Post and display data on the same page using PHP (MVC) AJAX - php

I'm trying to post input data and display it on the same page (view_group.php) using AJAX but I did not understand how it works with MVC, I'm new with MVC if anyone could help me it would be very helpful for me.
view_group.php
<script type = "text/javascript" >
$(document).ready(function() {
$("#submit").click(function(event) {
event.preventDefault();
var status_content = $('#status_content').val();
$.ajax({
type: "POST",
url: "view_group.php",
data: {
postStatus: postStatus,
status_content: status_content
},
success: function(result) {}
});
});
}); </script>
if(isset($_POST['postStatus'])){ $status->postStatus($group_id); }
?>
<form class="forms-sample" method="post" id="form-status">
<div class="form-group">
<textarea class="form-control" name="status_content" id="status_content" rows="5" placeholder="Share something"></textarea>
</div>
<input type="submit" class="btn btn-primary" id="submit" name="submit" value="Post" />
</form>
<span id="result"></span>
my controller
function postStatus($group_id){
$status = new ManageGroupsModel();
$status->group_id = $group_id;
$status->status_content = $_POST['status_content'];
if($status->postStatus() > 0) {
$message = "Status posted!";
}
}

first in the ajax url you must set your controller url , then on success result value will be set on your html attribute .
$.ajax({
type: "POST",
url: "your controller url here",
data: {
postStatus: postStatus,
status_content: status_content
},
success: function(result) {
$('#result).text(result);
}
});
Then on your controller you must echo the result you want to send to your page
function postStatus($group_id){
$status = new ManageGroupsModel();
$status->group_id = $group_id;
$status->status_content = $_POST['status_content'];
if($status->postStatus() > 0) {
$message = "Status posted!";
}
echo $status;
}

Related

$_POST data is not passing to my function in Wordpress via AJAX

I have created an AJAX function in Wordpress. The function is called on form submission. The function is run, but it is not receiving any of the form data that I have submitted. What am I missing?
PHP Function
I have added the PHP function here, which is called successfully via AJAX. This form creates a new user successfully, but only when I create the variables manually (eg. see $new_user_data['user_login'] = 'This Text Works';). For some reason, the $_POST data isn't coming through to the function.
add_action("wp_ajax_register_user", __NAMESPACE__ . "\\register_user");
add_action("wp_ajax_nopriv_register_user", __NAMESPACE__ . "\\register_user");
function register_user() {
// NONCE VERIFICATION
if ( !wp_verify_nonce( $_REQUEST['nonce'], "rtr_register_nonce")) {
exit("Oops! This is embarassing!");
}
// Get all post data for the user.
$new_user_data = array();
$new_user_data['first_name'] = sanitize_text_field($_POST['first-name']);
$new_user_data['last_name'] = sanitize_text_field($_POST['last-name']);
$new_user_data['user_email'] = $_POST['email'];
$new_user_data['user_pass'] = sanitize_text_field($_POST['password']);
$new_user_data['user_login'] = 'This Text Works';
$new_user_data['role'] = 'subscriber';
// Create the User
$registered_user = wp_insert_user( $new_user_data );
$result['user'] = $registered_user;
// AJAX CHECK
if(!empty($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest') {
$result = json_encode($result);
echo $result;
} else {
header("Location: ".$_SERVER["HTTP_REFERER"]);
}
die();
}
JQuery
function registerUser(){
var nonce = $('#regForm').attr("data-nonce");
var formData = $('#regForm').serialize();
$.ajax({
url: rtr_register_user.ajaxUrl,
type: 'post',
dataType: 'json',
data : {action: 'register_user', nonce: nonce, formData: formData},
success: function (response) {
console.log(response);
$('#regForm').html('Your form has been submitted successfully');
},
});
}
function nextPrev(n) {
// This function will figure out which tab to display
var x = document.getElementsByClassName("form-tab");
// Exit the function if any field in the current tab is invalid:
if (n === 1 && !validateForm()) {
return false;
}
// Hide the current tab:
x[currentTab].style.display = "none";
// Increase or decrease the current tab by 1:
currentTab = currentTab + n;
// if you have reached the end of the form... :
if (currentTab >= x.length) {
//...the form gets submitted:
//document.getElementById("regForm").submit();
registerUser();
return false;
}
// Otherwise, display the correct tab:
showTab(currentTab);
}
$('#nextBtn').click(function () {
nextPrev(1);
});
$('#prevBtn').click(function () {
nextPrev(-1);
});
Form
<?php
$nonce = wp_create_nonce("rtr_register_nonce");
$link = admin_url('admin-ajax.php?action=register_user&nonce='.$nonce);
?>
<form id="regForm" <?php echo 'data-nonce="' . $nonce . '"'; ?> action="<?php echo $link; ?>" method="post" enctype="multipart/form-data">>
<div class="my-3 text-center">
<span class="form-step">1</span>
<span class="form-step">2</span>
</div>
<div class="form-tab">
<p><input name="first-name" placeholder="First Name" oninput="this.className = ''"></p>
<p><input name="last-name" placeholder="Last Name" oninput="this.className = ''"></p>
<p><input name="dob" type="date" oninput="this.className = ''"></p>
</div>
<div class="form-tab">
<p><input name="email" type="email" placeholder="Email" oninput="this.className = ''"></p>
<p><input name="password" type="password" placeholder="Password" oninput="this.className = ''"></p>
</div>
<div style="overflow:auto;">
<div style="float:right;">
<button type="button" class="btn btn-brand" id="prevBtn">Previous</button>
<button type="button" class="btn btn-brand" id="nextBtn">Next</button>
</div>
</div>
</form>
Seems you are not triggering registerUser() check following script works fine for me
jQuery(document).ready(function($) {
jQuery('body').on('click', '#nextBtn', function() {
registerUser();
});
});
function registerUser(){
var nonce = jQuery('#regForm').attr("data-nonce");
var formData = jQuery('#regForm').serialize();
jQuery.ajax({
url: ajaxurl,
type: 'post',
dataType: 'json',
data : {action: 'register_user', nonce: nonce, formData: formData},
success: function (response) {
console.log(response);
$('#regForm').html('Your form has been submitted successfully');
},
});
}
add method="post" to your 'form' - 'get' is the default https://stackoverflow.com%2Fquestions%2F2314401%2Fwhat-is-the-default-form-http-method&usg=AOvVaw1dKc3hW4K6r5SwQurLztBw
The "user_login" is a username of the user so probably it doesn't accepts space too.
See also WP Insert Post
Please try passing some username such as "custom_user" and see the result.
Hope this might work.
Ok it was a bit of help from everyone here. But yes, I was calling the AJAX correctly, but not actually submitting the form. I added a .on(submit) to the form and then added a listener to the form to perform the AJAX call on submit. Here's the amendments below.
function nextPrev(n) {
var x = document.getElementsByClassName("form-tab");
if (n === 1 && !validateForm()) {
return false;
}
x[currentTab].style.display = "none";
currentTab = currentTab + n;
if (currentTab >= x.length) {
// ADDED THIS SUBMIT() HERE
document.getElementById("regForm").submit();
return false;
}
// Otherwise, display the correct tab:
showTab(currentTab);
}
// ADDED AN EVENT LISTENER TO TRIGGER THE AJAX CALL HERE
$('#regForm').on('submit', function () {
var nonce = $('#regForm').attr("data-nonce");
var formData = $('#regForm').serialize();
$.ajax({
url: rtr_register_user.ajaxUrl,
type: 'post',
dataType: 'json',
data: {
action: 'register_user',
nonce: nonce,
formData: formData
},
success: function (response) {
console.log(response);
$('#regForm').html('Your form has been submitted successfully');
},
});
});

How to post data (form) without page refresh PHP, jQuery, Ajax

Please help me how to submit form (comments) without page refresh for
HTML markup:
<form id="commentf" method="post">
<img width="40px" height="40px" src="uploads/<?php echo $row['photo']; ?>">
<textarea class="textinput"id="comment" rows="1" name="comments" placeholder="Comment Here......"></textarea>
<button type="submit" id="comq"name="compost" class="butn2">post comment</button>
</form>
PHP code (pnf.php):
if(isset($_POST["compost"]))
{
$comment=$_POST['comments'];
{
$reslt_user= mysqli_query($connection,"SELECT * FROM tbl_users,`queries` where id='".$_SESSION['id']."' AND qid= '".$qid."' ");
$row_lat_lng= mysqli_fetch_array($reslt_user);
$stmt = mysqli_query($connection,"INSERT INTO comments set uid='".$_SESSION['id']."',comments='".$comment."',reply='".$reply."' ,
qid= '".$qid."' ");
}
if($stmt)
{
echo "hello world";
}
}
jQuery and Ajax:
$(document).ready(function()
{
$("#comq").click(function() {
var comment=$("#comment").val();
$.ajax({
type: "POST",
url:"pnf.php",
data: {
"done":1,
"comments":comment
},
success: function(data){
}
})
});
});
I have tried many times and don't know what mistake I made, Ajax and jQuery are not working, please anyone help - thanks in advance
You have made couple of mistakes.
First:: You should put button type="button" in your HTML form code
Second:: You have made a syntax error. $("#comment").vol(); should be replaced with $("#comment").val(); in your jQuery AJAX
As you mentioned that you have to send request without refreshing page I modified your JS-code with preventing default submitting form:
$(document).ready(function () {
$("#commentf").submit(function (e) {
e.preventDefault();
var comment = $("#comment").val();
$.ajax({
type: "POST",
url: "pnf.php",
data: {
"done": 1,
"comments": comment
},
success: function (data) {
}
})
});
});
Javascript
$('form').on('submit', function(event){
event.preventDefault();
event.stopPropagination();
var dataSet = {
comment: $('#comment').val();
}
$.ajax({
url: "link.to.your.api/action=compost",
data: dataSet,
method: 'post',
success: function(data){
console.log('request in success');
console.log(data);
},
error: function(jqXHR) {
console.error('request in error');
console.log(jqXHR.responseText');
}
});
});
PHP
$action = filter_input(INPUT_GET, 'action');
swicht($action){
case 'compost':
$comment = filter_input(INPUT_POST, 'comment');
{
$reslt_user= mysqli_query($connection,"SELECT * FROM tbl_users,`queries` where id='".$_SESSION['id']."' AND qid= '".$qid."' ");
$row_lat_lng= mysqli_fetch_array($reslt_user);
$stmt = mysqli_query($connection,"INSERT INTO comments set uid='".$_SESSION['id']."',comments='".$comment."',reply='".$reply."' ,
qid= '".$qid."' ");
}
if(!$stmt)
{
http_response_code(400);
echo 'internal error';
}
echo 'your data will be saved';
break;
default:
http_response_code(404);
echo 'unknown action';
}
you have to prevent the submit on the form (look in javascript).
after that you send the request to the server and wait for success or error.
in php try to do it with a switch case. and try to not touch super globals directly, use filter_input function.
hope this helps
Modified JQuery Code...
$( document ).ready(function() {
console.log( "ready!" );
$("#comq").click(function() {
var comment=$("#comment").val();
console.log('comment : '+comment);
$.ajax({
type: "POST",
url:"pnf.php",
data: {
"done":1,
"comments":comment
},
success: function(data){
}
})
});
});
HTML Code
<form id="commentf" method="post">
<textarea class="textinput" id="comment" rows="1" name="comments" placeholder="Comment Here......"></textarea>
<input type="button" id="comq" name="compost" class="butn2" value="Post Comment">
</form> </div>
<form id="commentf" method="post">
<img width="40px" height="40px" src="uploads/<?php echo $row['photo']; ?>">
<textarea class="textinput"id="comment" rows="1" name="comments" placeholder="Comment Here......"></textarea>
<button type="button" id="comq"name="compost" class="butn2">post comment</button>
</form>
script
$(document).ready(function()
{
$("#comq").click(function() {
var comment=$("#comment").val();
$.ajax({
type: "POST",
url:"pnf.php",
data: {
"done":1,
"comments":comment
},
success: function(data){
}
})
});
});
PHP code (pnf.php):
comment=$_POST['comments'];
$reslt_user= mysqli_query($connection,"SELECT * FROM tbl_users,`queries` where id='".$_SESSION['id']."' AND qid= '".$qid."' ");
$row_lat_lng= mysqli_fetch_array($reslt_user);
$stmt = mysqli_query($connection,"INSERT INTO comments set uid='".$_SESSION['id']."',comments='".$comment."',reply='".$reply."' ,
qid= '".$qid."' ");
if($stmt)
{
echo "hello world";
}
if you are using jquery make sure to include jquery libraries before your script file.
latest jquery cdn minified version
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.0/jquery.min.js"></script>
example
<script src="https://code.jquery.com/jquery-3.2.1.min.js" type="text/javascript"></script>
<script src="yourjsfile.js" type="text/javascript"></script>

php can't get multiple ajax submit data

I have a form and it have 2 submit button.
<form name="posting" id="posting" method="post" action="posting_bk.php" role="form">
<input type="text" name="title" id="title" class="form-control" required="required">
....some form fields...
<input class="btn btn-home" type="submit" name="publish" id="publish" alt="Publish" value="Preview and Post" />
<input class="btn btn-home" type="submit" name="save" id="save" onclick="return confirm('Are you sure you want to Submit.')" alt="Save" value="Save as Draft" /></center>
</form>
i am using ajax to send/receive data.
$('#posting input[type="submit"]').on("click", function(e) {
e.preventDefault;
var btn = $('#publish');
var el = $(this).attr('id');
$.ajax({
type: 'post',
url: $('form#posting').attr('action'),
cache: false,
dataType: 'json',
data: {
data: $('form#posting').serialize(),
action: el
},
beforeSend: function() {
$("#validation-errors").hide().empty();
},
success: function(data) {
if (data.success == false) {
var arr = data.errors;
$.each(arr, function(index, value) {
if (value.length != 0) {
$("#validation-errors").append('<div class="alert alert-danger"><strong>' + value + '</strong><div>');
}
});
$("#validation-errors").show();
btn.button('reset');
} else {
$("#success").html('<div class="alert alert-success">Basic details saved successfully. <br> If you want to edit then please goto Edit. <div>');
$('#title').val('');
}
},
error: function(xhr, textStatus, thrownError) {
alert('Something went to wrong.Please Try again later...');
btn.button('reset');
}
});
return false;
});
this is my php file. posting_bk.php
if ($_POST['action'] == 'publish') {
if($title == 'test'){
array_push($res['errors'], 'data received by php.');
}else{
array_push($res['errors'], 'No data received by php.');
}
$res['success'] = true;
echo json_encode($res);
}
elseif ($_POST['action'] == 'save') {
array_push($res['errors'], 'Save button clicked.');
$res['success'] = true;
echo json_encode($res);
}
All the time if i click on publish button i am getting
No data recived by php
When I check in firebug it is showing data under post.
Like this
action publish
data title=test&
I am not sure what am i doing wrong here. Please advise.
Change the AJAX call to use:
data: $('form#posting').serialize() + '&action=' + el,\
Then access the parameter using
$title = $_POST['title'];
The way you're doing it, the form data is being nested a level down in the POST data, so you would have had to do:
$data = parse_str($_POST['data']);
$title = $data['title'];

Input validation through AJAX

I have the following AJAX in my index.php:
$(document).ready(function() {
$('.buttono').click(load);
});
function load() {
$.ajax({
url: 'http://localhost/Generator/js/ajaxRequest.php'
}).done(function(data) {
$('#content').append(data);
});
}
HTML (part of index.php):
<form method="POST" action="">
<input type="text" name="input">
<input type="submit" name="submit" class="buttono" value="Convert">
</form>
<div id='content'></div>
And in my ajaxRequest.php I have the following PHP snippet:
if ($_POST['input'] == 'dog') {
echo 'Status 1';
} else if ($_POST['input'] == 'cat') {
echo 'Status 2';
}
How can I perform the PHP check through AJAX? So that if I click the submit button and have typed 'dog', to return the string Status 1?
Well what I see in your code is that:
first you have not specified your request method,
second you have not set $_POST['dog']
I would have gone with this ajax:
$.ajax({
type : "POST",
url : 'to/url',
data : { input : $("input[name='input']").val() },
success : function(data){
// do whatever you like
}
});
What you have to do is make the user fill out the form and then instead of clicking a type="submit" button just make them click a regular button. Then when that person clicks the regular button submit. You can do this by:
<!-- HTML -->
<form method="POST">
<input type="text" id="type"/>
<button id="submit">Sumbit</button>
</form>
<!-- JS -->
$(document).ready(function(){
$('#submit').click(onSubmitClicked);
});
function onSubmitClicked(){
var data = {
"input": $('#type').val()
};
$.ajax({
type: "POST",
url: "url/To/Your/Form/Action",
data: data,
success: success
});
function success(data){
if(data == 'status 1'){
//Do something
}
}
}
Try this:
in you php file:
$res = array();
if ($_POST['input'] == 'dog') {
$res['status'] = '1';
} elseif ($_POST['input'] == 'cat') {
$res['status'] = '2';
}
echo json_encode($res);
Then in your jquery:
function load(){
$.ajax({
type : "POST",
data : { input : $("input[name='input']").val() },
url:'http://localhost/Generator/js/ajaxRequest.php'
}).done(function(data){
$('#content').append(data.status);
});
}

jquery/php form in modal window

I have a form in a modal window. When I submit the form through ajax I don't get the success message. My aim is to see the message created in the php file in the modal after submitting the form. Here is the code:
<p><a class='activate_modal' name='modal_window' href='#'>Sign Up</a></p>
<div id='mask' class='close_modal'></div>
<div id='modal_window' class='modal_window'>
<form name="field" method="post" id="form">
<label for="username">Username:</label><br>
<input name="username" id="username" type="text"/><span id="gif"><span>
<span id="user_error"></span><br><br>
<label for="email">Email:</label><br>
<input name="email" id="email" type="text"/><span id="gif3"></span>
<span id="email_error"></span><br><br>
<input name="submit" type="submit" value="Register" id="submit"/>
</form>
</div>
The modal.js
$('.activate_modal').click(function(){
var modal_id = $(this).attr('name');
show_modal(modal_id);
});
$('.close_modal').click(function(){
close_modal();
});
$(document).keydown(function(e){
if (e.keyCode == 27){
close_modal();
}
});
function close_modal(){
$('#mask').fadeOut(500);
$('.modal_window').fadeOut(500);
}
function show_modal(modal_id){
$('#mask').css({ 'display' : 'block', opacity : 0});
$('#mask').fadeTo(500,0.7);
$('#'+modal_id).fadeIn(500);
}
The test.js for the registration of the user
$(function() {
$('#form').submit(function() {
$.ajax({
type: "POST",
url: "test.php",
data: $("#form").serialize(),
success: function(data) {
$('#form').replaceWith(data);
}
});
});
});
And the PHP FILE
<?php
$mysqli = new mysqli('127.0.0.1', 'root', '', 'project');
$username = $_POST['username'];
$email = $_POST['email'];
$mysqli->query("INSERT INTO `project`.`registration` (`username`,`email`) VALUES ('$username','$email')");
$result = $mysqli->affected_rows;
if($result > 0) {
echo 'Welcome';
} else {
echo 'ERROR!';
}
?>
Try putting the returncode from your AJAX call into
$('#modal_window')
instead of in the form
$('#form')
BTW: Why not use the POST or GET method of jQuery? They're incredibly easy to use...
Try something like this.
First write ajax code using jquery.
<script type="text/javascript">
function submitForm()
{
var str = jQuery( "form" ).serialize();
jQuery.ajax({
type: "POST",
url: '<?php echo BaseUrl()."myurl/"; ?>',
data: str,
format: "json",
success: function(data) {
var obj = JSON.parse(data);
if( obj[0] === 'error')
{
jQuery("#error").html(obj[1]);
}else{
jQuery("#success").html(obj[1]);
setTimeout(function () {
jQuery.fancybox.close();
}, 2500);
}
}
});
}
</script>
while in php write code for error and success messages like this :
if(//condition true){
echo json_encode(array("success"," successfully Done.."));
}else{
echo json_encode(array("error","Some error.."));
}
Hopes this help you.

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