How To Target A Specific PHP Function Using Ajax - php

I am creating a simple comment program. i want to use ajax to refresh the comments and total number of comments. Keeping both functions in a single file is not working for me.
here is my code:
HTML:
<h3></h3>
<ul>
</ul>
PHP: include.php
//for comments
function main(){
try {
$query = connect()->prepare("SELECT * FROM comments WHERE article_id = 1627359589");
$query->execute();
echo json_encode($query->fetchAll());
} catch (PDOException $e) {
die("database connection failed");
}
}
// for total comment
function totalComment(){
$sql ="SELECT * FROM comments WHERE article_id = 1627359589";
$stmt = connect()->prepare($sql);
$stmt->execute();
$num = $stmt->rowCount();
echo json_encode($num);
}
main();
totalComment();
AJAX:
// for comment
setInterval(displayComments, 2000);
function displayComments(){
$.ajax({
url: "include.php",
type: "POST",
dataType: "JSON",
success: function(data){
for (var i = 0; i < data.length; i++) {
$("ul").append("<li>"+data[i] + "</li>")
}
}
})
}
// for total comments
setInterval(total, 2000);
function total(){
$.ajax({
url: "include.php",
success: function(data){
$("h3").html(data);
}
})
}

Set the PHP to process a named POST item and use switch to determine which function to use:
<?php
if( $SERVER['REQUEST_METHOD']=='POST' && isset( $_POST['cmd'] )){
function main(){
try {
$query = connect()->prepare("SELECT * FROM comments WHERE article_id = 1627359589");
$query->execute();
echo json_encode($query->fetchAll());
} catch (PDOException $e) {
die("database connection failed");
}
}
function totalComment(){
$sql ="SELECT * FROM comments WHERE article_id = 1627359589";
$stmt = connect()->prepare($sql);
$stmt->execute();
$num = $stmt->rowCount();
echo json_encode($num);
}
switch( $_POST['cmd'] ){
case 'comments':
main();
break;
case 'total':
totalComment();
break;
case 'banana':
banana();
break;
}
}
?>
In your javascript set the same parameter in each request but with a different value: data:{cmd:'banana'}, etc
function displayComments(){
$.ajax({
url: "include.php",
type: "POST",
data:{cmd:'comments'},
dataType: "JSON",
success: function(data){
for (var i = 0; i < data.length; i++) {
$("ul").append("<li>"+data[i] + "</li>")
}
}
})
}
function total(){
$.ajax({
url: "include.php",
data:{cmd:'total'},
success: function(data){
$("h3").html(data);
}
})
}
setInterval(displayComments, 2000);
setInterval(total, 2000);

Related

display data from database using ajax,mysql,php

Currently, I made script, which after onclick event,sending question to the database and showing data in console.log( from array ). This all works correctly, but.. I want to show data from array in the different position in my code. When I try to use DataType 'json' and then show some data, then it display in my console.log nothing. So, my question is: How to fix problem with displaying data? Is it a good idea as you see?
Below you see my current code:
$(document).ready(function(){
$(".profile").click(function(){
var id = $(this).data('id');
//console.log(id);
$.ajax({
method: "GET",
url: "../functions/getDataFromDB.php",
dataType: "text",
data: {id:id},
success: function(data){
console.log(data);
}
});
});
});
:
public function GetPlayer($id){
$id = $_GET['id'];
$query = "SELECT name,surname FROM zawodnik WHERE id='".$id."'";
$result = $this->db->query($query);
if ($result->num_rows>0) {
while($row = $result->fetch_assoc()){
$this->PlayerInfo[] = $row;
}
return $this->PlayerInfo;
}else {
return false;
}
}
:
$info = array();
$id = $_GET['id'];
$vv = new AddService();
foreach($vv->GetPlayer($id) as $data){
$info[0] = $data['name'];
$info[1] = $data['surname'];
}
echo json_encode($info);
I think it would be better to change the line fetch_all in mysqli to rm -rf. That information in the DB is all obsolete, or completely not true.
Try this:
<!DOCTYPE html>
<html>
<head>
</head>
<body>
<button class="profile" data-id="1">Click</button>
<script
src="https://code.jquery.com/jquery-3.3.1.min.js"
integrity="sha256-FgpCb/KJQlLNfOu91ta32o/NMZxltwRo8QtmkMRdAu8="
crossorigin="anonymous"></script>
<script>
$(document).ready(function(){
$(".profile").click(function(){
var id = $(this).data('id');
console.log(id);
$.ajax({
method: "GET",
url: "../functions/getDataFromDB.php",
dataType: "json",
data: {id:id},
success: function(data){
console.log(data);
$.each(data, function(idx, item) {
console.log(item.surname);
});
}
});
});
});
</script>
</body>
</html>
PHP side:
<?php
class AddService {
public function GetPlayer($id) {
if (filter_var($id, FILTER_VALIDATE_INT) === false) {
return false;
}
$query = "SELECT name, surname FROM zawodnik WHERE id={$id}";
$result = $this->db->query($query);
if ($result->num_rows <= 0) {
return false;
}
// assumming you are using mysqli
// return json_encode($result->fetch_all(MYSQLI_ASSOC));
// or
WHILE ($row = $result->fetch_assoc()) {
$data[] = $row;
}
return json_encode($data);
}
}
if (isset($_GET['id'])) {
$id = $_GET['id'];
$vv = new AddService();
// you don't need foreach loop to call the method
// otherwise, you are duplicating your results
echo $vv->GetPlayer($id);
}

Ajax data type JSON not working

I am trying to insert data using AJAX JSON but it's not working. I tried without JSON and it works, but an alert box shows with some HTML code.
HTML:
Short Break
AJAX:
$(document).ready(function() {
$('#sbreak').on('click', function() {
var name = $("SBreak").val();
$.ajax({
type: "POST",
dataType: 'json',
url: "brkrequest.php",
data: {
sname: name
}
cache: false,
success: function(server_response) {
if (server_response.status == '1') //if ajax_check_username.php return value "0"
{
alert("Inserted ");
} else if (server_response == '0') //if it returns "1"
{
alert("Already Inserted");
}
},
});
return false;
});
});
PHP: :
session_start();
date_default_timezone_set('Asia/Kolkata');
$sname=$_POST['sname'];
$sname= $_SESSION['myusername'];
$reqdate = date("Y-m-d H:i:s");
include("connection.php");
//Insert query
$query = sprintf("SELECT * FROM `breakqueue` WHERE (`sname` ='$sname')");
$result = mysql_query($query);
if(mysql_num_rows($result) > 0){
$data['status']= '1';//If there is a record match Already Inserted
}
else { // if there is no matching rows do following
$query = mysql_query("INSERT INTO `breakqueue`(`id`, `sname`, `btype`, `reqdate`, `apdate`, `status`) VALUES ('','$sname','Sbreak','$reqdate','','Pending')");
$data['status']= '0';//Record Insered
}
echo json_encode($data);
}
use it in php
header('Content-Type:application/json');
and write
success: function(server_response){
console.log(typeof server_response);
...
for finding response type,
if type of server_response isn't object
use it for convert it to object :
server_response = JSON.parse(server_response);
php Code:
session_start();
//Here added...
header('Content-Type:application/json');
date_default_timezone_set('Asia/Kolkata');
$sname=$_POST['sname'];
$sname= $_SESSION['myusername'];
$reqdate = date("Y-m-d H:i:s");
include("connection.php");
//Insert query
$query = sprintf("SELECT * FROM `breakqueue` WHERE (`sname` ='$sname')");
$result = mysql_query($query);
if(mysql_num_rows($result) > 0){
$data['status']= '1';//If there is a record match Already Inserted
}
else{ // if there is no matching rows do following
$query = mysql_query("INSERT INTO `breakqueue`(`id`, `sname`, `btype`, `reqdate`, `apdate`, `status`) VALUES ('','$sname','Sbreak','$reqdate','','Pending')");
$data['status']= '0';//Record Insered
}
echo json_encode($data);
}
Javascript Code:
$(document).ready(function()
{
$('#sbreak').on('click', function(){
var name = $("SBreak").val();
$.ajax({
type: "POST",
dataType:'json',
url: "brkrequest.php",
data: {sname: name}
cache: false,
success: function(server_response){
//TODO:REMOVE IT After seeing. alert or console.log for seeing type
alert(typeof server_response);
if(typeof server_response){
server_response = JSON.parse(server_response);
}
if(server_response.status == '1')//if ajax_check_username.php return value "0"
{
alert("Inserted ");
}
else if(server_response == '0')//if it returns "1"
{
alert("Already Inserted");
}
},
});
return false;

Using AJAX to return JSON from PHP

Apologies if this is a repeat question, but any answer I have found on here hasn't worked me. I am trying to create a simple login feature for a website which uses an AJAX call to PHP which should return JSON. I have the following PHP:
<?php
include("dbconnect.php");
header('Content-type: application/json');
$numrows=0;
$password=$_POST['password'];
$username=$_POST['username'];
$query="select fname, lname, memcat from members where (password='$password' && username='$username')";
$link = mysql_query($query);
if (!$link) {
echo 3;
die();
}
$numrows=mysql_num_rows($link);
if ($numrows>0){ // authentication is successfull
$rows = array();
while($r = mysql_fetch_assoc($link)) {
$json[] = $r;
}
echo json_encode($json);
} else {
echo 3; // authentication was unsuccessfull
}
?>
AJAX call:
$( ".LogIn" ).live("click", function(){
console.log("LogIn button clicked.")
var username=$("#username").val();
var password=$("#password").val();
var dataString = 'username='+username+'&password='+password;
$.ajax({
type: "POST",
url: "scripts/sendLogDetails.php",
data: dataString,
dataType: "JSON",
success: function(data){
if (data == '3') {
alert("Invalid log in details - please try again.");
}
else {
sessionStorage['username']=$('#username').val();
sessionStorage['user'] = data.fname + " " + data.lname;
sessionStorage['memcat'] = data.memcat;
storage=sessionStorage.user;
alert(data.fname);
window.location="/awt-cw1/index.html";
}
}
});
}
As I say, whenever I run this the values from "data" are undefined. Any idea where I have gone wrong?
Many thanks.

jQuery .ajax query to insert data inside a database is not working

I am building a website that uses jQuery/AJAX to send data to a php page, and from there insert it into a database. For some reason, the code isn't inserted and I get no response at all.
my javascript:
function insert_data(){
var title = debate_title.value;
var subtitle = debate_sub.value;
var sides = debate_sides.value;
$(function() {
$.ajaxSetup({
error: function(jqXHR, exception) {
if (jqXHR.status === 0) {
window.location.replace('errors/noConnection.html');
} else if (jqXHR.status == 404) {
window.location.replace('errors/noConnection.html');
} else if (jqXHR.status == 500) {
window.location.replace('errors/noConnection.html');
} else if (exception === 'parsererror') {
window.location.replace('errors/noConnection.html');
} else if (exception === 'timeout') {
window.location.replace('errors/noConnection.html');
} else if (exception === 'abort') {
window.location.replace('errors/noConnection.html');
} else {
window.location.replace('errors/noConnection.html');
}
}
});
});
$.ajax({
type: "POST",
url: "post_debate.php",
data: { post_title: title, post_sub: subtitle, post_sides: sidesm, ajax: 1 },
dataType: "json",
timeout: 5000, // in milliseconds
success: function(data) {
if(data!==null){
window.location.replace('show_debate.php?id=' + data);
}else{
window.location.replace('errors/noConnection.html');
}
}
});
}
My PHP code (post_debate.php):
<?php
require('connect.php');
$title = $_POST['post_title'];
$subtitle = $_POST['post_sub'];
$sides = $_POST['post_sides'];
$ajax = $_POST['ajax'];
$date = new DateTime();
$timeStamp = $date->getTimeStamp();
if($ajax==1){
$query = mysql_query("INSERT INTO debates VALUES('','$title','$subtitle','$sides','0','0','$timeStamp')");
$get_data = mysql_query("SELECT id FROM debates WHERE title='$title', subtitle='$subtitle', sides='$sides', timestamp='$timeStamp'");
while($id=mysql_fetch_array($get_data)){
$final_id = $id['id'];
}
exit($final_id);
}else{
die("404 SERVER ERROR");
}
?>
Thanks!
EDIT - NOT SOLVED YET
My new PHP code:
<?php
header("content-type: application/json");
require('connect.php');
$title = $_POST['post_title'];
$subtitle = $_POST['post_sub'];
$sides = $_POST['post_sides'];
$ajax = $_POST['ajax'];
$date = new DateTime();
$timeStamp = $date->getTimeStamp();
if($ajax==1){
$query = mysql_query("INSERT INTO debates VALUES('','$title','$subtitle','$sides','0','0','$timeStamp')");
$get_data = mysql_query("SELECT id FROM debates WHERE title='$title', subtitle='$subtitle', sides='$sides', timestamp='$timeStamp'");
while($id=mysql_fetch_array($get_data)){
$final_id = $id['id'];
}
print (json_encode(array("Id"=>$final_id)));
}else{
die("404 SERVER ERROR");
}
?>
my new Javascript .ajax:
$.ajax({
type: "POST",
url: "post_debate.php",
data: { post_title: title, post_sub: subtitle, post_sides: sides, ajax: 1 },
dataType: "json",
timeout: 5000, // in milliseconds
success: function(data) {
if(data!==null){
window.location.replace('show_debate.php?id=' + data['Id']);
}else{
window.location.replace('errors/noConnection.html');
}
}
});
Your code is expecting JSON as a response...
dataType: "json",
(Documentation Here)
But you're returning a non-json value without an appropriate content-type header.
Try changing your PHP script from
exit($final_id);
to (untested)
header("content-type: application/json");
print (json_encode(array(
"Id"=>$final_id
)));
Also, put a breakpoint on your success callback in your Javascript code (using Firebug or a similar tool) and examine what data contains. It should now be an associative array so you can do
window.location.replace('show_debate.php?id=' + data['Id']);
Improvement:
Instead of doing a SELECT to get the recently inserted Id, use mysql_insert_id(). Something like this...
$query = mysql_query("INSERT INTO debates VALUES('','$title','$subtitle','$sides','0','0','$timeStamp')");
$final_id = mysql_insert_id();
print (json_encode(array("Id"=>$final_id)));
Also, an alternate way to test what your PHP is returning if you can't see the response in your development tool is to browse to the page directly (You'd have to change all your $_POST to $_REQUEST)

Parsing values in JSON

I am trying to pass some values to my PHP page and return JSON but for some reason I am getting the error "Unknown error parsererror". Below is my code. Note that if I alert the params I get the correct value.
function displaybookmarks()
{
var bookmarks = new String();
for(var i=0;i<window.localStorage.length;i++)
{
var keyName = window.localStorage.key(i);
var value = window.localStorage.getItem(keyName);
bookmarks = bookmarks+" "+value;
}
getbookmarks(bookmarks);
}
function getbookmarks(bookmarks){
//var surl = "http://www.webapp-testing.com/includes/getbookmarks.php";
var surl = "http://localhost/Outlish Online/includes/getbookmarks.php";
var id = 1;
$.ajax({
type: "GET",
url: surl,
data: "&Bookmarks="+bookmarks,
dataType: "jsonp",
cache : false,
jsonp : "onJSONPLoad",
jsonpCallback: "getbookmarkscallback",
crossDomain: "true",
success: function(response) {
alert("Success");
},
error: function (xhr, status) {
alert('Unknown error ' + status);
}
});
}
function getbookmarkscallback(rtndata)
{
$('#pagetitle').html("Favourites");
var data = "<ul class='table-view table-action'>";
for(j=0;j<window.localStorage.length;j++)
{
data = data + "<li>" + rtndata[j].title + "</li>";
}
data = data + "</ul>";
$('#listarticles').html(data);
}
Below is my PHP page:
<?php
$id = $_REQUEST['Bookmarks'];
$articles = explode(" ", $id);
$link = mysql_connect("localhost","root","") or die('Could not connect to mysql server' . mysql_error());
mysql_select_db('joomla15',$link) or die('Cannot select the DB');
/* grab the posts from the db */
$query = "SELECT * FROM jos_content where id='$articles[$i]'";
$result = mysql_query($query,$link) or die('Errant query: '.$query);
/* create one master array of the records */
$posts = array();
for($i = 0; $i < count($articles); $i++)
{
if(mysql_num_rows($result)) {
while($post = mysql_fetch_assoc($result)) {
$posts[] = $post;
}
}
}
header('Content-type: application/json');
echo $_GET['onJSONPLoad']. '('. json_encode($posts) . ')';
#mysql_close($link);
?>
Any idea why I am getting this error?
This is not json
"&Bookmarks="+bookmarks,
You're not sending JSON to the server in your $.ajax(). You need to change your code to this:
$.ajax({
...
data: {
Bookmarks: bookmarks
},
...
});
Only then will $_REQUEST['Bookmarks'] have your id.
As a sidenote, you should not use alert() in your jQuery for debugging. Instead, use console.log(), which can take multiple, comma-separated values. Modern browsers like Chrome have a console that makes debugging far simpler.

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