Develop Reference

echo is adding a new line to what i output in PHP

I am sending echoing some data to be received in Javascript however when i debug it, it seems that a new line has been added.
PHP
<?php
header("Content-Type: application/json; charset=UTF-8");
require './connection.php';
$obj = json_decode($_POST["x"], false);
$usernamequery = "SELECT * FROM User WHERE username='$obj->newUser'";
$result = mysqli_query($db, $usernamequery);
$row = mysqli_fetch_assoc($result);
if($row["Username"] == null){
$updatequery = "UPDATE User SET User='$obj->newUser' WHERE username ='$obj->username'";
$result = mysqli_query($db, $updatequery);
echo "valid";
} else{
echo "invalid";
}
?>
JS
///// USERNAME
$(document).ready(function () {
$("#userSubmitForm").on("click", function(e) {
$username = document.getElementById("user").value;
$newUser = document.getElementById("newUser").value;
user = $newUser;
obj = { "username":$username, "newUser":$newUser};
dbParam = JSON.stringify(obj);
xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (this.readyState === 4 && this.status === 200) {
validity = this.responseText;
if (validity === "valid"){
$('#usernameModal .modal-header .modal-title').html("Result");
$('#usernameModal .modal-body').html("Your Username Has Been Changed to '$newUser'");
$("#passSubmitForm").remove();
$("#userCloseForm").remove();
window.setTimeout(redirect,3000);
} else{
$('#error').html("This Username Already Exists"); ;
}
}
};
What is happening is responseText will be receive "valid"/"Invalid" as "valid[newline]"/"invalid[newline]"

As stated at http://php.net/manual/en/function.echo.php that can't be a "problem" of the echo. There must be some newline-character after your -tags

A simple solution would be to just trim your response text like this: var validity = this.responseText.trim(); in order to strip it from unwanted space/tab/newline characters.

Using Php, mySql, ajax, (xmlhttp) and JS,how can i create a php multidimensional send it to JS and use it?

I looked a lot on the internet and wasn't able to find the answer i need, so here i come to you.
What i have : A database which look like this :
name latitude longitude
---- --------- ----------
foo 13.323 -51.356
foo 54.698 2.487
What i want to do : I need to retrieve the latitude and longitude from a mysqli request done with php and use it in a function that i defined.
My problem : I'm trying to use xmlrequest but it apparently doesn't work.
The code : JS :
var selI = document.getElementById("nameIti");
selI.onchange = function(){
var val = this[this.selectedIndex].getAttribute("value");
showMark(val);
}
function showMark(str){
var xhr;
if(str==""){
return;
}
if(window.XMLHttpRequest){
xhr=new XMLHttpRequest();
}
else{
xhr=new ActiveXObject("Microsoft.XMLHTTP");
}
xhr.onreadystatechange=function(){
if(xhr.readyState==4 && xhr.status ==200){
var object = JSON.parse(xhr.responseText);
for(var a in object){
newMark(v['lat'], v['lng']);
document.getElementById("pi").innerHTML=JSON.parse(xrh.responseText); // This is a test to display any kind of result.
}
}
}
xhr.open("GET", "getpos.php?q="+str, true);
xhr.send();
}
PHP :
<?php
$nom = $_GET['q'];
include("connexion.php");
$con = connect_LIF4();
$req1= "SELECT Latitude, Longitude FROM etape LEFT JOIN itineraires ON NomLieu=nomEtape WHERE nomIti LIKE '%$nom%'";
$result1 = mysqli_query($con, $req1);
$data = array();
while($row = mysqli_fetch_array($result1){
$data['lat'] = $row['Latitude'];
$data['lng'] = $row['Longitude'];
$resp[] = $data;
}
echo json_encode($resp);
mysqli_close($con);
?>
I tried to use newMark(lat, lng)(Which i coded and works fine) with random values, in showMark outside the onreadystatechange and it works, but i need to use it with the values retrieved from the php.

One problem with your PHP is that
while($row = mysqli_fetch_array($result1){
is missing the second brace. It should be:
while($row = mysqli_fetch_array($result1)){
Also the URL in the ajax request should be the full URL, not just getpos.php
Thirdly you have written xrh.responseText (should be xhr).
Basically there's loads of syntax errors in your code - you should use the javascript console to debug the front end ones, and PHP logging or error display for the back end ones. You should only need help here once you've debugged all obvious syntax errors.
EDIT - below is a working example (although I haven't done the MySQL part)
JS + HTML:
<span id='pi'></span>
<select id='nameIti'>
<option value='foo'>foo</option>
<option value='bar'>bar</option>
</select>
<script>
function newMark(lat,lng) {
console.log(lat);
console.log(lng);
}
var selI = document.getElementById("nameIti");
selI.onchange = function(){
var val = this[this.selectedIndex].getAttribute("value");
showMark(val);
}
function showMark(val){
var str=val;
var xhr;
// if(str==""){
// return;
// }
if(window.XMLHttpRequest){
xhr=new XMLHttpRequest();
}
else{
xhr=new ActiveXObject("Microsoft.XMLHTTP");
}
xhr.onreadystatechange=function(){
if(xhr.readyState==4 && xhr.status ==200){
var result = JSON.parse(xhr.responseText);
console.log(result);
for(var a in result){
newMark(result[a]['lat'], result[a]['lng']);
document.getElementById("pi").innerHTML = result[a]['lat'] + ', ' + result[a]['lng'];
}
}
}
// xhr.open("GET", "getpos.php?q="+str, true);
xhr.open("GET", "getpos.php?q="+str, true);
xhr.send();
}
</script>
PHP:
<?php
$nom = $_GET['q'];
$data = array();
if($nom == 'foo') {
$data['lat'] = '5.12';
$data['lng'] = '0.34';
$resp[] = $data;
}
else if($nom == 'bar') {
$data['lat'] = '2.34';
$data['lng'] = '1.34';
$resp[] = $data;
}
echo json_encode($resp);
?>

I can't retrieve JSON data array from php page to another

<?php
require_once './db_connect.php';
$db = new DB_Connect();
$db->connect();
$data = json_decode($_POST['myData']);
$array=json_decode($_REQUEST['question']);
if(isset($_POST['myData'])){
$obj = json_decode($_POST['myData']);
//some php operation
$q = "insert into questions(question)
values ('". $obj."')";
$result = mysql_query($q) or die(mysql_error());
}
?>
I want to retrieve the JSON data that is being sent from another php page to this page , but I just can't ,,why is that ?
here's the other page
function validateForm()
{
var q = document.forms["form1"]["question"].value;
var T = document.forms["form1"]["title"].value;
if (T == null || T == "")
{
alert("please type you form title first");
return false;
}
if (q == null || q == "")
{
document.getElementById("question").style.color="black";
alert("please enter your question");
return false;
}
question.push(q);
//alert(JSON.stringify(question));
var xhr = new XMLHttpRequest();
xhr.open('post', 'create_form.php',true);
// Track the state changes of the request
xhr.onreadystatechange = function(){
// Ready state 4 means the request is done
if(xhr.readyState === 4){
// 200 is a successful return
if(xhr.status === 200){
alert(xhr.responseText); // 'This is the returned text.'
}else{
alert('Error: '+xhr.status); // An error occurred during the request
}
}
}
// Send the request to send-ajax-data.php
xhr.send({myData:JSON.stringify(question)}); //+encodeURI(JSON.stringify(question))
// addField();
return true;
}
can someone please help me ??
I'm tried to solve this using jquery ajax , it's just the same ,, that's why i tried to use only javascript to solve this

try this
$json = file_get_contents('php://input');
$obj = json_decode($json, TRUE);
instead of this
$data = json_decode($_POST['myData']);
$array=json_decode($_REQUEST['question']);
if(isset($_POST['myData'])){
$obj = json_decode($_POST['myData']);

Ajax not working in visual studio for windows(Blend)

I am using ajax to login user but this script is not working. When i call login function to execute nothing happens..
function login() {
var login = new XMLHttpRequest;
var e = document.getElementById("email").value;
var p = document.getElementById("password").value;
var vars = "email=" + e + "&password=" + p;
login.open("POST", "http://example.com/app/login.php", true);
login.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
login.onreadystatechange = function(){
if (login.readyState == 4 && login.status == "200") {
var response = login.responseText;
//document.getElementById("status").innerHTML = response;
}
document.getElementById("status").innerHTML = "Loggin in....";
login.send(vars);
if (response == "Sucess") {
window.location.replace("/logged.html");
}
else {
document.getElementById("status").innerHTML == "Login Failed";
}
}
}
login.php contains following codes
require_once('../db_/connection.php');//Holds connection to database
$email = mysqli_real_escape_string($con, $_POST['email']);//Sanitizing email
$pass = md5($_POST['password']);//Hashing password
$sql = "SELECT id FROM users WHERE email='$email' AND password='$pass' LIMIT 1";
$query = mysqli_query($con, $sql);
$check = mysqli_num_rows($query);
if($check < 1){
echo "fail";
//mysqli_close($con);
exit();
}
else{
echo "Sucess";
//mysqli_close($con);
exit();
}
Calling login function does not execute the code.

Line 2: The parentheses are missing after new XMLHttpRequest
Line 15: Your if (response == "Sucess") { ... } is misspelled and it gets executed before the ajax request is returned, because it's asynchronous
I made a working version at JSFiddle for you: http://jsfiddle.net/eRv83/
You might find the MDN referrence helpfull: https://developer.mozilla.org/en-US/docs/Web/API/XMLHttpRequest

Ajax post request login issue

We are having a issue when trying to login. Is we send our username and password over a XMLhttprequest as post there parameters do not seem to be send with them and therefor we are unable to login.
The code is as following:
Javascript file
$("#submit").click(function(){
console.log("click");
usernm= document.getElementById("username").value;
passwd= document.getElementById("password").value;
var send2 = "username=" + usernm + "&password=" + passwd;
var request = new XMLHttpRequest;
request.open('POST' , "myurl.com/login.php",true);
request.dataType=('jsonp');
request.setRequestHeader("Content-type","application/x-www-form- urlencoded");
request.onreadystatechange = function() {//Call a function when the state changes.
if(request.readyState == 4 && request.status == 200) {
alert(request.responseText);
}
}
request.send(send2);
The login.php is this
require_once 'connect.php';
session_start();
$uName = ($_GET['username']);
$pWord = ($_GET['password']);
$login = "SELECT Username,Password FROM User WHERE Username = '$uName' and Password='$pWord'";
$res = mysql_query($login);
$num_row = mysql_num_rows($res);
$row=mysql_fetch_assoc($num_row);
if( $num_row == 1 ) {
echo "true";
}
else {
echo "false";
}

You collapsed both POST and GET method,
Modify this one,
$uName = ($_POST['username']);
$pWord = ($_POST['password']);

$uName = ($_POST['username']);
$pWord = ($_POST['password']);