Develop Reference

How to check if submitting form with jQuery if successful?

I have following codes as form, and I wanted to check if the form is been successfully submitted.
I was wondering how I can check on the console. I want to check if the form is been successfully submitted so I can display another form.
<form id="signup" data-magellan-target="signup" action="http://app-service-staging.com" class="epic_app-signup" method="POST">
<div class="grid__column " style="width: 100%;">
<input type="text" name="first_name" placeholder="Name" required/>
</div>
<div class="grid__column " style="width: 100%;">
<input type="text" name="password" placeholder="Password" required/>
</div>
<div class="grid__column " style="width: 100%;">
<input type="text" name="confimred-password" placeholder="Confirmed password" />
</div>
<div class="grid__column " style="width: 100%;">
<input type="date" name="startdate" id="startdate" min="2019-12-16">
</div>
</div>
<button type="submit" class="grid__column" style="width: 50%;"></button>
</div>
</div>
</div>
</form>
and the script,
$('.epic_app-signup').on('submit', function(e) {
e.preventDefault();
var formData = $('.epic_app-signup').serializeArray();
var jsonData = {};
formData.forEach(function(item, index) {
jsonData[item.name] = item.value;
});
console.log('data\n', jsonData);
$.ajax({
url: 'http://app-service-staging.com/api/auth/register',
type:'POST',
data: jsonData,
contentType: 'application/json'
}).done(function(data, textStatus, jqXHR) {
if (textStatus === 'success') {
}
});
});
});

you can do this by various ways currently you are not using ajax request if you want to achieve this without ajax let follow these steps
when user click on submit button your form is submitted received form information(you define the path in action attribute where form submitted) after processing successfully redirect toward a new form
second solution use jquery ajax request
//first form
<form action='test.php' id='form_1' method='post'>
<label>Full Name</label>
<input type='text' name='full_name'>
<input type='submit'>
</form>
//second form
<form action='test.php' id='form_2' method='post' style='display:none'>
<label>Father Name</label>
<input type='text' name='father_name'>
<input type='submit'>
</form>
use jquery cdn
<script src='https://code.jquery.com/jquery-git.js'></script>
<script>
$("#form_1").submit(function(e) {
e.preventDefault(); // avoid to execute the actual submit of the form.
var form = $(this);
var url = form.attr('action');
$.ajax({
type: "POST",
url: url,
data: form.serialize(), // serializes the form's elements.
success: function(data)
{
alert("form submitted successfully");
$('#form_1').hide();
$('#form_2').show();
},
error:function(data){
alert("there is an error kindly check it now");
}
});
return false;
});
</script>

Add image uploading function inside this existing ajax code

My code here works fine except image uploading. It inserts all data in database .
<input type="file" name="image2" class="file" id="imgInp"/>
But after adding file type input in php it is showing
Notice: Undefined index: image2 in C:\xampp\htdocs\upload\submit.php on line 18
How can I add image uploading function in my existing code.
<div id="form-content">
<form method="post" id="reg-form" enctype="multipart/form-data" autocomplete="off">
<div class="form-group">
<input type="text" class="form-control" name="txt_fname" id="lname" placeholder="First Name" required /></div>
<div class="form-group">
<input type="text" class="form-control" name="txt_lname" id="lname" placeholder="Last Name" required /></div>
<div class="form-group">
<input type="text" class="form-control" name="txt_email" id="lname" placeholder="Your Mail" required />
</div>
<div class="form-group">
<input type="text" class="form-control" name="txt_contact" id="lname" placeholder="Contact No" required />
</div>
// here is the problem
<input type="file" name="image2" class="file" id="imgInp"/>
//here is the problem
<hr />
<div class="form-group">
<button class="btn btn-primary">Submit</button>
</div>
</form>
</div>
<script type="text/javascript">
$(document).ready(function() {
// submit form using $.ajax() method
$('#reg-form').submit(function(e){
e.preventDefault(); // Prevent Default Submission
$.ajax({
url: 'submit.php',
type: 'POST',
data: $(this).serialize() // it will serialize the form data
})
.done(function(data){
$('#form-content').fadeOut('slow', function(){
$('#form-content').fadeIn('slow').html(data);
});
})
.fail(function(){
alert('Ajax Submit Failed ...'); });
});
</script>
submit.php
<?php
$con = mysqli_connect("localhost","root","","table" ) or die
( "unable to connect to internet");
include ("connect.php");
include ("functions.php");
if( $_POST ){
$fname = $_POST['txt_fname'];
$lname = $_POST['txt_lname'];
$email = $_POST['txt_email'];
$phno = $_POST['txt_contact'];
$post_image2 = $_FILES['image2']['name']; // this line shows error
$image_tmp2 = $_FILES['image2']['tmp_name'];
move_uploaded_file($image_tmp2,"images/$post_image2");
$insert =" insert into comments
(firstname,lastname,email,number,post_image) values('$fname','$lname','$email','$phno','$post_image2' ) ";
$run = mysqli_query($con,$insert);
?>

You can use FormData, also I suggest you can change the elements id of the form, now all of them have ('lname') Try this with your current:
In yout HTML, put an ID to your file input
<input type="file" name="image2" id="name="image2"" class="file" id="imgInp"/>
And change the id of the other input.
In your JavaScript:
var frmData = new FormData();
//for the input
frmData.append('image2', $('#image2')[0].files[0]);
//for all other input
$('#reg-form :input').each(function(){
if($(this).attr('id')!='image2' ){
frmData.append($(this).attr('name'), $(this).val() );
}
});
$.ajax( {
url: 'URLTOPOST',
type: 'POST',
data: frmData,
processData: false,
contentType: false
}).done(function( result ) {
//When done, maybe show success dialog from JSON
}).fail(function( result ) {
//When fail, maybe show an error dialog
}).always(function( result ) {
//always execute, for example hide loading screen
});
In your PHP code you can access the image with $_FILE and the input with $_POST

FormData() works on the modern browsers.If you want for older browser support use malsup/form plugin
Your Form
<form method="post" action="action.php" id="reg-form" enctype="multipart/form-data" autocomplete="off">
Javscript
<script type="text/javascript">
var frm = $('#reg-form');
frm.submit(function (ev) {
var ajaxData = new FormData(frm);
$.ajax({
type: frm.attr('method'),
url: frm.attr('action'),
data: ajaxData,
contentType: false,
cache: false,
processData:false,
success: function (data) {
alert('ok');
}
});
ev.preventDefault();
});
In php extract($_POST) to get all input data and $_FILE for files

upload audio mp3 file using ajax in codeigniter

good day. I have a code which can upload image file but I don't know if this code is also work for uploading media file such as mp3. My project need to upload a media file but my code didn't work.
view
form
<form name="uploadform" id="uploadform" method="POST" enctype="multipart/form-data" >
<div class="form-group">
<label for="Title">Song Title</label>
<input type="text" class="form-control" id="title" placeholder="Title">
</div>
<div class="form-group">
<label for="Artist">Artist/Singer</label>
<input type="text" class="form-control" id="artist" placeholder="Artist/Singer">
</div>
<div class="form-group">
<label for="lyrics">Lyrics</label>
<textarea class="form-control" id="lyrics" placeholder="Lyrics"></textarea>
</div>
<div class="form-group">
<label for="Artist">Audio</label>
<input type="file" class="form-control" name="file" id="file" accept="audio/mp3">
</div>
<div class="form-group">
<span class="input-group-btn">
<button class="btn btn-primary" id="btn">UPLOAD</button>
</span>
</div>
</form>
view
javascript
$('#btn').click(function() {
var title = document.getElementById('title').value;
var artist = document.getElementById('artist').value;
var lyrics = document.getElementById('lyrics').value;
var file = $('#file').val();
$.ajax({
type: "post",
url: "<?php echo base_url('Admin/upload/')?>",
cache: false,
mimeType: "multipart/form-data",
contentType: false,
processData: false,
data: {
"title" : title,
"artist" : artist,
"lyrics" : lyrics,
"file" : file,
},
success: function(data){
try{
console.log(data);
}catch(e) {
alert('Exception while request..');
}
},
error: function(){
alert('Error while request..');
}
});
});
controller
Admin.php
public function upload()
{
$title = $this->input->post('title');
$artist = $this->input->post('artist');
$lyrics = $this->input->post('lyrics');
$attachment_file=$_FILES["file"];
$output_dir = "uploads/";
$fileName = $_FILES["attachment_file"]["name"];
move_uploaded_file($_FILES["attachment_file"]["tmp_name"],$output_dir.$fileName);
echo "File uploaded successfully";
}
that code gave me an error message.
<h4>A PHP Error was encountered</h4>
<p>Severity: Notice</p>
<p>Message: Undefined index: file</p>
<p>Filename: controllers/Admin.php</p>
<p>Line Number: 35</p>
I do not know why the file is undefined index since the 'file' is exist on form tag.
my code is not working for uploading mp3. How can I make this problem works?

Google Form Response not working from the website

I'm using a Google Form's value to integrate with my website where I want to submit the form and store data in google sheet as form responses. I'm using AJAX to redirect to another page instead of google form submit page. But whenever I'm trying to submit it's redirecting to my page accurately but datas are not saved in google sheet. Here are my codes,
<strong>Full Name</strong>
<input type="text" name="Fullname" class="form-control" id="Fullname" />
<strong>Email Address</strong>
<input type="text" name="Email" class="form-control" id="Email" />
<strong>Subject</strong>
<input type="text" name="Subject" class="form-control" id="Subject" />
<strong>Details</strong>
<textarea name="Details" rows="8" cols="0" class="form-control" id="Details"></textarea><br />
<button type="button" id="btnSubmit" class="btn btn-info" onclick="postContactToGoogle()">Submit</button>
<script type="text/javascript" src="https://code.jquery.com/jquery-1.11.3.min.js"></script>
<script>
function postContactToGoogle() {
var email = $('#Email').val();
var fullname = $('#FullName').val();
var subject = $('#Subject').val();
var details = $('#Details').val();
$.ajax({
url: "https://docs.google.com/forms/d/abcdefgh1234567xyz/formResponse",
data: {
"entry_805356472": fullname,
"entry_1998295708": email, "entry_785075795":
subject, "entry_934055676": details
},
type: "POST",
dataType: "xml",
statusCode: {
0: function () {
window.location.replace("Success.html");
},
200: function () {
window.location.replace("Success.html");
}
}
});
}
</script>
How can I save the data in google sheet? Am I missing something in my code? Need this help badly? Thanks.

You can directly copy the form from google view form page like shown in the demo, and then do the following changes in the AJAX call as shown below.
And now once you submit data it is visible in google forms, view responses.
$(function(){
$('input:submit').on('click', function(e){
e.preventDefault();
$.ajax({
url: "https://docs.google.com/forms/d/18icZ41Cx3-n1iW7yTOZdiDx9a6mySWHOy9ryd1l59tM/formResponse",
data:$('form').serialize(),
type: "POST",
dataType: "xml",
crossDomain: true,
success: function(data){
//window.location.replace("youraddress");
//console.log(data);
},
error: function(data){
//console.log(data);
}
});
});
});
<div class="ss-form"><form action="https://docs.google.com/forms/d/18icZ41Cx3-n1iW7yTOZdiDx9a6mySWHOy9ryd1l59tM/formResponse" method="POST" id="ss-form" target="_self" onsubmit=""><ol role="list" class="ss-question-list" style="padding-left: 0">
<div class="ss-form-question errorbox-good" role="listitem">
<div dir="auto" class="ss-item ss-text"><div class="ss-form-entry">
<label class="ss-q-item-label" for="entry_1635584241"><div class="ss-q-title">What's your name
</div>
<div class="ss-q-help ss-secondary-text" dir="auto"></div></label>
<input type="text" name="entry.1635584241" value="" class="ss-q-short" id="entry_1635584241" dir="auto" aria-label="What's your name " title="">
<div class="error-message" id="1979924055_errorMessage"></div>
<div class="required-message">This is a required question</div>
</div></div></div>
<input type="hidden" name="draftResponse" value="[,,"182895015706156721"]
">
<input type="hidden" name="pageHistory" value="0">
<input type="hidden" name="fvv" value="0">
<input type="hidden" name="fbzx" value="182895015706156721">
<div class="ss-item ss-navigate"><table id="navigation-table"><tbody><tr><td class="ss-form-entry goog-inline-block" id="navigation-buttons" dir="ltr">
<input type="submit" name="submit" value="Submit" id="ss-submit" class="jfk-button jfk-button-action ">
</td>
</tr></tbody></table></div></ol></form></div>
<!-- jQuery (necessary for Bootstrap's JavaScript plugins) -->
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>

If you go to your form on Google and click on 'View Live Form', and then view source on the form, you'll see that the fields you want to upload have aname in the form entry.12345678; the id is in the form entry_12345678. You need to use the name value. Try this:
<script>
function postContactToGoogle() {
var email = $('#Email').val();
var fullname = $('#FullName').val();
var subject = $('#Subject').val();
var details = $('#Details').val();
$.ajax({
url: "https://docs.google.com/forms/d/abcdefgh1234567xyz/formResponse",
data: {
"entry.805356472": fullname,
"entry.1998295708": email,
"entry.785075795": subject,
"entry.934055676": details
},
type: "POST",
dataType: "xml",
statusCode: {
0: function () {
window.location.replace("Success.html");
},
200: function () {
window.location.replace("Success.html");
}
}
});
}
</script>

Ajax - How to submit one by one input value - Codeigniter

Sorry for my english is not so good. And i'm newbie :)
I want to update one-by-one input value with ajax in Codeigniter, but it not work right.. only one save button (one form) work, others form not work .. please help me edit below code
Here's the demo code:
View:
<script>
$(function(){
$(".submit45").click(function(){
dataString = $("#prod_upd").serialize();
$.ajax({
type: "POST",
url: "<?=PREFIX?>admin/update/change_ppx3/",
data: dataString,
success: function(data){
console.log(data);
document.getElementById('dd').innerHTML=data;
}
});
return false;
});
});
</script>
<?$i=0;if(count($PPX) > 0)foreach($PPX as $item){$i++;?>
<form name="prod_upd" id="prod_upd" method="post" >
<input type="text" name="p_ppx" id="p_ppx" size="8" value="<?= number_format($item['p_ppx'],0,'','')?>" class="i_ppx">
<input type="hidden" name="ids_p" id="ids_p" size="8" value="<?=$item['id']?>" class="i_ppx">
<input type="button" name="sub" id="sub" class="submit45" value="Save4" />
<div id="dd" style="float: left;">hello</div>
</form>
<?}else{?>
<div class="no_data">Nothing here</div>
<?}?>
Controller:
function change_ppx3(){
$id_p = $_POST['ids_p'];
$rs = $this->ppx->get_ppx_by_id($id_p);
$ppx_value = $_POST['p_ppx'];
$this->ppx->update_ppx(array("id"=>$id_p),array("ppx_r"=>$ppx_value));
if($_POST['p_ppx']):
echo "done: ";
print_r($_POST['ids_p']);
echo "-";
print_r($_POST['p_ppx']);
return true;
endif;
}

because every form has the same id="prod_upd".
test this
<script>
$(function(){
$(".prod_upd").submit(function(){
var $this = $(this), dataString = $this.serialize();
$.ajax({
type: "POST",
url: "<?=PREFIX?>admin/update/change_ppx3/",
data: dataString,
success: function(data){
console.log(data);
$this.find('.dd').html(data);
}
});
return false;
});
});
</script>
<?$i=0;if(count($PPX) > 0)foreach($PPX as $item){$i++;?>
<form name="prod_upd" class="prod_upd" method="post" >
<input type="text" name="p_ppx" size="8" value="<?= number_format($item['p_ppx'],0,'','')?>" class="i_ppx">
<input type="hidden" name="ids_p" size="8" value="<?=$item['id']?>" class="i_ppx">
<input type="submit" class="submit45" value="Save4" />
<div class="dd" style="float: left;">hello</div>
</form>
<?}else{?>
<div class="no_data">Nothing here</div>
<?}?>