How to use server validation & client-side validation with jQuery Tools? - php

Ok, so I have the following form in a jQuery overlay:
<div class="profile_about_edit_container" id="profile_about_edit_container">
<form id="profile_edit_form" name="profile_edit_form" method="post" action="validation.php">
<label>First Name:</label>
<input type="text" name="firstname" maxlength="50" size="30">
<label>Last Name:</label>
<input type="text" name="lastname" maxlength="50" size="30">
<button type="submit" class="save">Save</button>
<button class="close">Cancel</button>
</form>
</div>
This is displayed using an <a> with class="profile_popups_form" and the following Javascript:
$(document).ready(function() {
$(".profile_popups_form").overlay({
});
});
This shows correctly, and validation.php then echo's an array of error messages like so:
if (count($errors) > 0) {
echo json_encode($errors);
}
But now I'm trying to use jQuery client & server validation on this form.
I tried this:
$(document).ready(function(){
var form = $("#profile_edit_form");
$("#profile_edit_form").submit(function(e) {
e.preventDefault();
var input = $("#profile_edit_form").validator();
$.getJSON(form.attr("action") + "?" + form.serialize(), function(json) {
if (json !== '') {
input.data("validator").invalidate(json);
}
else
$('#profile_edit_form').unbind('submit').submit();
});
});
With the objective of submitting the form and displaying this array of error messages in the normal way jQuery Tools Validation does. But I'm having no luck.
Am I approaching this right? If so, what am I doing wrong? I'm not sure if it's the Javascript causing the issue, or if I'm approaching this right logically. I can find literally no resources explaining how to use JQuery Tools Validation with PHP successfully.
Currently the array is just displayed on the screen as if you echo'd text.
I used the following resource to get the code for returning the array:
http://www.abdullahyahya.com/2012/06/20/javascript-client-side-form-validation-using-server-side-form-validation/

Try doing an ajax request to a php file and get back the response from server. The client side can be done with various ways; from HTML5 tags to plain regex
data = $(this).serializeArray();
//we serialized the data of the form and then we do the ajax request
$.ajax({
url: 'validate.php',
type: 'post',
data: data,
dataType : 'json',
success: function (dat) {
if(dat.error==0){
//dat.error is my returned error from the php file
}else{
//handle the errors
}
}
},
error: function(){
//that's if something is wrong with sent data
alert('failure');
}
});

Related

How to troubleshoot an AJAX call via jQuery to a PHP page [duplicate]

This question already has an answer here:
How to get php page to receive ajax post from html page
(1 answer)
Closed 3 years ago.
I am trying to get help for the second time in two days. This post is my previous question.
I have a very simple form that has one input field in it for First Name. The form was created using Bootstrap 4. Here is the bare HTML code:
<form action="form.php" method="POST">
<label for="firstName">First name</label>
<input type="text" class="form-control" name="firstName" id="firstName" placeholder="First name">
<button class="btn btn-primary" type="submit">Submit form</button>
</form>
I am trying to capture the value of the firstName input field and send it as string to my PHP page via the jQuery ajax method. At the PHP page, all what I wanted to do at the moment is to save it to a variable and echo the value of that variable back to the HTML page without refreshing the page.
I failed to do so after I tried everything that I know. Not sure what I am doing wrong and where to go from here.
Here is my ajax call:
<script>
$(document).ready(function() {
$('form').submit(function(event) {
var formData = {
var formData = $("form").serialize();
console.log(formData);
$.ajax({
type: 'POST',
url: 'form.php',
data: formData,
dataType: 'json',
encode: true
})
.done(function(data) {
console.log(data);
});
event.preventDefault();
});
});
</script>
And,
<?php
if (isset($_POST['firstName'])) {
$ajaxData = $_POST['firstName'];
echo $ajaxData;
}
?>
There is an error in the Javascript.
This line var formData = { should not be there.
As you are specifying that you are expecting json returned from the server in dataType: 'json' you could update the php part to:
if (isset($_POST['firstName'])) {
header('Content-Type: application/json');
echo json_encode($_POST['firstName']);
exit;
}

Sending Input from a Form via Ajax Request

After following the example and answers by the following threads
jQuery AJAX submit form
submitting a form via AJAX
I have built a similar test form to get to learn the ajax request on submit. Your guess was right, it doesn't work for me (no alert popping up).
My testajax.php with the form:
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.4/jquery.min.js"></script>
<script src="../test.js"></script>
<form name="feedback" id="idForm" action="(myurl)/testajax.php" method="post">
<input id="name" type="text">
<input type="submit" name="feedbacksent" value="Send" />
</p>
</form>
My test.js:
// this is the id of the form
$("#idForm").submit(function(e) {
var url = "(myurl)/testajaxinput.php"; // the script where you handle the form input.
e.preventDefault(); // avoid to execute the actual submit of the form.
alert("bla"); // does not work either
$.ajax({
type: "POST",
url: url,
data: $("#idForm").serialize(), // serializes the form's elements.
success: function(data)
{
alert(data); // show response from the php script.
}
});
});
My testajaxinput.php that should handle the input:
if (isset($_POST['feedbacksent'])){
echo "<h1>".WORKS."</h1>";
}
Try this :
if (isset($_POST['feedbacksent'])){
echo "<h1>".WORKS."</h1>";
return true;
}
Then try your alert and also check have you got any error in console.

ajax form submit -> receive respone from php

I've been reading multiple threads about similar cases but even now I'm still unable to do it correctly.
What I want to do
Basically, i.e. I have form which allows user to change his login (simply query to database).
PHP script looks like that:
if(isset($_POST['login'])) {
$doEdit = $user->editData("login", $_POST['login']);
if($doEdit) {
$result = displayInfobox('success', 'Good!');
} else {
$result = displayInfobox('warning', 'Bad!');
}
} else {
$error = 'Bad!';
echo $error;
}
displayInfobox is just a div with class i.e. success and content - Good!.
Right now I would like to send this form by AJAX and display $result without reloading page.
HTML:
<form id="changeLogin" method="post" class="form-inline" action="usercp.php?action=editLogin">
<label for="login">Login:</label><br />
<div class="form-group ">
<input type="text" class="form-control" name="login" id="login" required>
<input type="submit" value="ZmieƄ" class="btn btn-primary">
</div>
</form>
And finnally - my jquery/ajax:
$("#changeLogin").submit(function(e) {
var postData = $(this).serializeArray();
var formURL = $(this).attr("action");
$.ajax({
url: formURL,
type: "POST",
data: postData,
success: function(result) {
alert(result);
},
error: function(response) {}
});
e.preventDefault();
});
$("#changeLogin").submit();
If I leave "success" blank, it works -> form is submitted by ajax, login changed, but I do not see the result message. Otherwise whole page get reloaded.
Also, when I hit F5 form is being submited once again (even in Ajax).
I cant add comments because i do not have enough reputation but...
You should delete the last line with $("#changeLogin").submit();
And then in your php script file you should echo the result so you can get this result in ajax request. After that in your success method you have to read the result and (for example) append it somewhere to show the success or error box
I think you can use a normal button instead of submit button,just onclick can be an ajax request, the form should not be submitted,good luck.

sending form data to php using ajax

I Have an requirement to pass form data to php using ajax and implement it in php to calculate the sum , division and other arithmetic methods I am a new to ajax calls trying to learn but getting many doubts....
It would be great help if some one helps me out with this
index.html
<script type="text/javascript">
$(document).ready(function(){
$("#submit_btn").click(function() {
$.ajax({
url: 'count.php',
data: data,
type: 'POST',
processData: false,
contentType: false,
success: function (data) {
alert('data');
}
})
});
</script>
</head>
<form name="contact" id="form" method="post" action="">
<label for="FNO">Enter First no:</label>
<input type="text" name="FNO" id="FNO" value="" />
label for="SNO">SNO:</label>
<input type="text" name="SNO" id="SNO" value="" />
<input type="submit" name="submit" class="button" id="submit_btn" value="Send" />
</form>
In count.php i want to implement
<?php
$FNO = ($_POST['FNO']);
$SNO=($_post['SNO']);
$output=$FNO+$SNO;
echo $output;
?>
(i want to display output in count.php page not in the first page index.html)
Thanks for your help in advance.
You can use a simple .post with AJAX. Take a look at the following code to be able to acheive this:
$('#form').submit(function() {
alert($(this).serialize()); // check to show that all form data is being submitted
$.post("count.php",$(this).serialize(),function(data){
alert(data); //check to show that the calculation was successful
});
return false; // return false to stop the page submitting. You could have the form action set to the same PHP page so if people dont have JS on they can still use the form
});
This sends all of your form variables to count.php in a serialized array. This code works if you want to display your results on the index.html.
I saw at the very bottom of your question that you want to show the count on count.php. Well you probably know that you can simply put count.php into your form action page and this wouldn't require AJAX. If you really want to use jQuery to submit your form you can do the following but you'll need to specify a value in the action field of your form:
$("#submit_btn").click(function() {
$("#form").submit();
});
I have modified your PHP code as you made some mistakes there. For the javscript code, i have written completely new code for you.
Index.html
<html>
<head>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js"></script>
</head>
<body>
<form name="contact" id="contactForm" method="post" action="count.php">
<label for="FNO">Enter First no:</label>
<input type="text" name="FNO" id="FNO" value="" />
<label for="SNO">SNO:</label>
<input type="text" name="SNO" id="SNO" value="" />
<input type="submit" name="submit" class="button" id="submit_btn" value="Send" />
</form>
<!-- The following div will use to display data from server -->
<div id="result"></div>
</body>
<script>
/* attach a submit handler to the form */
$("#contactForm").submit(function(event) {
/* stop form from submitting normally */
event.preventDefault();
/* get some values from elements on the page: */
var $form = $( this ),
//Get the first value
value1 = $form.find( 'input[name="SNO"]' ).val(),
//get second value
value2 = $form.find( 'input[name="FNO"]' ).val(),
//get the url. action="count.php"
url = $form.attr( 'action' );
/* Send the data using post */
var posting = $.post( url, { SNO: value1, FNO: value2 } );
/* Put the results in a div */
posting.done(function( data ) {
$( "#result" ).empty().append( data );
});
});
</script>
</html>
count.php
<?php
$FNO = $_POST['FNO'];
$SNO= $_POST['SNO'];
$output = $FNO + $SNO;
echo $output;
?>
There are a few things wrong with your code; from details to actual errors.
If we take a look at the Javascript then it just does not work. You use the variable data without ever setting it. You need to open the browser's Javascript console to see errors. Google it.
Also, the javascript is more complicated than is necessary. Ajax requests are kind-of special, whereas in this example you just need to set two POST variables. The jQuery.post() method will do that for you with less code:
<script type="text/javascript">
$(document).ready(function(){
$("#form").on("submit", function () {
$.post("/count.php", $(this).serialize(), function (data) {
alert(data);
}, "text");
return false;
});
});
</script>
As for the HTML, it is okay, but I would suggest that naming (i.e. name="") the input fields using actual and simple words, as opposed to abbreviations, will serve you better in the long run.
<form method="post" action="/count.php" id="form">
<label for="number1">Enter First no:</label>
<input type="number" name="number1" id="number1">
<label for="number2">Enter Second no:</label>
<input type="number" name="number2" id="number2">
<input type="submit" value="Calculate">
</form>
The PHP, as with the Javascript, just does not work. PHP, like most programming languages, are very picky about variables names. In other words, $_POST and $_post are not the same variable! In PHP you need to use $_POST to access POST variables.
Also, you should never trust data that you have no control over, which basically means anything that comes from the outside. Your PHP code, while it probably would not do much harm (aside from showing where the file is located on the file system, if errors are enabled), should sanitize and validate the POST variables. This can be done using the filter_input function.
<?php
$number1 = filter_input(INPUT_POST, 'number1', FILTER_SANITIZE_NUMBER_INT);
$number2 = filter_input(INPUT_POST, 'number2', FILTER_SANITIZE_NUMBER_INT);
if ( ! ctype_digit($number1) || ! ctype_digit($number2)) {
echo 'Error';
} else {
echo ($number1 + $number2);
}
Overall, I would say that you need to be more careful about how you write your code. Small errors, such as in your code, can cause everything to collapse. Figure out how to detect errors (in jQuery you need to use a console, in PHP you need to turn on error messages, and in HTML you need to use a validator).
You can do like below to pass form data in ajax call.
var formData = $('#client-form').serialize();
$.ajax({
url: 'www.xyz.com/index.php?' + formData,
type: 'POST',
data:{
},
success: function(data){},
error: function(data){},
})

Making Ajax Form Use Native Wordpress

I have a form that uses Ajax to send a serialized forms data to send to a php file called contact-submit
i know i should wrap the contents of that page in a function...and add it to the functions.php file
i plan on calling the function MyContactForm
but i dont know the proper syntax to serialize the form and post the data to the function
heres what i have so far ...*keep in mind i left out the form fields...because we are focusing on the script part of this ...if the form is serialized...should grab everything
html
<form id="contactform" action="<?php echo home_url('contact-submit'); ?>" method="post">
<input class="textbox required" type="text" name="name2" id="name" value="Your Name" />
<input class="submit" value="Send" type="submit" alt="Send message" name="submit" />
</form>
script
jQuery("#postform").validate();
var AjaxSubmit = function(){
var btnText=jQuery('#contactform .submit').val();
// inform client that data is been sent:
jQuery('#contactform .submit').val('Sending...');
jQuery('#contactform .submit').attr('disabled', true);
jQuery.ajax({
type: 'POST',
url: jQuery('#contactform').attr('action'),
data: jQuery('#contactform').serialize(),
// successful POST - display result in success div:
success: function(msg){
jQuery('#contactform .form, #contactform .contacthead').slideUp(500,function(){
jQuery('#contactform div.success').removeClass('hiddne').fadeIn(500);
});
},
error: function(response) {
jQuery('#contactform .submit').val(btnText);
jQuery('#contactform div.error').html(response.statusText).slideDown(500);
}
});
}
jQuery("#contactform").validate({
submitHandler: AjaxSubmit
});
});
My Question is.....what is the proper syntax for serializing a form and passing the data to a php function?

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