ajax form submit -> receive respone from php - 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="Zmień" 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.

Related

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 not receiving POST variables

I've read at least 20 of these similar questions and can't figure out what's wrong with my code.
I've got a login form as shown below:
<div class="login-form">
<form class="login-form-container" action="./" method="post" autocomplete="off">
<input id="login-user" type="text" placeholder=" Username"/>
<input id="login-pass" type="password" placeholder=" Password"/>
<div class="login-button-container">
<input id="login-button" type="submit" value="Log in"/>
</div>
</form>
</div>
My login.js script is as below:
$(document).ready(function(){
$("#login-button").click(function(e){
var username = $("#login-user").val();
var password = $("#login-pass").val();
$.ajax({url: "../includes/index.php", //.js file is in a diff directory
data: {'login':true,'name':username,'pwd':password},
type: "POST",
success: function(html) {
if (html=='true') {
//window.location="dashboard.php";
alert('success');
}
else {
alert('failure');
}
}
});
return false;
});
});
This is supposed to get sent back to 'index.php', which has included a separate php file that handles the login information. The reason for this is so I can update the page with the user's successful login without refreshing.
My other php file is simply checking for the post data:
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
var_dump($_POST);
if (isset($_POST['login']) && $_POST['login']) {
} else if (isset($_POST['reg']) && $_POST['reg']) { //this is for registration, which will be used once this login is working
} else {
echo 'Neither login/reg were set';
}
The var dump shows:
array(0) { }
So the data is not being received. However, it clearly reaches the php page. I don't know what's wrong, any suggestions?
Edit: I figured I would add, that adding an alert at the top of login.js with the username/password does print the entered values. So the login.js is being reached, and the data is accessible there. But it doesn't go beyond that.
Edit2: Here is my updated HTML and login.js code:
index.php:
<form class="login-form-container" action="./" method="post" autocomplete="off">
<input id="login-user" type="text" placeholder=" Username"/>
<input id="login-pass" type="password" placeholder=" Password"/>
<div class="login-button-container">
<input id="login-button" type="button" value="Log in"/>
</div>
</form>
login.js
$(document).ready(function(){
alert(1);
$("#login-button").click(function(e){
alert(0);
//e.preventDefault();
var username = $("#login-user").val();
var password = $("#login-pass").val();
$.ajax({url: "../includes/index.php",
data: {'login':true,'name':username,'pwd':password},
type: "POST",
success: function(html) {
if (html=='true') {
//window.location="dashboard.php";
alert('success');
}
else {
alert('failure');
}
}
});
return false;
});
});
The solution was to change my
$("#login-button").click(function(e){
to
$(document).on("submit",".login-form-container",function(e) {
which will handle the form submit, and this includes pressing the button as well as hitting enter (ideal situation).
Now the data is being sent successfully and I can handle the actual login/registration SQL, so thank you to #Ashokkumar M. Prajapati for helping to come to the conclusion that the .js script was not actually firing upon submit (not that the rest of you didn't help!)
please replace type: "POST" with method: "POST"
you mentioned it wrong in jquery ajax function.
Default is "GET" in case you don't specify it. So, if you had checked both $_POST & $_GET then you would have found your data in $_GET.
try var_dump($_REQUEST) if it don't show up anything then your client is not sending data to server.
also, check net panel of you browser developer console. so, you will know what data are passed to server.

Jquery Post ajax variables wont enter PHP

I have the following code:
$(document).on('submit', function (event) {
event.preventDefault();
console.log('submitting');
hostingURL = '/<?echo $this->CI->uri->uri_string();?>';
$('form').attr('action', hostingURL);
var form = $(this);
$.ajax({
type: "POST",
url: hostingURL,
data: form.serialize()
}).done(function (data) {
$(document).unbind("submit");
$("#maincenter").html(data);
console.log('posted');
}).fail(function (data) {
alert('failed');
});
});
this is getting the requested url, and change the form to the action needed, then posting it to the server, and returning data.
server sided i have:
echo print_r($_POST);
echo "<br> type: ".$_SERVER['REQUEST_METHOD']." <br>";
It seems like whatever i do in my form wont acually submit it. The data back is correct, but from the server side code response i do get: Array ( ) 1 type: GET no matter what im doing.
So my question is: How can i acually make the form submit the post data?
The jquery function will be used on all my forms, so i need to get every field in the form to be submitted to.
Below is one example form.
example form:
<form method="POST">
<textarea class="textarea col-12" style="height:150px;" id="messageholder" name="profiltekst" placeholder="Profiltekst">
<? echo $userprofile->profiletext ?>
</textarea>
<div class="bottomform">
<input type="submit" class="makecenter" name="updateprofile" value="Endre profiltekst">
</div>
</form>

Form submit using AJAX is not working

PHP/Ajax newbie here...I am trying to save the contents of a textarea into MySQL via Ajax. Although the data IS being saved correctly, Ajax isn't quite working. Basically, the page is "reloaded/refreshed" after the data is saved, unlike Ajax. Can you please tell me what is that I am doing wrong?
index.html:
<form action="save.php" method="post" id="source-form">
<span><input type="submit" value="Save" /></span>
<div>
<textarea id="editor" name="editor" >
</textarea>
</div>
</form>
javascript:
$(document).ready(function() {
$("#source-form").submit(function(){
$.ajax({
url:"save.php",
type:"post",
data:$(this).serialize(),
success: alert('saved');
});
});
save.php
<?php
// connect to the database
include('connect-db.php');
// get form data, making sure it is valid
$submit_date = date('Y-m-d H:i:s');
$content = mysql_real_escape_string(htmlspecialchars($_POST['editor']));
//build query
mysql_query("INSERT source SET submit_date='$submit_date',content='$content'")
or die(mysql_error());
header('Location: index.html');
?>
Any help on this is appreciated. Thank you.
EDIT:
For folks running into the same issue or something similar...here's a great solution from:
http://jquery.malsup.com/form/#getting-started
change your <form action="save.php" method="post" id="source-form">
to <form method="post" id="source-form">
remove the action you already declared your url in the ajax
remove header('Location: index.html'); since your should not redirect since your are using ajax. remember that if you are using ajax you dont need to refresh the page just let it receive a confirmation that the result was successful
Add a return false; to forum submit so that the form submit is cancelled:
$("#source-form").submit(function(){
$.ajax({
url:"save.php",
type:"post",
data:$(this).serialize(),
success: alert('saved');
});
return false;
});
You have to stop the default behavior of the submit button (form posting).Otherwise the form will be submitted again and you will see the page loading again. You use the preventDefault function
$("#source-form").submit(function(e){
e.preventDefault();
$.ajax({
url:"save.php",
type:"post",
data:$(this).serialize(),
success: alert('saved');
});
});
If you're only using the form for a ajax submit, I would suggest removing the form from the HTML completely and using a click event. I've stripped things down to a minimum in the HTML:
HTML
<input id="save-text" type="submit" value="Save" />
<textarea id="editor" name="editor" ></textarea>
JavaScript
$(document).ready(function() {
$("#save-text").click(function(){
$.ajax({
url:"save.php",
type:"post",
data: $("#editor").serialize(),
success: alert('saved');
});
});

Problem using jQuery-AJAX to submit form to PHP and display new content in div without refreshing

I am trying to use jQuery-AJAX to submit the data in my form to my controller (index.php) where it is processed by PHP and inserted via PDO into the database if valid. Once the code is inserted into the database, the div where the form previously existed should be replaced by the contents of another page (newpage.php). The original page should not be refreshed upon submitting of the form, only the div where the form previously existed should be refreshed. There is a particular problem with my code, although I can't seem to find where the issue is at:
Here is my jQuery:
<script type="text/javascript">
function processForm() {
var action= $('#action').val();
var data = $('#data').val();
var dataString = 'action=' + action + '&data=' + data;
$.ajax ({
type: "POST",
url: ".",
data: dataString,
success: function(){
$('#content_main').load('newpage.php');
}
});
}
</script>
Here is my HTML: (As a side note, I noticed that when I take the "return false;" out of the HTML, that the form will submit to my database, but the whole page also reloads - and is blank. When I leave the "return false;" in the HTML, the newpage.php loads correctly into the div, but the data does not make it into the database)
<form action="" method="post" onsubmit="processForm();return false;">
<input type="hidden" name="action" value="action1" />
<input type="text" name="data" id="data" />
<input type="submit" value="CONTINUE TO STEP 2" name="submit" />
</form>
Here is my PHP:
<?php
$action = $_POST['action'];
switch ($action) {
case 'action1':
$data = $_POST['data'];
pdo ($data);
exit;
}
?>
I feel like I am making a silly mistake somewhere, but I just can't put my finger on it. Thanks for any assistance you can provide!
SOLUTION (via Jen):
jQuery:
<script type="text/javascript">
function processForm() {
var dataString = $("#yourForm").serialize();
$.ajax ({
type: 'POST',
url: ".",
data: dataString,
success: function(){
$('#content_main').load('newpage.php');
}
});
return false;
});
</script>
HTML:
<form id="yourForm">
<input type="hidden" name="action" value="action1" />
<input type="text" id="data" name="data" />
<input type="submit" value="CONTINUE TO STEP 2" name="submit" />
</form>
PHP:
<?php
$action = $_POST['action'];
switch ($action) {
case 'action1':
$data = $_POST['data'];
pdo ($data);
exit;
}
?>
What I learned: Use the .serialize() jQuery method if it is an option, as it will save a bunch of time writing out the var for each form value and .serialize() does not typically make mistakes sending the info to php.
Give this a try:
$("#yourForm").submit(function(){
// Could use just this line and not vars below
//dataString = $("#yourForm").serialize();
// vars are being set by selecting inputs by id but id not set in form fields
var action= $('#action').val(); // value of id='action'
var data = $('#data').val(); // value of id='data'
var dataString = 'action=' + action + '&data=' + data;
// dataString = 'action=&data=' so nothing is posted to db
// because input fields cannot be found by id
// fix by adding id fields to form fields
$.ajax ({
type: "POST",
url: ".",
data: dataString,
success: function(){
$('#content_main').load('newpage.php');
}
});
return false;
});
And change the form to (I added the id attributes):
<form id="yourForm">
<input type="hidden" id="action" name="action" value="action1" />
<input type="text" id="data" name="data" id="data" />
<input type="submit" value="CONTINUE TO STEP 2" name="submit" />
</form>
Seems like you need an explanation rather than a fix of code. Here is a brief explanation for the 2 cases:
When you take out return false; the code will treat your form as a normal HTML form that will be submitted to the server via action="", which leads to nowhere. The Javascript, however, also does its job in this case but because the page is redirected to nowhere, then it turns blank at the end.
When you put return false; back to the form, the form will catch the event handler and know that this form will be returned FALSE to submit. That's why you can see how your Javascript code does the job. However, one thing you should notice is that your jQuery AJAX function needs to POST (or GET) to a processing file, not '.'
Considering this reply based on no knowledge of your real situation. You need to look back over your code and see how you can edit it. Would be happy to reply if you have any questions.
Hope this small hint helps (:

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