I'd like to have a div on my web page that is based off of the data in my MySQL database. When the data in the database changes, the content in the div should change as well.
Example: Let's say I have a field in the first row of a table called "server_statistics" in my MySQL database which keeps track of a server being online or offline. When the server goes offline, the MySQL field changes from 1 to 0. When it goes online, it goes from 0 to 1.
So I want to use Javascript to display the server status on my webpage and update it without refreshing the page.
I thought I'd be able to do it like this:
<script type="text/javascript">
var int = self.setInterval("update()", 1000);
function update() {
var statusElement = document.getElementById("status");
var status = "<?php
$con = mysql_connect("localhost", "root", "");
mysql_select_db("database_name", $con);
$row = mysql_fetch_array(mysql_query("SELECT * FROM server_statistics LIMIT 1"));
mysql_close($con);
echo $row[0]; ?>";
if (status == 0) {
statusElement.style.color = "red";
statusElement.innerHTML = "offline";
}
else {
statusElement.style.color = "green";
statusElement.innerHTML = "online";
}
}
</script>
But this doesn't work. The page needs to be refreshed for it to update and I don't know why...
I've read I can use JQuery but I have no knowledge of this language at all.
I tried this:
<script type="text/javascript">
var int = self.setInterval("update()", 1000);
function update() {
$('#status').load('load.php');
}
</script>
and then in load.php I had this:
<?php
echo "test";
?>
But nothing happened after 1 second passed. I'm probably doing it wrong because as I said, I don't know anything about this JQuery/AJAX stuff.
So how can I accomplish retrieving a field from a MySQL database at every specified interval and then automatically update an HTML element accordingly? It would be ideal to only update the HTML when a change occurred in the database but for now, I just want it to change every few seconds or minutes...
Thanks in advance.
What you need to do is utilize AJAX. Your page will need to make a request to the server each time it wants to check the status. There are two ways to do it: manually refresh the page yourself, or use an AJAX call.
AJAX really isn't all that difficult, you start by creating a separate PHP page check_status.php with the following in it:
<?php
$con = mysql_connect("localhost", "root", "");
mysql_select_db("database_name", $con);
$row = mysql_fetch_array(mysql_query("SELECT * FROM server_statistics LIMIT 1"));
mysql_close($con);
echo $row[0]; ?>
Then, in your HTML page, the easiest way to do this would be to use jQuery:
var statusIntervalId = window.setInterval(update, 1000);
function update() {
$.ajax({
url: 'check_status.php',
dataType: 'text',
success: function(data) {
if (parseInt(data) == 0) {
$("#status").css({ color: "red" }).text("offline");
} else {
$("#status").css({ color: "green" }).text("online");
}
}
}
}
That's it really. Every second, it will call the update() method which makes an AJAX call and inserts the result back into your HTML.
instead of var int = self.setInterval("update()", 1000); try using self.setInterval(update, 1000); and place it after the update function.
Related
I am trying to work implement ajax due to maximum site load which PHP causes. But I am not aware of where I am making a mistake here. it is an anchor tag, when it is clicked the status of the particular row should be changed to a string which is hard coded.
PHP WAY
USERDETAIL.PHP
Next
Then it triggers This (IGNORE SQL INJECTION)
if(isset($_GET['changeStatus'])){
$id = $_GET['changeStatus'];
$user=$_SESSION['user'];
$sql = "select * from productOrder where id = ".$id;
$result = mysqli_query($conn, $sql);
if(mysqli_num_rows($result) > 0){
$row = mysqli_fetch_assoc($result);
$sql = "update productOrder set prodStatus = 'Ready', By='".$user."' where id=".$id;
if(mysqli_query($conn, $sql)){
header("Location:USERDETAIL.php");
}
}
}
According to this way, it works neat, but the userdetail.php would refresh anyways which is a lot time consuming. Then tried AJAX way is below.
Next
and that hits to
$(document).ready(function() {
$(".changeStatus").click(function(event){
event.preventDefault();
var status = "Ready";
var id = $(this).attr('data-id');
$.ajax({
url : 'action.php',
method : 'POST',
data : {status : status , id : id},
dataType: 'html',
success : function(response){
console.log(response);
}
});
});
});
and in the action.php it is (IGNORE SQL INJECTION AGAIN)
if(isset($POST['prodStatus'])){
$status = $_POST['prodStatus'];
$id = $_POST['id'];
$sql = "update productOrder set prodStatus= '$status' where id=".$id;
$result = mysqli_query($conn, $sql);
if($result){
return 'Updated';
}
}
The output is nothing happens. in the console it is just adding int values. I know I am making a mistake, or understood AJAX in a wrong way. it is just one button click and the string in SQL should be updated without an input text / modal. Please suggest what should be improved?
Also instead of having a seperate action php for these actions, can I do all these in userdetail.php itself with Ajax? is it possible?
Thanks in advance.
As B_CooperA pointed out, $POST should be $_POST.
Also, in $.ajax script data object your property name is status and in action.php you are checking it by prodStatus.
Furthermore, you should check the errors PHP is throwing in your script by enabling error reporting: error_reporting(E_ALL);
It's better to separate ajax calls from your view files. You can create a Class to handle all of your ajax calls (You should also consider authenticating your calls as per your use cases).
I'm making a website. and one of the features i want it to have is to have a simple game that connects 2 players together. my problem is I don't know how to make it so both player are in the same "room" because on room holds only 2 players.
On way i approached this is once one player joined, he gets a "wait for next player" message and waits while sending to the database that one player have joined. how do i make it keep checking for the next 3 minutes if the next player joined?
UPDATE
First here is the code so far:
<html>
<title>SiteName (test)</title>
<head>
<?php
$servername = "localhost";
$u
sername =
$password =
$dbname =
try
{
$conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
// set the PDO error mode to exception
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$stmt = $conn->prepare('SELECT * FROM game');
$stmt->execute(array('gameID' => $gameID));
while($row = $stmt->fetch()) {
print_r($row);
echo "<br />\n";
}
}
catch(PDOException $e)
{
echo $sql . "<br>" . $e->getMessage();
}
?>
<button onclick="myFunction()" id="w">Look for Game</button><br>
<script>
function myFunction() {
var elem = document.getElementById("w").innerHTML = "Wait";
var counter = 10;
var label= document.getElementById("lbl");
var counter = 10;
var clabel= document.createElement("p");
clabel.innerHTML = "You can download the file in 10 seconds.";
var id;
label.parentNode.replaceChild(clabel, label);
id = setInterval(function() {
counter--;
if(counter < 0) {
clabel.parentNode.replaceChild(label, clabel);
clearInterval(id);
} else {
clabel.innerHTML = "You can download the file in " + counter.toString() + " seconds.";
}
}, 1000);
}
</script>
<?php
$conn = null;
?>
</body>
</html>
Am trying to make it so that if the first player joined, he will be waiting (i have it for 10 seconds here as a test) until the other joins. the way am trying to do it is to have a field in the database will know if the a player is in that page and await the next player. I read something about long polling but not sure how to implement it for my case.
Any feed back would be helpful, Thank you
PHP is not the best language to do this in, but if you still want to do it.
Look into using Ratchet (http://socketo.me/), which is a PHP websocket library. A websocket is full duplex, meaning that a connection between the server and client is kept open. Game state and player actions can then be communicated through this.
http://socketo.me/docs/hello-world is an example you can learn from.
first you will want javascript or some client side code to handle this. as php will execute on the server side then display to the user. if you use ajax with javascript you can get the client side and server side to work together.
you will want to use a while loop, in this loop you will set a timeout.
in the while loop you can call the ajax script you want untill you get your result you want. I'm assuming you plan on making this a turn by turn game for the players. you will want a table that sets "true" to if player 1 or player 2 are in the game. if both are turn then the game begins.
Hope this logic helps
guys im trying to make a simple commenting system, that when i click the submit the fields will automatically save in the database then fetch it using ajax. but im getting all the data repeatedly instead of getting the recently added data in the database here is the code:
<div id="wrap-body">
<form action="" method="post">
<input type="text" name="username" id="username">
<input type="text" name="msg" id="msg">
<input type="button" id="submit" value="Send">
</form>
<div id="info">
</div>
</div>
<script>
$(document).ready(function (){
$('#submit').click(function (){
var username = $('#username').val();
var msg = $('#msg').val();
if(username != "" && msg != ""){
$.ajax({
type: 'POST',
url: 'get.php',
dataType: 'json',
data:{ 'username' : username , 'msg' : msg},
success: function (data){
var ilan=data[0].counter;
var i = 0;
for(i=0;i<=ilan;i++){
$('#info').append("<p> you are:"+data[i].username+"</p> <p> your message is:"+data[i].mesg);
}
}
});
}
else{
alert("some fields are required");
}
});
});
</script>
PHP:
<?php
$host='localhost';
$username='root';
$password='12345';
$db = 'feeds';
$connect = mysql_connect($host,$username,$password) or die("cant connect");
mysql_select_db($db) or die("cant select the".$db);
$username = $_POST['username'];
$msg = $_POST['msg'];
$insert = "INSERT INTO info(user_name,message) VALUES('$username','$msg')";
if(#!mysql_query($insert)){
die('error insertion'.mysql_error());
}
$get = "SELECT * FROM info ";
$result=mysql_query($get)or die(mysql_error());
$inside_counter = mysql_num_rows($result);
$data=array();
while ($row = mysql_fetch_array($result))
{
$data[] = array(
'username'=>$row['user_name'],
'mesg'=>$row['message'],
'counter'=>$inside_counter
);
}
echo json_encode($data);
?>
SELECT *
FROM "table_name"
ORDER BY "id" desc
LIMIT 1
This is a SQL query to get last record from table. It return last inserted according to id. id should be a auto increment field. I think this will be helpful for you.
Its because you are returning all the row in the table again to the ajax call via
$get = "SELECT * FROM info ";
if you do return all of them again you will have to test if they are not already there before appending them with the jQuery
Perhaps only return the newly inserted row.
or
The jQuery already knows the data it sent to the server, just return success true or false, and then append it if true with the jQuery, no need to return the data back again to the client side
EDIT
I'm not going to write code for you but perhaps these suggestions may help
mysql_query($insert)
link here for reference - http://php.net/manual/en/function.mysql-query.php
will return true of false depending on if the insert statement was successful
you could potentially also check that it acutally inserted a row by calling
mysql_affected_rows()
link here for reference - http://php.net/manual/en/function.mysql-affected-rows.php
then assuming that the insert was successful (i.e. true from mysql_query($insert) or mysql_affected_rows() > 0) simply return successful (i.e. something like
echo json_encode(true);
then on the client side you could do something like the following:
(jQuery Ajax success function only:)
success: function (data){
if (data) {
$('#info').append("<p> you are:"+username +"</p> <p> your message is:"+msg);
}
else
{
alert('There has been an error saving your message');
}
}
using the variables that you just used to send the request
(Please note code samples provided here are only for example, not intended to be used verbatim)
Couple of points though. Is your intention to post the data via ajax only? (i.e. no page refresh) as if that is the case, you will need to return false from you form submit event handler to stop the page full posting. Also you do not appear to have any logic, or are not worried about complete duplicates being entered (i.e. same username, same message) - not sure if you want to worry about this.
I would also potentially look at tracking all the messages via ids (although I don't know your DB structure), perhaps using
mysql_insert_id()
link here FYI - http://php.net/manual/en/function.mysql-insert-id.php
That way each message can have a unique id, may be easier to establish duplicates, even if the username and message are identical
There are numerous ways to deal with this.
Anyway hope that helps
PS - using the mysql_ - group of commands is generally discouraged these days, use the mysqli_ range of function instead
Is it possible to refresh a divs content only if new content is added to the database?
I'm using this for "recent posts" that appear in the side menu.
Right now the div is set refresh every 10 seconds.
Is it somehow possible to check it a new post was added to the db and then add only that posts data to the div?
I'm using MySql, php and jquery to do all of this.
Yes, you can do it by comparing the text content.
function updateHtmlContent() {
var htmlContent=$('#divToUpdate').text();
$.ajax({
type: "GET",
datatype: "text",
//php function that should give you the text back
url: 'dataBaseFunction.php',
success: function(data) {
if (htmlContent != data) {
$('#divToUpdate').text(data);
}
}
});
}
//compare and upate every 20 seconds
setInterval(updateHtmlContent(), 20000);
Hope this help!
YES IT IS POSSIBLE
The below code will get the recent photos that has been added:
AFTER THE TIMER.PHP WAS LAODED (or your wordpress blog page)
AND ONLY AFTER A NEW PHOTO IS ADDED (you can use it for recent posts, comments, or anything)
This can be used to create a live blog for example, where the user will see all the recent comments, images or posts, or everything, even if he doesn't reload the current page. And it won't consume a lot of bandwidth as the content will be reloaded ONLY IF THERE IS NEW ADDED (it will only send a tiny check in the database otherwise).
I've just finished working out a solution for that, and I have to share it with you. I was doing it for Wordpress, i needed a function that gets the photos that has been added ONLY after the user loaded the page. It could be done by simply refreshing the div every 5 seconds, but imagine if there were 20 photos, and they had to be refreshed every 5 seconds... that's a lot of MB of data. So we will refresh them ONLY when a new photo is added (you can apply it to anything, from posts to comments or users, etc.).
There are 3 PHP files: timer.php, check.php and display.php.
The timer.php will load the check.php every 5 seconds, to check if new content is added. Notice the -6 extraction from the time of the current load of check.php.
The date-time of the timer.php will be passed (by check.php?tim=**print date) through the check.php (by **display.php?timm='. $tizz .') and to the display.php, so we can use it for a reference in our database (that will be the time from where we will load all the images, if new images are added).
If any questions, just ask. It has to work for you too, as it works for me.
ENJOY! :)
Below are the 3 PHP files, just customize them for your needs:
THE TIMER.PHP (or your wordpress blog pages):
<script type="text/javascript" src="jquery.js"></script>
<script type="text/javascript">
$(document).ready(function(){
setInterval(function(){
$("#quotex a").load("check.php?tim=<?php print date("Y-m-d H:i:s"); ?>");
}, 5000);
});
</script>
<div id="quote"><a></a></div>
<div id="quotex"><a></a></div>
THE CHECK.PHP :
<?php
// Connects to your Database
mysql_connect("localhost", "username", "password") or die(mysql_error());
mysql_select_db("database name") or die(mysql_error());
// SQL query
$strSQL = "SELECT * FROM wp_posts WHERE post_mime_type LIKE 'image/jpeg' ORDER BY `wp_posts`.`id` DESC LIMIT 1";
// Execute the query (the recordset $rs contains the result)
$rs = mysql_query($strSQL);
// Loop the recordset $rs
// Each row will be made into an array ($row) using mysql_fetch_array
while($row = mysql_fetch_array($rs)) {
$atime = $row['post_date'];
$tizz = $_GET['tim'];
$dsff = $row['post_date'];;
$duff = date("Y-m-d H:i:s");
//convert the date-time into seconds so we can extract 6 seconds from it
$six = strtotime(date("Y-m-d H:i:s"))-6;
//convert the latest image date-time too from database so we can compare it
$sox = strtotime("$dsff");
if ($six < $sox)
{
echo '<script type="text/javascript">$(document).ready( function(){ $("#quote a").load("display.php?timm='. $tizz .'"); } ); </script>';
}
}
// Close the database connection
mysql_close();
?>
THE DISPLAY.PHP:
<?php
$tipp = $_GET["timm"];
// Connects to your Database
mysql_connect("localhost", "username", "password") or die(mysql_error());
mysql_select_db("database name") or die(mysql_error());
// SQL query
$strSQL = "SELECT * FROM wp_posts WHERE post_mime_type LIKE 'image/jpeg' AND post_date > '$tipp' ORDER BY `wp_posts`.`id` DESC LIMIT 10";
// Execute the query (the recordset $rs contains the result)
$rs = mysql_query($strSQL);
// Loop the recordset $rs
// Each row will be made into an array ($row) using mysql_fetch_array
while($row = mysql_fetch_array($rs)) {
//guid is the column where the image url is located in the wordpress database table
$atime = $row['guid'];
echo "<img src='". $atime ."' /><br />";
}
// Close the database connection
mysql_close();
?>
Yes is possible, but you need to create a method in php and use Ajax to refresh your div.
If you update your question with code I can provide an example.
I think you are probably about 90% of the way there. I assume you are using jQuery to AJAX in the content every 10 seconds (PS That seems like a lot?)
I think you could solve your problem by making the backside function start with posts after the most recent that was found.
So if your query looks like this now:
SELECT a,b,c FROM foo
You should change it to
SELECT a,b,c FROM foo WHERE id > $last_found_id
You can store last_found_id in your Javascript and send it back when you call the function. If your result set is empty then you don't have to refresh the div.
This change will also probably require that you are pre/appending to the div instead of overwriting it completely.
look at the setInterval(function, time) in javascript to make a loop each "time" (in milliseconds),
and the function $.ajax() in jQuery to pass the information to the server..
also, you have to create the new content with .html() or .append()
I'm trying to get some values off a DB and then putting those values into javascript variables. I managed to do just that, the problem I'm having is when the values in the DB change the values of the variables don't. I figured the problem lies within my PHP, but I cant find it. Can you guys help me?
here's my code:
PHP
<?php
session_start();
if(!isset($_SESSION['u_name'])){
$_SESSION['u_name'] = '';
}
mysql_connect ("localhost", "root", "") or die ('Error: ' . mysql_error());
mysql_select_db('raffleiz_Main')or die ("cannot select DB :(");
$signups = mysql_query("SELECT * FROM `Rafflez_info`") or die ('Error: ' . mysql_error());
$row = mysql_num_rows($signups);
//pull all of the data and store it
for($p = 0; $p < $row; $p++){
$participants[$p] = mysql_result($signups, $p, "#_participants");
};
for($a = 0; $a < $row; $a++){
$max_participants[$a] = mysql_result($signups, $a, "max_participants");
};
?>
and my javascript function:
function progress(){
var signups = "<?php echo $participants[0]; ?>";
var maxP = "<?php echo $max_participants[0]; ?>";
alert (signups);
alert (maxP);
var pSignup = signups / maxP;
alert (pSignup);
var total = 550 * pSignup;
var theImg = document.getElementById('progress');
theImg.width = total;
alert (total);
};
I put the "alert" command there so that I could see the change in the values. right now the values don't change no matter what I change them to in the DB.
PHP is a server-side language, meaning that when it served the script from the server to the client, there is no going back.
JavaScript is a client-side language, thus can receive values from a server-side language such as PHP. The values received are then client-side only, a copy if you will.
You can use the XMLHttpRequest API to request a script from a server, updating the local client-side values.
I recommend using the jQuery $.ajax function to easily achieve that.
Here's a nice tutorial from Nettuts to get your started.
You need to call your php page and ask it if values are changes. You can easily use jQuery AJAX to achieve that. Use JSON for easier data transfer. If you want to check if values are changes every 3 seconds for example you can do this:
var interval = window.setInterval(function(){
jQuery.ajax({
url: 'your php file address',
success: function(data){
progress(data)
}
});
}, 3000);
You will need to change you proccess and php code. This is just to give you the idea.