I'm trying to implement the invisible reCaptcha onto a website. But I can not get it working. Here is what I'm doing:
header
<!-- Invisible reCaptcha -->
<script src="https://www.google.com/recaptcha/api.js" async defer></script>
form.php
<form id="contact-form" class="contact-form" action="#contact">
<p class="contact-name">
<input id="contact_name" type="text" placeholder="Full Name" value="" name="name" />
</p>
<p class="contact-email">
<input id="contact_email" type="text" placeholder="Your E-Mail Address" value="" name="email" />
</p>
<p class="contact-message">
<textarea id="contact_message" placeholder="Your Message" name="message" rows="15" cols="40"></textarea>
</p>
<p class="contact-submit">
<a type="submit" id="contact-submit" class="submit" href="#">Send Your Email</a>
</p>
<div id="recaptcha" class="g-recaptcha" data-sitekey="6LceN0sUAAAAAPvMoZ1v-94ePuXt8nZH7TxWrI0E" data-size="invisible" data-callback="onSubmit"></div>
<div id="response">
</div>
</form>
script.js
// contact form handling
$(function() {
$("#contact-submit").on('click',function() {
$contact_form = $('#contact-form');
var fields = $contact_form.serialize();
var test = grecaptcha.execute();
console.log(fields);
console.log(test);
$.ajax({
type: "POST",
url: "assets/php/contact.php",
data: fields,
dataType: 'json',
success: function(response) {
if(response.status){
$('#contact-form input').val('');
$('#contact-form textarea').val('');
}
$('#response').empty().html(response.html);
}
});
return false;
});
});
contact.php
private function validateFields(){
// Check reCaptcha
if(!$this->captcha){
echo "Please fill out the reCaptcha correctly";
}
$response=file_get_contents("https://www.google.com/recaptcha/api/siteverify?secret=".$secretkey."&response=".$this->captcha."&remoteip=".$_SERVER['REMOTE_ADDR']);
if(intval($responseKeys["success"]) !== 1) {
echo "You are a bot! GO AWAY!";
}
The backend (contact.php) is working fine, if g-recaptcha-response is not null. However my Problem is that g-recaptcha-response (in var fields and test) is always empty when I try to do it. When I show the recaptcha on the form and fill it out, the g-recapcha-response is not empty and everything works fine.
I know that I must invoke the grecaptcha.execute() somewhere, but even if I do, the variable is empty. How do programmaticaly call the this?
I appreciate every help! Thank you in advance!
You are missing the onSubmit() callback function.
To rearrange your js to utilize the function, this would be your new js block:
<script>
// this block must be defined before api.js is included
function onSubmit(token) {
var fields = $('#contact-form').serializeArray(); // get your form data
fields.push({name: "g-recaptcha-response", value: token});// add token to post
$.ajax({
type: "POST",
url: "assets/php/contact.php",
data: fields,
dataType: 'json',
success: function(response) {
if(response.status) {
$('#contact-form input').val('');
$('#contact-form textarea').val('');
}
$('#response').empty().html(response.html);
}
});
grecaptcha.reset();// to reset their widget for any other tries
}
</script>
<script src="https://www.google.com/recaptcha/api.js" async defer></script>
<script>
// this block can be defined anywhere
$(function() {
$("#contact-submit").on('click',function() {
// call grecaptcha.execute, which causes recaptcha to
// do its thing and then calls onSubmit with the token
grecaptcha.execute();// does not return anything directly
return false;
});
});
</script>
Related
In my webapp, the Ajax request is executed 3 times, and I have no idea why this is happening.
Can someone please help here?
My Javascript:
$(document).ready(function() {
console.log("ready!");
$('form').on('submit', function(e) { //
e.preventDefault();
// on form submission ...
console.log("the form has beeen submitted");
// grab values
valueOne = $('input[name="perfid"]').val();
valueTwo = $('input[name="hostname"]').val();
valueThree = $('input[name="iteration"]').val();
console.log(valueOne)
console.log(valueTwo)
console.log(valueThree)
$.ajax({
type: "POST",
url: "/",
dataType:'json',
data : { 'first': valueOne,'second': valueTwo,'third': valueThree},
success: function(data) {
var res = data.AVG;
var p = '<p><pre>'+res+'</pre></p>';
$('#result').append(p);
},
error: function(error) {
console.log(error)
}
});
}); });
And my HTML is:
<form role="form" method="post" onsubmit="return false;">
<div class="form-group">
<input type="text" class="input-medium" id="perfid" name="perfid" placeholder="Enter a Perf ID" required style="height:30px;">
<input type="text" class="input-medium" id="hostname" name="hostname" placeholder="Enter a HostName" style="height:30px;">
<input type="text" class="input-medium" id="iteration" name="iteration" placeholder="Enter a Iteration" required style="height:30px;">
<button type="submit" class="btn btn-default" style="height:30px;">Get Data</button>
</div>
</form>
I have written the code for only one AJAX POST request,
EDIT:
This is the console output:
Please make sure you have included the js file only once,
and add a return false at the end of the submit event callback
look at the selector
$('form').on('submit', function(e) {
if the page has 3 forms, the above selector will execute 3 times
Try to add id to the form like this. sorry about my bad english
I am getting problem to save my form data in the database. I am done small code on that which is shown below, when i enter data in form and click on my submit button it not work.
$(".btn").click(function(e) {
e.preventDefault();
var form = $("#frm");
$.ajax({
url: '/Form/Index',
data: form.serialize(),
type: 'POST',
success: function(data) {
}
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<form role="form" id="frm">
<div class="form-group">
<div class="col-sm-6 col-lg-12 col-md-12">
<div class="form-group">
<label for="name" style="color:black;">Product Name</label>
<input type="text" class="form-control" id="name"
placeholder="Product Name" style="color:black;">
</div>
<div class="form-group">
<label for="name" style="color:black;">Product Date</label>
<input type="text" class="form-control" id="Text1"
placeholder="Date" style="color:black;">
</div>
<div class="form-group">
<label for="name" style="color:black;">Product Price</label>
<input type="text" class="form-control" id="Text2"
placeholder="Date" style="color:black;">
</div>
</div>
</div>
<button type="submit" class="btn btn-default" id="ok" >Submit</button>
</form>
Above is my code please give me solution on that
As I've checked you code, client side code is working fine, The only problem I can imagine in this case is you url path.
make sure you are providing correct url path.
You should check if its hitting the that page or not.
Which Framework you are using. Different framework has different syntax to pass the value in URL. Check the path you are getting in the page source page view in URL parameter or you can check the error in console log after the submit. It may be not getting the correct path of your action.
Make sure ajax library loaded successfully, and try to have alert messages to have forward step where you reached, have this test:
$(".btn").click(function(e) {
e.preventDefault();
var form = $("#frm");
$.ajax({
url: '/Form/Index',
data: form.serialize(),
type: 'POST',
success: function(data) {
},
beforeSend: function() {
alert('before send alert')
},
error: function (request, status, error) {
alert(error);
},
});
});
if beforeSend not executed so your issue is related to ajax library.
use this :
$("#ok").click(function(e) {
// your code
}
Refer to id in javascript rather than class attribute.
If you refer class attribute than once it has click javascript perform preventDefault on that class so that if not refresh your page, The button is not working.
Put preventDefault function at last of your function.
Remove the type="submit" from button
You have to get the form submit with id and serialize the form data
`
$("#formid").submit(function(e) {
var url = "urlpathtohandlerequest";
$.ajax({
type: "POST",
url: url,
data: $("#formid").serialize(),
success: function(response)
{
alert(response);
}
});
e.preventDefault(); // stops default submit.
});
`
I have a modal form using bootstrap. The form contains some text inputs and a image input.
I submit the form with ajax, and all data is received at the PHP file correctly. Alas, the image isn't being uploaded.
What is my code problem?
<script>
$(document).ready(function () {
$("input#submit").click(function(){
$.ajax({
type: "POST",
url: "insert.php",
data: $('form.contact').serialize(),
success: function(msg){
$("#th").html(msg)
$("#form-content").modal('hide');
$("#pro").html(content);
},
error: function(){
alert("failure");
}
});
});
});
</script>
The form:
<form class="contact" name="contact" enctype="multipart/form-data">
<label for="inputNombre" class="sr-only">Título</label>
<input id="inputNombre" name="inputNombre" class="form-control" placeholder="Título" required="TRUE" autofocus="" type="text">
<br>
....
<div class="upload_pic1 inline">
<input id="imagen" name="imagen" type="file">
</div>
<div class="modal-footer">
Cerrar
<input id="submit" class="btn btn-success" type="submit" value="Crear">
</div>
</div>
</form>
EDIT:
insert.php
<?php
session_start();
if (!isset($_SESSION["name"]) && $_SESSION["name"] == "") {
// user already logged in the site
header("location: Login.html");
}
require_once('funt.php');
conectar('localhost', 'root', '', 'db');
if (isset($_POST['inputNombre'])) {
$nombre = strip_tags($_POST['inputNombre']);
....
//Here the var imagen
if(is_uploaded_file($_FILES['imagen']['tmp_name'])){
$rutaEnServidor='imagenes';
$rutaTemporal=$_FILES['imagen']['tmp_name'];
$nombreImagen=$_FILES['imagen']['name'];
$rutaDestino=$rutaEnServidor.'/'.$nombreImagen;
move_uploaded_file($rutaTemporal,$rutaDestino);
} else { //Always enter here, so is not uploaded
$rutaEnServidor='imagenes';
$rutaTemporal='/noPicture.png';
$rutaDestino=$rutaEnServidor.'/noPicture.png';
move_uploaded_file($rutaTemporal,$rutaDestino);
}
...
How can I change this and upload the picture with all data in form?
You can use FormData interface. Then you have to tell jQuery not to set content type, nor process data.
Check compatibility table for the FormData constructor first. It might suffice.
Otherwise read through this discussion, How can I upload files asynchronously?.
UPDATE - the contact form is found at this URL.
I am trying to get the following contact form to function, using this tutorial.
I manage to get everything to work as expected on my computer using apache webserver.
After uploading the files to an online website, the ajax function does not kick in.
I seems like the e.preventDefault(); stops working after the upload, and the form is redirected to a new site,and not just being processed on the site without the reload.
I have also been trying to use the return false; instead of e.preventDefault(); without any success.
Her is my code:
.html
<form method="post" action='mail/mail.php'>
<label>Name</label>
<input name="name" id="name" placeholder="Name.." required="true" class="input-field">
<label>Mail</label>
<input type="email" name="email" placeholder="Mail.." required="true" class="input-field">
<label>Msg</label>
<textarea name="message" id="message" class="textarea-field" required="true"></textarea>
<input type="submit" id="submit" name="submit" value="Send">
</form>
<div id="loading">
Sender melding...
</div>
<div id="success">
</div>
.js
$(function(){
$('form').submit(function(e){
var thisForm = $(this);
//Prevent the default form action
//return false;
e.preventDefault();
//Hide the form
$(this).fadeOut(function(){
//Display the "loading" message
$("#loading").fadeIn(function(){
//Post the form to the send script
$.ajax({
type: 'POST',
url: thisForm.attr("action"),
data: thisForm.serialize(),
//Wait for a successful response
success: function(data){
//Hide the "loading" message
$("#loading").fadeOut(function(){
//Display the "success" message
$("#success").text(data).fadeIn();
});
}
});
});
});
})
Please help!
That's because your JS is missing a closing });. Please check this demo to confirm that the default action is indeed prevented and the ajax does kick in. However, I was expecting a POST but instead I am seeing an OPTIONS request.
NOTE: Giving an element a name or id attribute value of submit is bad practice. You cannot for example use JavaScript to submit the form via default form submission -- this.submit() or $('form')[0].submit() without getting the error ...submit() is not a function .....
$(function() {
$('form').submit(function(e) {
var thisForm = $(this);
//Prevent the default form action
//return false;
e.preventDefault();
//Hide the form
$(this).fadeOut(function() {
//Display the "loading" message
$("#loading").fadeIn(function() {
//Post the form to the send script
$.ajax({
type: 'POST',
url: thisForm.attr("action"),
data: thisForm.serialize(),
//Wait for a successful response
success: function(data) {
//Hide the "loading" message
$("#loading").fadeOut(function() {
//Display the "success" message
$("#success").text(data).fadeIn();
});
}
});
});
});
});
}); // <==== MISSING THIS
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<form method="post" action='mail/mail.php'>
<label>Name</label>
<input name="name" id="name" placeholder="Name.." required="true" class="input-field">
<label>Mail</label>
<input type="email" name="email" placeholder="Mail.." required="true" class="input-field">
<label>Msg</label>
<textarea name="message" id="message" class="textarea-field" required="true"></textarea>
<input type="submit" id="submit" name="submit" value="Send">
</form>
<div id="loading">
Sender melding...
</div>
<div id="success">
</div>
Since you are submitting via AJAX anyway, you may find it easier to change your input type to button, and bind to click instead of form submit, to avoid the default submit behaviour you are trying to circumvent.
I'm using Ajax to submit the login form without refreshing the page. I've added a function to see whether the data returns 'error' (which comes up when the user enters an incorrect email/password). If it does not return 'error', the user has been logged in and will be transferred to the page within 2 seconds.
The problem is that my button acts like a double-click button and I cannot see why. This is my JS file:
$(function() {
$("#goLogin").click(function() {
$.ajax({
type: "POST",
url: "db-requests/db-login.php",
data: $("#loginForm").serialize(),
success: function(data,textStatus,jqXHR){ finishLogin(data,textStatus,jqXHR); }
});
});
});
function finishLogin( data , textStatus ,jqXHR ) {
if ( data == "error" ) {
$('.errorMsg').fadeIn(500).hide();
$('.succesMsg').fadeOut(300).hide();
} else {
$('.succesMsg').fadeIn(500).show();
$('.errorMsg').fadeOut(300).hide();
setTimeout("location.href = 'protected.php';",2000);
}
}
I've tried placing it between the document_ready tags, but that isn't working either.
Part of the HTML code:
<div class="login form">
<div class="login-header">Please Login</div>
<form method="post" id="loginForm" name="form">
<label for="email" class="short">Email*</label>
<input type="text" name="email" id="email" class="required" placeholder="" />
<label for="password" class="short">Password *</label>
<input type="password" name="password" id="password" class="required" placeholder="" maxlength="15" />
</form>
<div id="login-functions">
<div class="loginbtn-container">
<input type="submit" id="goLogin" name="goLogin" class="button green" value="Login" />
</div>
<div class="login form actions">
<p class="register account">Register an account</p>
<p class="request password">Lost your password?</p>
</div>
</div>
</div>
<div class="errorMsg">Incorrect. Please recheck your details</div>
<div class="succesMsg"><b>You've been logged in!</b> Please wait while we transfer you</div>
$('.errorMsg').fadeIn(500).hide();
$('.succesMsg').fadeOut(300).hide();
did you mean tto hide both? I see the click is working fine, though you should ideally do submit
Take your submit inside the form, and prevent normal form submit using preventDefault()
$("#goLogin").submit(function(e) {
e.preventDefault();
$.ajax({
type: "POST",
url: "db-requests/db-login.php",
data: $("#loginForm").serialize(),
success: function(data,textStatus,jqXHR){ finishLogin(data,textStatus,jqXHR); }
});
});
Please move your submit button inside the form closing tag first
<input type="submit" id="goLogin" name="goLogin" class="button green" value="Inloggen" />
The above button is placed after the </form> tag.
Because you click on input type submit and progress Ajax on it; it cause submit 2 times.
To avoid it, you can use as Zach Leighton said above ; or use as below
$("#goLogin").click(function(e) {
e.preventDefault();
$.ajax({
type: "POST",
url: "db-requests/db-login.php",
data: $("#loginForm").serialize(),
success: function(data,textStatus,jqXHR){ finishLogin(data,textStatus,jqXHR); }
});
});