Submit form using Ajax/Jquery and check result? - php

I am wanting to submit a form using ajax to go to my page 'do_signup_check.php'.
There it will check the email address the user entered against the database to see if there is a match.
If there is a match I want my ajax form to redirect the user to the login.php page.
If there isn't a match I want it to load my page 'do_signup.php'
for some reason the code seems to be doing nothing. Please can someone show me where I am going wrong?
My Ajax form:
<html lang="en">
<head>
<script src="https://www.google.com/recaptcha/api.js" async defer></script>
<script src="https://www.google.com/recaptcha/api.js?onload=onloadCallback&render=explicit"
async defer>
</script>
<?php include 'assets/config.php'; ?>
<script>
$(function () {
$('form').on('submit', function (e) {
e.preventDefault();
$.ajax({
type: 'post',
url: 'do_signup_check.php',
data:{"name":name,"email":email},
success: function () {
if(result == 0){
$('.signup_side').fadeOut(500).promise().done(function() {
$('.signup_side').load('do_signup.php',function(){}).hide().fadeIn(500);
});
}else{
$('.signup_side').fadeOut(500).promise().done(function() {
$('.signup_side').load('login.php',function(){}).hide().fadeIn(500);
}
});
});
});
</script>
</head>
<body>
<div class="sign_up_contain">
<div class="container">
<div class="signup_side">
<h3>Get Onboard.</h3>
<h6>Join the directory.</h6>
<form id="signup" action="" method="POST" autocomplete="off" autocomplete="false">
<div class="signup_row action">
<input type="text" placeholder="What's your Name?" name="name" id="name" class="signup" autocomplete="new-password" autocomplete="off" autocomplete="false" required />
<input type="text" placeholder="Got an Email?" name="email" id="email" class="signup" autocomplete="new-password" autocomplete="off" autocomplete="false" required />
<div class="g-recaptcha" style="margin-top:30px;" data-sitekey="6LeCkZkUAAAAAOeokX86JWQxuS6E7jWHEC61tS9T"></div>
<input type="submit" class="signup_bt" name="submit" id="submt" value="Create My Account">
</div>
</form>
</div>
</div>
</div>
</body>
</html>
My Do_Signup_Check.php page:
<?php
session_start();
require 'assets/connect.php';
$myName = $_POST["name"];
$myEmail = $_POST["email"];
$check = mysqli_query($conn, "SELECT * FROM user_verification WHERE email='".$myEmail."'");
if (!$check) {
die('Error: ' . mysqli_error($conn));
}
if (mysqli_num_rows($check) > 0) {
echo '1';
} else {
echo '0';
}
?>

Try to use input id 'submt' to trigger the form as such :
$("#submt").click(function(){
e.preventDefault();
//your logic

You have not included the jquery in your page, the function is also missing the closing parentheses.
Include the jquery at the top
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
Replace your function with the following code:-
<script>
$(function () {
$('form').on('submit', function (e) {alert(123);
e.preventDefault();
$.ajax({
type: 'post',
url: 'receive-callback.php',
data:{"name":name,"email":email},
success: function () {
if(result == 0){
$('.signup_side').fadeOut(500).promise().done(function() {
$('.signup_side').load('do_signup.php',function(){}).hide().fadeIn(500);
});
}else{
$('.signup_side').fadeOut(500).promise().done(function() {
$('.signup_side').load('login.php',function(){}).hide().fadeIn(500);
});
}
}
});
});
});
</script>

After receiving success in Ajax and doing all necessary logic (fadeOut etc) just relocate user to login.php:
window.location.href = 'path/to/login.php';
Otherwise, in error function (it is goes just after success function) do the next:
window.location.href = 'path/to/do_signup.php';
Upd:
In case you want to have some piece of code from another file, you can use jQuery method load.
$('#some-selector').load('path/to/course/login.php');
It's a very basic example of using 'load', go google around it to explore more.

Please Try this code.
$.ajax({
type: 'post',
url: 'do_signup_check.php',
data:{"name":name,"email":email},
success: function (result) {
if(result == 0){
window.location.href = 'do_signup.php';
}else{
window.location.href = 'login.php';
}
});
My Do_Signup_Check.php page:
<?php
session_start();
require 'assets/connect.php';
$myName=$_POST["name"];
$myEmail=$_POST["email"];
$check = mysqli_query($conn, "SELECT * FROM user_verification WHERE email='".$myEmail."'");
if (!$check) {
die('Error: ' . mysqli_error($conn));
}
if(mysqli_num_rows($check) > 0){
echo json_encode(array('result'=>'1'));
exit;
}else{
echo json_encode(array('result'=>'0'));
exit;
}
?>
I hope this will work for you.

Related

Submit form using ajax: Was working, now not working?

I have the following HTML form in signup.php:
<form id="signup" action="" method="POST" autocomplete="off" autocomplete="false">
<div class="signup_row action">
<input type="text" placeholder="What's your Name?" name="name" id="name" class="signup" autocomplete="new-password" autocomplete="off" autocomplete="false" required />
<input type="text" placeholder="Got an Email?" name="email" id="email" class="signup" autocomplete="new-password" autocomplete="off" autocomplete="false" required />
<div class="g-recaptcha" style="margin-top:30px;" data-sitekey="6LeCkZkUAAAAAOeokX86JWQxuS6E7jWHEC61tS9T"></div>
<input type="submit" class="signup_bt" name="submit" id="submt" value="Create My Account">
</div>
</form>
I am trying to submit the form using ajax, without page refresh:
<!-- include files -->
<?php include 'assets/config.php';?>
<?php if(isset($_SESSION["CUSTOMER_ID"])){
header('Location: myaccount.php'); } ?>
<script>
$(function () {
$('form').on('submit', function (e) {
e.preventDefault();
$.ajax({
type: 'post',
url: 'do_signup_check.php',
data:{"name":name,"email":email},
success: function () {
if(result == 0){
$('.signup_side').fadeOut(500).promise().done(function() {
$('.signup_side').load('do_signup.php',function(){}).hide().fadeIn(500);
});
}else{
$('.signup_side').fadeOut(500).promise().done(function() {
$('.signup_side').load('assets/login.php',function(){}).hide().fadeIn(500);
}
});
});
});
</script>
I am posting the form to do_signup_check.php and running a query to see if the user is already registered. echo 1 for a positive result and 0 for a negative result:
Do_Signup_Check.php:
<?php
session_start();
require 'assets/connect.php';
$myName=$_POST["name"];
$myEmail=$_POST["email"];
$check = mysqli_query($conn, "SELECT * FROM user_verification WHERE email='".$myEmail."'");
if (!$check) {
die('Error: ' . mysqli_error($conn)); }
if(mysqli_num_rows($check) > 0){
echo '1';
}else{
echo '0';
}
?>
If the result is 0 then the ajax should load my page do_signup.php.
But alas it is not getting this far. It was working and then i switched off the computer and came back to it and now it won't work.
Please can someone show me where I've gone wrong?
if(result == 0){ here result is not using in success function:
you must need to pass resultant variable here:
success: function () {
as:
success: function (result) {
Now, you can use your condition if(result == 0){
Second, i suggest you to pass dataType: 'html' in your ajax request.
Edit:
You are using <?php if(isset($_SESSION["CUSTOMER_ID"])){ line in your code, if you are not using session_start() in your code then this check will not work.
For this line data:{"name":name,"email":email}, i didnt see name and email in your code, where you define these 2 variables which you are using in your ajax params.

ajax and php code in the same page

I am doing form validation through ajax and php and it works fine.
Now there are two files i.e., ajax.html and codeValidate.php.
html form is in ajax and it checks the code validation using another file i.e. codeValidation.php.
Now I would like to have both in the same file and name it as Ajax.php. The code is as below.
//Ajax file
<html>
<form id="form_submit">
<input id="coupon" name="coupon" type="text" size="50" maxlength="13" />
<input id="btn_submit" type="button" value="Submit">
</form>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script type="text/javascript">
$("form_submit").submit(function(e){ e.preventDefault(); return false;});
$("#btn_submit").click(function(e) { e.preventDefault();
var coupon = $("#coupon").val();
// validate for emptiness
if(coupon.length < 1 ){ window.location.href="success.html"; }
else{
$.ajax({
url: './codeValidate.php',
type: 'POST',
data: {coupon: coupon},
success: function(result){
console.log(result);
if(result){ window.location.href="success.html"; }
else{ alert("Error: Failed to validate the coupon"); }
}
});
}
});
</script>
</html>
My working php file :-
<?php
require("dbconfig.php");
if(isset($_POST['coupon']) && !empty($_POST['coupon']) ){
$coupon = trim($_POST['coupon']);
$checkcoupon = "SELECT couponCode FROM cc WHERE couponCode='".$coupon."'";
$results_coupon = mysqli_query($dbc,$checkcoupon);
if(mysqli_num_rows($results_coupon)) { echo true; }
else { echo false; }
}
?>

Pop-up message after submitting a form with php

I'm trying to get a pop-up message saying if it was successfully submitted or not without having to go to a different page.
Now chrome gives me the pop-up message but it redirects me to a blank page after.
Here is my current code.
<?php
include "header.php";
include "conexao.php";
echo "<h1 align='center'>Pagina para alterar produtos</h1><div class='container'><hr>";
$referencia=$_GET['id'];
$sql = "SELECT * ";
$sql = $sql . " FROM tb_produto ";
$sql = $sql . " WHERE pr_codigo='".$referencia."'";
$produtos = $db->query($sql);
foreach ($produtos as $produto) {
$referencia = $produto["pr_codigo"];
$nome = $produto["pr_descricao"];
$preco = $produto["pr_preco"];
$disponivel = $produto["disponivel"];
}
echo "<h2>Referencia: ".$referencia."</h2>";
echo "<h2>Nome: ".$nome."</h2><hr>";
?>
<form action="confirmaAlterar.php">
<div class="form-group">
<label>Referencia</label>
<input class="form-control" type="text" name="referencia" value="<?php echo $referencia?>">
</div>
<div class="form-group">
<label>Nome</label>
<input class="form-control" type="text" name="nome" value="<?php echo $nome?>">
</div>
<div class="form-group">
<label>Preço</label>
<input class="form-control" type="text" name="preco" value="<?php echo $preco?>">
</div>
<button class="btn btn-primary">Alterar</button>
</form>
Here is where it submits the information of the form.
<?php
include "header.php";
include "conexao.php";
$nome=$_GET['nome'];
$referencia=$_GET['referencia'];
$preco=$_GET['preco'];
$sql="UPDATE tb_produto SET pr_descricao='".$nome;
$sql.="', pr_preco=".$preco;
$sql.= " WHERE pr_codigo='".$
try{
$comando=$db->prepare($sql);
$comando->execute();
echo "<script type='text/javascript'>alert('submitted successfully!')</script>";
header( "refresh2;Location:index.php" );
}
catch (PDOException $e){
echo "A";
}
To pass values using ajax. Form:
<form id="form">
<input type="text" value="test" name="akcija">
</form>
All inputs fields values in your form will be passed.
Ajax:
jQuery(function(){
$('#form').on('submit', function (e) { //on submit function
e.preventDefault();
$.ajax({
type: 'post', //method POST
url: 'yoururl.php', //URL of page where u place query and passing values
data: $('#form').serialize(), //seriallize is passing all inputs values of form
success: function(){ //on success function
$("#input").attr("disabled", true); //example
$("#input").removeClass('btn-primary').addClass('btn-success');//example
},
});
}
});
And on the ajax page you can get values by
$akcija = $_POST['akcija']
for this Problem you must use ajax method .
1- create html form and all input Required .
<form id="contactForm2" action="/your_url" method="post">
...
</form>
2- add jQuery library file in the head of html page .
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js">
</script>
...
3- add this method Under the jQuery library
<script type="text/javascript">
var frm = $('#contactForm2');
frm.submit(function (ev) {
$.ajax({
type: frm.attr('method'),
url: frm.attr('action'),
data: frm.serialize(),
success: function (data) {
if(data == 'pass')
alert('ok');
else(data == 'fail')
alert('no');
}
});
ev.preventDefault();
});
</script>
4- in your_url .php file
<?php
$a = ok ;
if( $a == 'ok' ){
echo 'pass';
}else{
echo 'fail';
}
?>
this top answer is easy management form with jquery , but you need managment Complex form better use this library http://jquery.malsup.com/form/

Form still redirect to PHP with preventDefault

EDIT:
This was related to a typo elsewhere in my Javascript. I had forgotten to check the Javascript console. Thank you for your comments.
This is my first post on this site. I have been reading it for a long while though.
I am working on a login form utilizing jQuery, AJAX, and PHP. Several times now I have run into the problem where I am redirected to the PHP page where I see the echoed data I wanted returned. I have tried to figure this out but I am stuck.
EDIT:
I did include jQuery:
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
<script src="http://code.jquery.com/ui/1.10.3/jquery-ui.js"></script>
<script src="http://malsup.github.com/jquery.form.js"></script>
HTML:
<form name="login" id="loginForm" action="login.php" method="post">
<label for="usernameInput">Username: </label>
<input type="text" name="usernameInput" id="usernameInput" placeholder="Username" autofocus required>
<label for="passwordInput">Password: </label>
<input type="password" name="passwordInput" placeholder="Password" required>
<input type="submit" name="loginSubmit" value="Log In">
</form>
jQuery:
function login () {
$('#loginForm').on('submit', function(e){
e.preventDefault();
var formObject = $(this);
var formURL = formObject.attr("action");
$.ajax({
url: formURL,
type: "POST",
data: formObject.serialize(),
dataType: 'json',
success: function(data)
{
$("#loginDiv").remove();
if(data.new) {
$("#setupDiv").show();
} else {
statusUpdate();
/* EDIT: Changed from dummy text 'continue()' */
}
},
error: function(jqXHR, textStatus, errorThrown)
{
$("#loginDiv").append(textStatus);
}
});
});
}
Call:
$(document).ready(function() {
login();
});
PHP:
// Main
if (isset($_POST['usernameInput'], $_POST['passwordInput']))
{
require "hero.php";
// Starts SQL connection
$sql = getConnected();
$userArray = validateUser($sql);
if ( $userArray['id'] > 0 ) {
sessionSet($userArray);
$userArray['user'] = (array) unserialize($userArray['user']);
$userArray = json_encode($userArray);
echo $userArray;
exit();
}
else
{
echo 'Username does not exist';
}
}
else
{
echo "Please enter a username and password.";
}
I know I have not included everything, but here's the output:
{"id":"11","name":"st5ph5n","new":true,"user":[false]}
So everything up to $userArray is working as expected. Why is this not staying on index.html and instead redirecting to login.php?
Thank you for any responses.

Can a submit button work without refresh AJAX?

I have a problem with my very simple chat. The page is constanly refreshing with AJAX with an timeout of 750ms. If I press or use enter to submit my 'reaction', the page refreshes: is there an way to remove that, so that you can instantly see what you've posted?
You can see the chat at my website: chat
The code:
Index.php
<!DOCTYPE HTML>
<?php include 'config.php'; ?>
<html>
<head>
<script type="text/javascript" src="jquery-1.7.1.js">
function submitenter(myfield,e)
{
var keycode;
if (window.event) keycode = window.event.keyCode;
else if (e) keycode = e.which;
else return true;
if (keycode == 13)
{
myfield.form.submit();
return false;
}
else
return true;
}
</script>
<title>JavaScript Chat</title>
<link href="style.css" rel="stylesheet" type="text/css"/>
</head>
<body>
<div class="container">
<div id="chatwindow">
</div>
<div class="inputMessage">
<form method="post">
enter code here
<hr></hr>
<textarea name="message" rows="1" cols="55"></textarea><br/>Fill username here<br/>
<input type="text" value="" name="username" />
<input type="submit" value="verstuur" name="submit" onKeyPress="return submitenter(this,event)" />
</form>
<?php include 'send.php'; ?>
</div>
<script type="text/javascript">
$(document).ready(function(){
setInterval ( "get()", 750 );
});
function get(){
$.ajax({
type: 'GET',
url: 'chat.php',
success: function(data){
$("#chatwindow").html(data);
}
});
}
</script>
</div>
</body>
</html>
chat.php
<?php
include 'config.php';
$result = mysql_query("select * from Message");
while($row = mysql_fetch_array($result))
{
echo '<p>' . $row['username'] . " : " . $row['message'] . '</p>';
}
?>
send.php
<?php
if(isset($_POST['submit']))
{
if (!empty($_POST['username']))
{
if(!empty($_POST['message']))
{
$message = mysql_real_escape_string(htmlentities($_POST['message']));
$username = mysql_real_escape_string(htmlentities($_POST['username']));
$query = "INSERT INTO Message (`username`,`message`) VALUES ('".$username."','".$message."')";
mysql_query($query);
}
else
{
echo '<script type="text/javascript">alert(\'Je kan niet niks sturen\')</script>';
}
}
else
{
echo '<script type="text/javascript">alert(\'Vul een gebruikresnaam in!\')</script>';
}
}
?>
if my question is not clear say it please.
And is there a topic/question/post about good spacing? google translated it as "indent".
Thanks
Replace
<form method="post">
With
<form onsubmit="event.preventDefault()" method="post">
You may also use your callback function here like:
<form onsubmit="event.preventDefault();return submitenter(this,event);" method="post">
Working demo: http://jsfiddle.net/usmanhalalit/5RCwF/2/
Add e.preventDefault(); in JS.
Your desired action is to prevent the onSubmit action as the other answers have mentioned. Currently your script isn't quite ready to block submit as you don't have an ajax post method.
You need ajax functionality for the submission side of the application still. For this you can use jQuery post().
You want to create something like
function send() {
$.post(); // Fill in appropriate post() code.
return false;
}
And then call it from your event handlers like onsubmit="return send()" and in place of myfield.form.submit() in your keypress handler.

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