascript">
var predefined_val = '<?php echo $orderStatus; ?>';// your predefined value.
$(document).ready(function () {
setInterval(function () {
$.ajax({
type: "POST",
url: "autorefresh.php", //put relative url here, script which will return php
data: {orderId: <?php echo $orderId; ?>}, // if any you would like to post any data
dataType: "json",
success: function (response) {
var data = response; // response data from your php script
if (predefined_val !== data.orderStatus) {
window.location.href = window.location.href;
}
}
});
}, 5000);// function will run every 5 seconds
});
</script>

I have tested this by creating two files(autorefresh.php,index.php) and test db with table and it is working for me. I think the below code would be helpful, If not please share you code, i will check and fix it.
autorefresh.php
// Create connection
$db = new mysqli("localhost", "root", "","test");
$orderId = $_POST["orderId"];
$query = "SELECT * FROM orderinhomeonlinecall WHERE orderId='$orderId'";
$result = mysqli_query($db, $query);
while($row = mysqli_fetch_array($result))
{
$orderStatus = $row['orderStatus'];
$data = array(
'orderStatus' => $orderStatus
);
echo json_encode($data);
}
?>
index.php
<?php
$orderStatus ='pending';
$orderId =1;
?>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script>
<script type="text/javascript">
var predefined_val = '<?php echo $orderStatus; ?>';// your predefined value.
$(document).ready(function () {
setInterval(function () {
$.ajax({
type: "POST",
url: "autorefresh.php", //put relative url here, script which will return php
data: {orderId: <?php echo $orderId; ?>}, // if any you would like to post any data
dataType: "json",
success: function (response) {
var data = response; // response data from your php script
if (predefined_val !== data.orderStatus) {
window.location.href = window.location.href;
}
}
});
}, 5000);// function will run every 5 seconds
});
</script>

How to use AJAX to call php file from javascript file

I want to call php file from javascript, and this php file will update id=1
like this way:
javascript:
if(lastTemp >= document.getElementById("TempSet").value){
var jsonData2 =$.ajax({
url: "setpp.php",
dataType: "json",
async: false
}).responseText;
var obj2 = JSON.parse(jsonData2);
console.log(obj2);
}
else {
}
php file:
<?php
$DATABASE_HOST = 'localhost';
$DATABASE_USER = 'use';
$DATABASE_PASS = 'pass';
$DATABASE_NAME = 'database';
// Try and connect using the info above.
$db = mysqli_connect($DATABASE_HOST, $DATABASE_USER, $DATABASE_PASS,
$DATABASE_NAME);
if (!$db){
die("Connection Failed: ". mysqli_connect_error());
}
$db_update = "UPDATE setpoint_control SET status='ON' WHERE id=1";
$result = mysqli_query($db, $db_update);
?>
<?php
$data = array();
if(mysqli_num_rows($result)>0){
while($row = mysqli_fetch_array($result)){
array_push($data, $row['status']);
}
}
echo json_encode($data);
?>
the code is executed and the status in database table is changed but I got error in console : SyntaxError: JSON.parse: unexpected character at line 4 column 2 of the JSON data
How can I solve this issue which I think I need to rewrite json_encode but I don't know how?

$.ajax({
type: 'post',
dataType: 'json',
cache: false,
url: 'setpp.php',
success: function (response) {
$.each(response, function(i, item) {
alert(item);
});
},
error: function () {
alert("error");
},
});
example php answer setpp.php
if (mysqli_num_rows($result) > 0) {
while ($row = mysqli_fetch_array($result)) {
array_push($data, $row['status']);
}
die(json_encode($data));
} else {
$answer = array(
'No Records'
);
die(json_encode($answer));
}
I think the problem is the value returned by setpp.php.
remember to die(), otherwise the php answer will not be correct

fullcalendar ajax success does not working

I am making reservation website using fullcalendar. I insert event into my DB using Ajax. However, the problem is that success function was not working, but every data was successfully inserted into DB. Also I changed success to error, but still nothing happened.
Here is code where I insert event into DB
select: function(start, end, allDay)
{
var teacher = prompt("Enter ResourceID");
if(teacher)
{
var start = $.fullCalendar.formatDate(start, "Y-MM-DD HH:mm:ss");
var end = $.fullCalendar.formatDate(end, "Y-MM-DD HH:mm:ss");
$.ajax({
url:"http://show981111.cafe24.com/login-system/addevent.php",
type:"POST",
data:{userName: '<?php echo $name; ?>' , newlyBookedDate:start, courseTeacher : teacher, userBranch:'<?php echo $userBranch; ?>', userID: '<?php echo $userID; ?>'},
success:function()
{
calendar.fullCalendar('refetchEvents');
alert("Added Successfully");
}
})
}
},
Here is where I get the php variables that I use above
<?php
session_start();
if ( $_SESSION['logged_in'] != 1 ) {
$_SESSION['message'] = "You must log in before viewing your profile page!";
header("location: error.php");
}
else {
// Makes it easier to read
$name = $_SESSION['userName'];
$userBranch = $_SESSION['userBranch'];
$userID = $_SESSION['userID'];
}
?>
This is code for addevent.php
<?php
// Values received via ajax
$userName = $_POST['userName'];
$newlyBookedDate = $_POST['newlyBookedDate'];
$courseTeacher = $_POST['courseTeacher'];
$userBranch = $_POST['userBranch'];
$userID = $_POST['userID'];
// connection to the database
try {
$bdd = new PDO('mysql:host=localhost;dbname=show981111', 'show981111', 'pass');
} catch(Exception $e) {
exit('Unable to connect to database.');
}
// insert the records
$sql = "INSERT INTO DAYSCHEDULE (newlyBookedDate, userID, userName, courseTeacher,userBranch ) VALUES (:newlyBookedDate, :userID, :userName, :courseTeacher, :userBranch)";
$q = $bdd->prepare($sql);
$q->execute(array(':newlyBookedDate'=>$newlyBookedDate, ':userID'=>$userID, ':userName'=>$userName, ':courseTeacher'=>$courseTeacher,':userBranch'=>$userBranch ));
?>

if your data inserted into DB then take the response from your controller/W to the success function parameter and see the log.
$.ajax({
url:"http://show981111.cafe24.com/login-system/addevent.php",
type:"POST",
data:{userName: '<?php echo $name; ?>' , newlyBookedDate:start, courseTeacher : teacher, userBranch:'<?php echo $userBranch; ?>', userID: '<?php echo $userID; ?>'},
success:function(response)
{
console.log(response);
calendar.fullCalendar('refetchEvents');
alert("Added Successfully");
}
});
if you are getting no error then in the success function should be
success:function(response)
{
console.log(response);
//calendar.fullCalendar('refetchEvents');
//if calendar is used as an id then
$("#calendar").fullCalendar('refetchEvents');
//or it is a class then
$(".calendar").fullCalendar('refetchEvents');
alert("Added Successfully");
}

Canvas will not save server side PHP

I have a canvas which I need to save to a directory and store the URL in a database.
When I save the file without storing the URL in the database it works fine, and vice versa.
However, when I put the two together and specify the PHP file through AJAX, for some reason it doesn't recognise the session variable?
When I try to call the "success" on AJAX, nothing shows up. I get no response.
This could possibly be an easy fix! I think I've been staring at this code for too long.
JavaScript:
function doodleSave() {
var canvas = document.getElementById("doodle-canvas");
var canvasData = canvas.toDataURL("image/png");
$.ajax({
url:'doodleupload.php',
type:'POST',
data:{ data:canvasData },
success: function(response){
alert(response);
//echo what the server sent back...
}
});
}
PHP:
<?php
session_start();
/* AUTOMATED VARIABLES */
$url = md5(uniqid(rand(), true));
$unique_user_id = $_SESSION['unique_user_id'];
$unique_post_id = md5(uniqid(rand(), true));
$timestamp = time();
$nature = "doodle";
$imageUrl = $upload_dir.$url.'.png';
$upload_dir = "images/external/doodles/";
$img = $_POST['data'];
$img = substr($img,strpos($img,",")+1);
$data = base64_decode($img);
$file = $upload_dir . $url . ".png";
$success = file_put_contents($file, $data);
echo $success ? $file : 'Unable to save the file.';
require_once 'php/connect.php';
try
{
$stmt = $pdo->prepare("INSERT INTO posts (unique_user_id, unique_post_id, nature, image_url, timestamp) VALUE (:unique_user_id, :unique_post_id, :nature, :image_url, :timestamp)");
$stmt->bindParam(":unique_user_id",$unique_user_id);
$stmt->bindParam(":unique_post_id",$unique_post_id);
$stmt->bindParam(":nature",$nature);
$stmt->bindParam(":image_url",$imageUrl);
$stmt->bindParam(":timestamp",$timestamp);
if($stmt->execute())
{
echo "File in database";
}
else
{
echo "Not in database";
}
}
catch(PDOException $e){
echo $e->getMessage();
}
?>

Move $upload_dir at the top, as you are calling it before you initialize it.
$upload_dir = "images/external/doodles/";
$url = md5(uniqid(rand(), true));
$unique_user_id = $_SESSION['unique_user_id'];
$unique_post_id = md5(uniqid(rand(), true));
$timestamp = time();
$nature = "doodle";
$imageUrl = $upload_dir.$url.'.png';

error in getting ajax response in java script

In my project am using ajax for sending message the problem is i can't get the response in the ajax function the function works perfectly before,Can't find exact cause of the issue help me to solve it
ajax
var str = {
message:message, department_id:department, email:email, username:name
};
$.ajax( {
type: "POST",
url:'<?php echo base_url(); ?>home/savemessage',
dataType:"json",
data: str,
success: function(msg) {
$('#sentchat') . hide();
$('#chatmessage') . show();
$('#userchat') . html(msg . dataid);
$('.chat-box-content') . hide();
$('#adminname span').html('waiting for admins reply');
var elem = document . getElementById('userchat');
elem.scrollTop = elem . scrollHeight;
}
});
php controller
function savemessage() {
extract($this->input->post());
$data['message_id'] = $this->session->userdata('msgid');
$data['username'] = $username;
$data['email'] = $email;
$data['department_id'] = $department_id;
$data['message'] = $message;
$data['datetime'] = date('Y-m-d H:i:s');
$data['status'] = 'new';
$data['message_by'] = '1';
$this->db->insert('message', $data);
echo json_encode($username);
exit;
}
I cant get the response in it help me to solve it

You use ajax to communicate with a PHP script, inside the PHP script you could have the content of the function you want to execute. For example in your code:
$.ajax( {
type: "POST",
url:'<?php echo base_url(); ?>home/savemessage.php',
dataType:"json",
data: {myData:str},
success: function(msg) {
$('#sentchat') . hide();
$('#chatmessage') . show();
$('#userchat') . html(msg . dataid);
$('.chat-box-content') . hide();
$('#adminname span').html('waiting for admins reply');
var elem = document . getElementById('userchat');
elem.scrollTop = elem . scrollHeight;
}
});
Then, on the server side the php script "savemessage.php" would receive the POST action, so you could have:
if(isset($_POST['myData']) && !empty($_POST['myData'])) {
$obj = $_POST['myData'];
//rest of your code
echo json_encode($username);
exit;
}
However from your code I cannot see $username defined, so that probably would return an error.