We Keep Coding

How to GET javascript data in PHP file without page reload

I am building my best attempt at a twitter clone and have run into a bit of a problem. I want to be able to click on a post and, without a page refresh, display that post in the overlay of the page (as you would on a twitter feed to look at replies, etc.).
In script.js, I check for a click and try to change the url.
$('body').on("click", ".chirp", function(){
var uid = $_GET['id'];
var pid = $(this).attr("id");
var pidSplit = pid.split("chirp");
var messageID = pidSplit[1];
var obj = {foo: "status"};
$('.chirpOverlay').addClass("active");
window.history.pushState(obj, "Status", "profile.php?id="+uid+"&status="+pid);
});
The javascript works as intended...but as I will soon find out, the victory is short-lived.
In profile.php, I attempt to GET the status id from the URL parameter.
<?php
$status_id = $_GET['status'];
$sql = $db->query("SELECT * FROM chirps WHERE id='$status_id'");
if (mysqli_num_rows($sql) > 0) {
$c = $sql->fetch_object();
}
?>
This doesn't work because, as I've learned, using 'window.history.pushState' only changes the url- but doesn't load the page. Thus the $_GET statement fails. I need a way to get the id of the post I click on into profile.php without a page refresh. Even if it means taking a different approach (instead of using a URL parameter).
PS: I tried to do an XMLHttpRequest as well- to no avail. :(
Thanks in advance!

$('body').on("click", ".chirp", function(){
var uid = $_GET['id'];
var pid = $(this).attr("id");
var pidSplit = pid.split("chirp");
var messageID = pidSplit[1];
var obj = {foo: "status"};
$('.chirpOverlay').addClass("active");
$.ajax({
url: "profile.php?id="+uid+"&status="+pid,
type: "GET",
data: obj,
dataType: "html",
success: function(data){
console.log(data);
}
});
});

You need to just get something up and going that works and then you can add more to it as you figure things out. This should give you a good starting place.
Here are your two files. Make sure they are both in the same directory.
You will need to make sure you have a jquery version loaded. Put this on whatever page you are calling the script.js from.
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
script.js
$(document).ready(function(){
$('body').click(function(){
var id; //define your id.
var pid; //define your pid.
var datastring = 'id=' + uid + '&status=' + pid;
console.log(datastring);
$.ajax({
url: 'profile.php',
type: 'POST',
cache: false,
data: datastring,
dataType: 'json',
success: function(data){
console.log('Made it to the success function: ' + data);
if (data) {
//It works, do something.
console.log(data);
} else{
//It does not work, do something.
console.log('Your ajax failed to get the info from your php page. Keep troubleshooting');
}
}
});
});
});
profile.php
<?php
/*
$status_id = $_POST['status']; //This needs to be sanitized and validated.
$sql = $db->query("SELECT * FROM chirps WHERE id='$status_id'"); //This is unsafe sql practice.
if (mysqli_num_rows($sql) > 0) {
$c = $sql->fetch_object();
}
echo json_encode($c); //This is what gets sent back to script.js
*/
echo 'You made it to your php page.';
?>
A few things:
You can not call any php variable from within your js. var uid = $_GET['id']; does not work.
Any value that you pass to the php page needs to be validated to make sure it is a legitimate value.
Your SQL query is prone to sql injections. Please read up on how to parameterize your queries. Good Mysqli Practices

I have finally found a AJAX-based solution to my problem.
I created a new php file called "chirp_open_ref.php" and added this ajax to script.js:
var datastring = 'status=' + messageID;
$.ajax({
url: "chirp_open_ref.php",
type: "POST",
data: datastring,
cache: false,
dataType: "text",
success: function(data){
$('.chirp-container').html(data);
}
});
Inside of 'chirp_open_ref.php':
<?php
require 'core.inc.php';
if (isset($_POST['status']) && isset($_SESSION['user_id'])){
$chirp_id = $_POST['status'];
$c = "";
$sql = $db->query("SELECT * FROM chirps WHERE id='$chirp_id'");
if (mysqli_num_rows($sql) > 0){
$c = $sql->fetch_object();
}
include'chirp.inc.php';
}
?>
'chirp.inc.php' is simply a template for the layout/structure of each post.
This works like a charm, but I am always open to any criticism of how I am performing this. Thanks for all the help guys!

Send variable to php via ajax

I'm trying to send a input value to php via ajax but I can't seem to get this right. I'm trying to create a datatable based on the user input.
This is my code:
<input class="form-control" id="id1" type="text" name="id1">
My javascript code:
<script type="text/javascript">
$(document).ready(function() {
var oTable = $('#jsontable').dataTable(); //Initialize the datatable
$('#load').on('click',function(){
var user = $(this).attr('id');
if(user != '')
{
$.ajax({
url: 'response.php?method=fetchdata',
data: {url: $('#id1').val()},
dataType: 'json',
success: function(s){
console.log(s);
oTable.fnClearTable();
for(var i = 0; i < s.length; i++) {
oTable.fnAddData([
s[i][0],
s[i][1],
s[i][2],
s[i][3],
s[i][4],
s[i][5],
s[i][6],
s[i][7]
]);
} // End For
},
error: function(e){
console.log(e.responseText);
}
});
}
});
});
</script>
My php script:
<?php
$conn = pg_connect(...);
$id1 = $_POST["id1"];
$result = pg_query_params($conn, 'SELECT * FROM t WHERE id1 = $1 LIMIT 20', array($id1));
while($fetch = pg_fetch_row($result)) {
$output[] = array ($fetch[0],$fetch[1],$fetch[2],$fetch[3],$fetch[4],$fetch[5],$fetch[6],$fetch[7]);
}
echo json_encode($output);
?>
I don't know a lot of js but my php is correct i test it. So i guess the problem is in the javascript code.
The problem is, my datatable is not being created based on the user input.
Thank you!

change
data: {url: $('#id1').val()},
to:
type: 'POST',
data: {id1: $('#id1').val()},
However the problem might be bigger. You might not be getting the correct data from PHP. You can debug by adding the error option to your ajax() call, like this:
$.ajax({
url: 'response.php?method=fetchdata',
type: 'POST',
data: {id1: $('#id1').val()},
dataType: 'json',
success: function(s){
},
error: function (xhr, status, errorThrown) {
console.log(xhr.status);
console.log(xhr.responseText);
}
});
Then check your browser's Console for the output, this should give you some type of error message coming from PHP.
My assumption is that since you are using dataType: 'json', the ajax request expects JSON headers back, but PHP is sending HTML/Text. To fix, add the correct headers before echoing your JSON:
header('Content-Type: application/json');
echo json_encode($output);

How do I get two arrays in Ajax call?

JS CODE:
$.ajax({
url: 'assignavailtrainers.php',
data: {action:'test'},
type: 'post',
success: function(data) {
}
});
PHP CODE:
<?php
$username = "trainerapp";
$password = "password";
$hostname = "localhost";
$link = #mysql_connect($hostname, $username, $password);
if(#mysql_select_db("trainer_registration"))
{
$select_query_num = #mysql_query("select program_id,facilitator_id,availability_status from program_facilitator where availability_status in (1,2)");
$select_query_name = #mysql_query("select facilitator_id,firstname,lastname,email_id from facilitator_details");
$num_rows = #mysql_num_rows($select_query_num);
$trainerdetails = [];
$traineravaildetails = [];
$i = 0;
$j = 0;
while($row = #mysql_fetch_assoc($select_query_num))
{
$trainerdetails[$i]['pgidi'] = $row['program_id'];
$trainerdetails[$i]['facilitatorid'] = $row['facilitator_id'];
$trainerdetails[$i]['avail_status'] = $row['availability_status'];
$trainerdetails[$i]['idi'] = $row['facilitator_id'];
$i++;
}
while($row1 =#mysql_fetch_assoc($select_query_name))
{
$traineravaildetails[$j]['facilitatorid'] = $row1['facilitator_id'];
$traineravaildetails[$j]['firstname'] = $row1['firstname'];
$traineravaildetails[$j]['lastname'] = $row1['lastname'];
$traineravaildetails[$j]['emailidvalue'] = $row1['email_id'];
$j++;
}
echo json_encode(array('result1'=>$trainerdetails,'result2'=>$traineravaildetails));
}
?>
Please help me with the code in the ajax success function area. I've tried using initChart2 but I get an error which says initChart2 is not defined. I don't seem to understand of how to get two arrays from PHP in ajax since I'm a newbie ajax. If someone can help me with the code along with explanation, it'd be great. And I also need to know how to differentiate outputs in ajax which are sent from PHP.

You have two choices:
First one is to simply parse received (text) data to JSON:
var jsonData = JSON.parse(data);
// or simply data = JSON.parse(data);
But the best one in my oppinion is to specify json dataType to the $.ajax() request:
$.ajax(
data: {action:'test'},
type: 'post',
dataType: 'json',
success: function(data) {
...
}
});
This way, $.ajax() will also check for the validity of the received JSON data and error callback will be called instead of success one in case of wrong JSON data received.
...also is important to note you missed to send the json content-type header in your php with:
header("Content-Type: application/json");
Sending this header the dataType: 'json' parameter is no longer (strictly) necessary because $.ajax() guesses it by default attending to the received content-type. But, personally, I prefer to do both.
See $.ajax() documentation.

You forgot:
header("Content-Type: application/json");
… in your PHP.
While you are outputting JSON, you are telling the browser that it is HTML (which is the default for PHP) so jQuery isn't converting it to a useful data structure.
Add that and then you should be able to access data.result1 and data.result2.

To get ajax data:
$.ajax({
url: 'assignavailtrainers.php',
data: {action:'test'},
type: 'post',
success: function(data) {
data.result1;
data.result2;
}
});
You can use console.log(data); to view data structure

Trying to pass POST from JavaScript to PHP

For a few days now I have been trying to pass a simple POST call to a php script from JavaScript. I've done countless amounts of searching online without any positive results.
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script type="text/javascript">
function successHandler(location) {
var dataString = '?lat=' + location.coords.latitude + '&long=' + location.coords.longitude + '&accuracy=' + location.coords.accuracy;
alert(dataString);
if (location.coords.latitude == '0') {
} else {
alert("AJAX made");
$.ajax({
type: "POST",
url: "updatepos.php",
data: dataString,
cache: false,
success: function(html) {
alert(html);
}
});
setTimeout(navigator.geolocation.getCurrentPosition(successHandler),15000);
}
}
function getLocation() {
navigator.geolocation.getCurrentPosition(successHandler);
}
getLocation();
</script>
Above is my JavaScript file. The datastring gets made, and it alerts it out to my browser. No problem there. The problem is, my variables don't get passed to PHP whatsoever.
Here is the PHP that is in the same directory as the JavaScript.
<?php
include 'wp-load.php';
/*global $current_user;
get_currentuserinfo();
echo $current_user->user_login;*/
include('dbconnect.php');
global $current_user;
get_currentuserinfo();
$lat = $_POST['lat'];
$long = $_POST['long'];
$accuracy = $_POST['accuracy'];
/*$lat = $_GET['lat'];
$long = $_GET['long'];
$accuracy = $_GET['accuracy'];*/
$query = "UPDATE ltc_users SET lat='$lat',accuracy=$accuracy,lon='$long' WHERE name='$current_user->user_login'";
mysqli_query($GLOBALS['DB'],$query);
echo mysqli_error($GLOBALS['DB']);
echo $lat;
echo $long;
echo $accuracy;
echo $current_user->user_login;
?>
I may note that before the script would return mysql syntax errors as it was echoed in php due to missing variables. The syntax works if I use the $_GET method and just type in the data into my browser address bar for testing. It just doesn't get the JavaScript variables for whatever reason.

You're passing a string to jquery. When you do that, the string is sent out as-is by jquery. Since it's just a bare string, and not a key:value pair, there's no key for PHP to glom onto and populate $_POST with.
In fact, you shouldn't ever have to manually build a string of key:value pairs for ajax - jquery will take an array/object and do it all for you:
var stuff = {
lat : location.coords.latitude,
long : location.coords.longitude,
accuracy : location.coords.accuracy
}
$.ajax({
data: stuff
});

Here is your code combined with #Marc B
<script src = "http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js" > </script>
<script type="text/javascript">
function successHandler(location) {
if (location.coords.latitude == '0') {
} else {
alert("AJAX made ");
$.ajax({
type: "POST",
url: "updatepos.php",
data: {
"lat":location.coords.latitude,
"long":location.coords.longitude,
"accuracy":location.coords.accuracy
},
dataType: "json",
async:true,
cache: false,
success: function(html) {
alert(html);
},error: function(a,b,c){
console.log(a.responseText);
}
});
setTimeout(function(){navigator.geolocation.getCurrentPosition(successHandler);},15000);
}
}
function getLocation() {
navigator.geolocation.getCurrentPosition(successHandler);
}
$(function(){
getLocation();
});
</script>
code in jsfiddle http://jsfiddle.net/y7f1vkej/ it seems to be working for me

mysql query executed even though fields are empty

I have created a simple tagging system for my schools websites for the students. Now the tagging system is working perfectly now i also have to save tags in a notifications table with respective article id to later notify the students which article they have been tagged in even that i managed to do. But now if by chance you want to remove the tags sometime realizing while typing the article you don't need to tag that person, then the first put tag also gets updated in the db.
//ajax code (attach.php)
<?php
include('config.php');
if(isset($_POST))
{
$u=$_POST['v'];
mysql_query("INSERT INTO `notify` (`not_e`) VALUES ('$u')");
}
?>
// tagsystem js code
<script type="text/javascript">
var id = '<?php echo $id ?>';
$(document).ready(function()
{
var start=/%/ig;
var word=/%(\w+)/ig;
$("#story").live("keyup",function()
{
var content=$(this).text();
var go= content.match(start);
var name= content.match(word);
var dataString = 'searchword='+ name;
if(go.length>0)
{
$("#msgbox").slideDown('show');
$("#display").slideUp('show');
$("#msgbox").html("Type the name of someone or something...");
if(name.length>0)
{
$.ajax({
type: "POST",
url: "boxsearch.php",
data: dataString,
cache: false,
success: function(html)
{
$("#msgbox").hide();
$("#display").html(html).show();
}
});
}
}
return false();
});
$(".addname").live("click",function()
{
var username=$(this).attr('title');
$.ajax({
type: "POST",
url: "attach.php",
data: {'v': username},
});
var old=$("#story").html();
var content=old.replace(word,"");
$("#story").html(content);
var E="<a class='blue' contenteditable='false' href='profile2.php?id="+username+"'>"+username+"</a>";
$("#story").append(E);
$("#display").hide();
$("#msgbox").hide();
$("#story").focus();
});
});
</script>

Looks like your problem appears on the if statement in php code:
even though $_POST['v'] is empty and the sql still get excuted.
There is the quote from another thread:
"
Use !empty instead of isset. isset return true for $_POST because $_POST array is superglobal and always exists (set).
Or better use $_SERVER['REQUEST_METHOD'] == 'POST'
"
Or in my opinion.
Just put
if ($_POST['v']){
//sql query
}
Hope it helps;)

<?php
include('config.php');
$u = $_POST["v"];
//echo $a;
if($u != '')
{
mysql_query("your insert query");
}
else
{
}
?>