I'v got a problem with my site registration form. Getting 'null' from server, using json format of exchange.
Code:
JS part of code:
$("#customer-form").submit(function() {
var form_data = {
customer_name: $("#customer_name").val(),
customer_dir: $("#customer_dir").val(),
confirm_dir: $("#confirm_dir").val(),
customer_fname: $("#customer_fname").val(),
customer_address: $("#customer_address").val(),
customer_inn: $("#customer_inn").val(),
customer_kpp: $("#customer_kpp").val(),
customer_ogrn: $("#customer_ogrn").val(),
customer_okpo: $("#customer_okpo").val(),
};
$.ajax({
type: "POST",
url: "customer_form.php",
data: form_data,
dataType: "json",
success: function(data) {
console.log(data.status);
},
error: function(xhr, status) {
console.log("Error"+" "+xhr.responseText +" "+ status)
}
});
});
PHP
<?php
$customer_fname=trim($_POST["customer_fname"]);
$customer_fname=strip_tags($customer_fname);
$confirm_dir=trim($_POST["confirm_dir"]);
$confirm_dir=strip_tags($confirm_dir);
$confirm_dir = md5($confirm_dir);
$customer_name=trim($_POST["customer_name"]);
$customer_name=strip_tags($customer_name);
$customer_dir = trim($_POST["customer_dir"]);
$customer_dir=strip_tags($customer_dir);
$customer_dir=md5($customer_dir);
$customer_address=trim($_POST["customer_address"]);
$customer_address=strip_tags($customer_address);
$customer_inn=trim($_POST["customer_inn"]);
$customer_inn=strip_tags($customer_inn);
$customer_kpp=trim($_POST["customer_kpp"]);
$customer_kpp=strip_tags($customer_kpp);
$customer_ogrn=trim($_POST["customer_ogrn"]);
$customer_ogrn=strip_tags($customer_ogrn);
$customer_okpo=trim($_POST["customer_okpo"]);
$customer_okpo=strip_tags($customer_okpo);
if($customer_dir==$confirm_dir){
$con = new mysqli("localhost", "root", "", "berezka");
$sql = "SELECT * FROM CUSTOMER where customer_name='".$customer_name."'";
$result = $con->query($sql);
if($row = $result->fetch_assoc()){
$response = array("status"=>"Login has been already given");
echo json_encode($response);
}
else{
$sql_2 = "INSERT INTO customer (`customer_id`, `customer_fname`, `customer_name`, `customer_dir`, `customer_address`, `customer_inn`, `customer_kpp`, `customer_ogrn`, `customer_okpo`) VALUES ('', '$customer_fname', '$customer_name', '$customer_dir', '$customer_address', '$customer_inn', '$customer_kpp', '$customer_ogrn', '$customer_okpo')";
mysqli_query($con, $sql_2);
$response = array("status"=>"Registration is ok");
echo json_encode($response);
// mysqli_free_result($result);
// mysqli_close($con);
}
}
else{
$response = array("status"=>"Password doesnt match");
echo json_encode($response);
}
?>
Here is my error in firefox firegub or Chrome console:
Uncaught TypeError: Cannot read property 'status' of null
I was trying to use JSON.Parse or etc. Tried to make json message by object,array or another format.
Please add header info:
header('Content-Type:application/json');
before sending the response.
Related
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;
i want to insert data to mysql database using php service and json but when i click nothing happens it shows no error no message and the data is not added to the data base help please
here is the save function
function save(){
var eml = document.getElementById("tbemail").value;
var mp = document.getElementById("tbmdp").value;
var data = {email: eml, mdp: mp}
$.ajax({
url:"http://localhost:800/test/insert.php",
type: 'POST',
data: data,
dataType: 'json',
success: function()
{alert("success");}
error: function()
{alert("fail");}
});
}
and this my php file insert.php
<?php
$json = $_POST['data'];
$new=json_decode($json, true);
$conn= mysqli_connect("localhost","root","") or die ("could not connect to mysql");
mysqli_select_db($conn,"bd") or die ("no database");
$sql = "INSERT INTO user (email,mdp) VALUES ($new['email'],$new['mdp'])";
if (mysqli_query($conn, $sql)) {
echo "created ";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
?>
You don't have "data" key in your $_POST array, you have "email" and "mdp", which you can access directly:
$email = mysqli_real_escape_string($_POST['email']);
$mdp = mysqli_real_escape_string($_POST['mdp']);
There is no json passed in this way, similarly when you have get string, you also don't need to parse it. Turn on error reporting, then you will see that $_POST['data'] is undefined.
BTW, use mysqli_real_escape_string to sanitize the input to prevent from injection.
"Insert.php" - > Not use for get data $json = $_POST['data'];
Only use this and try
$conn= mysqli_connect("localhost","root","") or die ("could not connect to mysql");
mysqli_select_db($conn,"bd") or die ("no database");
$email = $_POST['email'];
$mdp = $_POST['mdp'];
$new1 = json_encode($email);
$new2 = json_encode($mdp);
$sql = "INSERT INTO user ('email','mdp') VALUES ('".$new1."','".$new2."')";
$insert = mysqli_query($sql);
if ($insert) {
echo "created ";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
Your PHP code seems to be correct, but please try the jQuery AJAX code as follows:
function save(){
var eml = document.getElementById("tbemail").value;
var mp = document.getElementById("tbmdp").value;
var data = {email: eml, mdp: mp}
$.ajax({
url: "http://localhost:800/test/insert.php",
type: 'POST',
dataType: 'json',
data: JSON.stringify(data),
contentType: "application/json; charset=utf-8",
error: function () {
alert('fail');
},
success: function (data) {
alert('success');
}
});
}
In your data section has to be passed as JSON String, secondly you missed to in include the data contentType. Here content type is set as application/json, therefore pass the data as JSON string.
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.
I am new to php/ajax/jquery and am having some problems. I am trying to pass json to the php file, run some tasks using the json data and then issue back a response. I am using the facebook api to get log in a user,get there details, traslate details to json, send json toe the server and have the server check if the users id already exists in the database. Here is my javascript/jquery
function checkExisting() {
FB.api('/me', function(response) {
console.log('Successful login for: ' + response.id );
var json = JSON.stringify(response);
console.log(json);
$.ajax({
url: "php.php",
type: "POST",
data: {user: json},
success: function(msg){
if(msg === 1){
console.log('It exists ' + response.id );
} else{
console.log('not exists ' + response.id );
}
}
})
});
}
Here is my php file
if(isset($_POST['user']) && !empty($_POST['user'])) {
$c = connect();
$json = $_POST['user'];
$obj = json_decode($json, true);
$user_info = $jsonDecoded['id'];
$sql = mysql_query("SELECT * FROM user WHERE {$_GET["id"]}");
$count = mysql_num_rows($sql);
if($count>0){
echo 1;
} else{
echo 0;
}
close($c);
}
function connect(){
$con=mysqli_connect($host,$user,$pass);
if (mysqli_connect_errno()) {
echo "Failed to connect to Database: " . mysqli_connect_error();
}else{
return $con;
}
}
function close($c){
mysqli_close($con);
}
I want it to return either 1 or 0 based on if the users id is already in the table but it just returns a lot of html tags. . The json looks like so
{"id":"904186342276664","email":"ferrylefef#yahoo.co.uk","first_name":"Taak","gender":"male","last_name":"Sheeen","link":"https://www.facebook.com/app_scoped_user_id/904183432276664/","locale":"en_GB","name":"Tadadadn","timezone":1,"updated_time":"2014-06-15T12:52:45+0000","verified":true}
Fix the query part:
$sql = mysql_query("SELECT * FROM user WHERE {$_GET['id']}");
Or another way:
$sql = mysql_query("SELECT * FROM user WHERE ". $_GET['id']);
Then it's always better to use dataType in your ajax
$.ajax({
url: "php.php",
type: "POST",
data: {user: json},
dataType: "jsonp", // for cross domains or json for same domain
success: function(msg){
if(msg === 1){
console.log('It exists ' + response.id );
} else{
console.log('not exists ' + response.id );
}
}
})
});
Where is $jsonDecoded getting assigned in your PHP? Looks unassigned to me.
I think you meant to say:
$obj = json_decode($json, true);
$user_info = $obj['id'];
And your SELECT makes no sense. Your referencing $_GET during a POST. Maybe you meant to say:
$sql = mysql_query("SELECT * FROM user WHERE id = {$user_info}");
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)