I am posting a form to a php file using ajax but for some reason which I can't figure out, it is not using ajax but directly posting to php.
My form:
<form id="editform" name="editform" action="ajaxeditform.php" method="post">
<input type="hidden" name="aid" id="aid" readonly class="form-control col-md-7 col-xs-12"/>
<input type="text" name="number" id="tailnumber" readonly class="form-control col-md-7 col-xs-12"/>
<input type="text" name="type" id="type" class="form-control col-md-7 col-xs-12" placeholder="e.g. C172" />
<input type="text" name="colormarkings" id="colormarkings" class="form-control col-md-7 col-xs-12" />
<button type="submit" class="btn btn-success">Update info</button></div>
</form>
My jquery:
$('#editform').on('submit', function(e){
e.preventDefault();
$.ajax({
type: $(this).attr('method'),
url: $(this).attr('action'),
data: $(this).serialize(),
dataType: 'json',
cache: false,
})
.success(function(response) {
// remove all errors
$('input').removeClass('error').next('.errormessage').html('');
if(!response.errors && response.result) {
new PNotify({
title: 'Success',
text: 'Info updated successfully',
type: 'success'
});
} else {
$.each(response.errors, function( index, value) {
// add error classes
$('input[name*='+index+']').addClass('error').after('<div class="errormessage">'+value+'</div>')
});
}
});
});
Changes:-
Form opening tag is not proper, change like below:-
<form id="editform" name="editform" action="ajaxeditform.php" method="post">
try to change your jquery code first line like below:-
$('Form#editform').on('submit', function(e){ OR $('.btn-success').click(function(e){
Add jquery library also (check that it is added or not?). for example add this code before jQuery code:-
<script src="//code.jquery.com/jquery-1.12.0.min.js"></script>
add $(document).ready(function(){ in your code.like below:-
$(document).ready(function(e){
........ //your code
});
Related
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>
I'm having some problems with Jquery/Ajax.
This is my code for the form :
<form class="form-auth-small" method="POST" id="form" enctype="multipart/form-data">
<div class="form-group">
<label for="signin-email" class="control-label sr-only">Email</label>
<input type="email" class="form-control" name="email" value="samuel.gold#domain.com" placeholder="Email">
</div>
<div class="form-group">
<label for="signin-email" class="control-label sr-only">Name</label>
<input type="text" class="form-control" name="name" value="John">
</div>
<div class="form-group">
<label for="signin-email" class="control-label sr-only">Phone</label>
<input type="text" class="form-control" name="phone" placeholder="938434928">
</div>
<div class="form-group">
<label for="signin-email" class="control-label sr-only"></label>
<input type="password" class="form-control" name="password" placeholder="****">
</div>
<div class="form-group">
<input name="image" type="file" />
</div>
<button type="submit" id="submit" class="btn btn-primary btn-lg btn-block">Register</button>
</form>
And this is my code to AJAX/Jquery:
<script>
$(".submit").on("click", function(e) {
var form = $("#form");
// you can't pass Jquery form it has to be javascript form object var formData = new FormData(form[0]);
console.log(formData);
$.ajax({
type: "POST",
url: '/user/signup/',
data: formData,
contentType: false, //this is requireded please see answers above processData: false, //this is requireded please see answers above //cache: false, //not sure but works for me without this error: function (err) { console.log(err);
success: function(res) {
console.log(res);
},
});
});
</script>
When i do console.log to check that from form i don't receive any value, and when i check in network i dont see any HTTP callback.
Your code doesn't work because you were using the class selector token . (dot) instead of the id selector token # hashtag. Further details can be found on the documentation
Change this instruction
$(".submit").on("click", function(e) // …
to
$("#submit").on("click", function(e) // …
You dont have submit class. You have id="submit"
$("#submit").on("click", function(e) {
Use this and it should work.
Also change type="submit" to type="button" or in your js code you need to add
e.preventDefault();
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
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>
My code is working in chrome,safari but its not working in firefox.
Here is my code
<script>
$(document).ready(function () {
$("#loginform").on('submit',(function(e) {
e.preventDefault();
$('#loading').html("Loading....");
$.ajax({
url: "login.php",
type: "POST",
data: new FormData(this),
contentType:false,
cache: false,
processData:false,
async:false,
success: function(data)
{
$('#loading').hide();
$("#message").html(data);
}
});
return false;
}));
});
</script>
Can anyone solve this issue??? I have one more same script in the same page only url is different but its working fine,but this script is not working .Iam getting empty data .But in chrome,safari its working fine.
My Html Code :
<form role="form" method="post" id="loginform" action="">
<div class="form-group">
<label for="username">Username</label>
<input type="text" class="form-control" id="username" name="username" placeholder="Enter Username" required>
</div>
<div class="form-group">
<label for="password"> Password</label>
<input type="password" class="form-control" id="password" name="password" placeholder="Enter password">
</div>
<div class="checkbox">
<label><input type="checkbox" value="" name="user_rememberme" checked>Remember me</label>
</div>
<div id="loading"></div>
<div id="message"></div>
<button type="submit" name="login" class="btn btn-success pull-right">Login</button>
</form>
Never use async:false in ajax call unless you knows specifically wat you are doing.The problem is that async:false freezes the browser until ajax call is complete (either error or success).set it to true or remove it (by default it is true).Implement error block too and check if its an error from server side
success: function(data)
{
$('#loading').hide();
$("#message").html(data);
},error:function(data){
console.log(data)
}
Try this:
<script>
$(document).ready(function () {
$("#loginform").on('submit',(function(e) {
dta = $(this).serialize();
e.preventDefault();
$('#loading').html("Loading....");
$.ajax({
url: "login.php",
type: "POST",
data:dta,
success: function(data)
{
$('#loading').hide();
$("#message").html(data);
}
});
return false;
}));
});
</script>