My Form looks like this:
<form id="invite-form" method="post" action="" >
<input type="text" name=friends[0][first_name] />
<input type="text" name=friends[0][last_name] />
<input type="text" name=friends[0][email] />
<input type="text" name=friends[1][first_name] />
<input type="text" name=friends[1][last_name] />
<input type="text" name=friends[1][email] />
<input type="text" name=friends[2][first_name] />
<input type="text" name=friends[2][last_name] />
<input type="text" name=friends[2][email] />
<input type="submit" value="Invite" />
</form>
And here is my javascript code:
var friends = [];
jQuery("#invite-form").serializeArray().map(function(x){friends[x.name] = x.value;});
var formData = {
'friends' : JSON.stringify(friends),
'action' : 'invite-friends'
};
jQuery.ajax({
type : 'POST',
url : '/invitation.php',
data : formData,
dataType : 'json',
encode : true
}).done(function(data){
console.log(data);
if(data.success == true){
...
It doesn't work properly and the list of friends is not POSTed properly. How do I convert this kind of form to json data properly? I want to be able to use this data as a regular array, as if it were a regular form POSTed.
In this way you can submit form with serialize data and you many use
post or get request for this or any other method.
jQuery("#invite-form").submit(function(e) {
e.preventDefault();
let data = jQuery(this).serialize();
jQuery.ajax({
type: "get",
url: 'invitation.php',
data: data,
cache: false,
success: function(data){
console.log(data);
}
});
});
Related
I having 3 textbox and I want to send value enter over there to controller method
<input class="form-control" type="text" id="id" name="id">
<input class="form-control" type="text" id="id1" name="id1">
<input class="form-control" type="text" id="id2" name="id2">
#Html.ActionLink("Send", "MethodNameInController", "ColtrollerName", new { id= $('#id').val(),
id1= $('#id1').val(), id2= $('#id2').val()})
and in controller
public ActionResult MethodNameInController(int id, string id1, string id2)
{
//some text here
}
but it's sending null value
If you want to stay on the same view or redirect to another url after, instead of using #Html.ActionLink() try passing the values to the controller using an ajax call.
function submitAjax(){
//purely for readability
var id = $('#id').val();
var id1 = $('#id1').val();
var id2 = $('#id2').val();
//ajax call
$.ajax({
url: "/ControllerName/MethodNameInController",
data: { id: id, id1: id1, id2: id2 },
success: function (result) {
//handle something here
}
});
};
<input class="form-control" type="text" id="id" name="id">
<input class="form-control" type="text" id="id1" name="id1">
<input class="form-control" type="text" id="id2" name="id2">
<button type="button" onclick="submitAjax()">submit</button>
If the method does some processing and then returns you to a different view, you could do a form submit.
function submitForm(){
var $form = $("#idForm");
//optional validation
$form.submit();
};
<form id="idForm" class="form-horizontal" asp-controller="ControllerName" asp-action="MethodNameInController" >
<input class="form-control" type="text" id="id" name="id">
<input class="form-control" type="text" id="id1" name="id1">
<input class="form-control" type="text" id="id2" name="id2">
<!-- type="button" prevents the from from submitting so you can do validation checks in the js -->
<button type="button" onclick="submitForm()">submit</button>
</form>
I got the solution using ajax just want to know is there any direct solution
$(function () {
$('#receipt').unbind('click');
$('#receipt').on('click', function () {
$.ajax({
url: "/Controller/Method",
type: 'POST',
contentType: "application/json; charset=utf-8",
dataType: 'json',
data: JSON.stringify({
fromDate: $("#fromDate").val(),
toDate: $("#toDate").val(),
id: $("#id").val()
}),
async: false
});
});
});
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
HTML CODE
<form id="details">
<div>
<input type="text" placeholder="Email ID" id="email"></input>
<input type="text" placeholder="Mobile Number" id="mobile"></input>
</div>
<h5>Data</h5>
<div>
<div>
<input type="text" placeholder="Name" id="name1"></input>
<input type="text" placeholder="Age" id="age1"></input>
</div>
<div>
<input type="text" placeholder="Name" id="name2"></input>
<input type="text" placeholder="Age" id="age2"></input>
</div>
</div>
</form>
JS CODE
$(document).ready(function() {
$("form").on('submit', function(e) {
// Prepare data
var form = $("form");
var formData = new FormData(document.getElementById("details"));
e.preventDefault();
$.ajax(form.attr('action'), {
type: 'POST',
data: formData,
contentType: false,
cache: false,
processData:false,
dataType: "json",
success: function(result) {
// Success Code
},
error: function(result) {
// Failure Code
},
timeout: 5000,
});
});
});
The codepen link for my code is http://codepen.io/anon/pen/VKzGRG
I want to send data like
{
"email" : "xyz#gmail.com",
"mobile" : "9898989898",
"data" : [
{
"name":"xyz",
"age":45
},
{
"name":"xyz",
"age":45
}
]
}
I tried sending the data using jQuery.
The problem is that it's sending only one name and age.
Also, in my project, I'm dynamically adding a Name and Age box using jQuery and a button.
How can I send the data using AJAX post method using jQuery?
you have missed the name of input that's why data is not posting
see the below working code.
//HTML code
<form name="details" action="t.php" id="details">
<div>
<input type="text" name="email" placeholder="Email ID" id="email"></input>
<input type="text" name="mobile" placeholder="Mobile Number" id="mobile"></input>
</div>
<h5>Data</h5>
<div>
<div>
<input type="text" name="data1[]" placeholder="Name" id="name1"></input>
<input type="text" name="data1[]" placeholder="Age" id="age1"></input>
</div>
<div>
<input type="text" name="data2[]" placeholder="Name" id="name2"></input>
<input type="text" name="data2[]" placeholder="Age" id="age2"></input>
</div>
</div>
<input type="submit" value="send">
</form>
//ajax code
$(document).ready(function() {
$("form").on('submit', function(e) {
// Prepare data
var form = $("form");
var formData = new FormData(document.getElementById("details"));
console.log(formData);
e.preventDefault();
$.ajax(form.attr('action'), {
type: 'POST',
data: formData,
contentType: false,
cache: false,
processData:false,
dataType: "json",
success: function(result) {
// Success Code
},
error: function(result) {
// Failure Code
},
timeout: 5000,
});
});
});
To post data with jQuery and Ajax, you can do something like this
var myObj = {
"email" : "xyz#gmail.com",
"mobile" : "9898989898",
"data" : [
{
"name":"xyz",
"age":45
},
{
"name":"xyz",
"age":45
}
]
};
$.post( "test.php", myObj)
.done(function( data ) {
alert( "Data Loaded: " + data );
});
That will post the data and alert the response of the request.
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>
I'm trying to POST some JSON data to a local host and I keep getting a 404 Not Found error which is strange because the php file is located in the correct location as specified in the script. I would appreciate any feedback from anyone who has experience with this. Am I getting this error because the the server can not locate the ajax.php file for some unknown reason?
<div class="container">
<div class="header">
<h3 class="text-muted">AJAX JSON Data</h3>
</div>
<div id="data-div">
<form method="post" action="api/ajax.php" class="ajax">
<p><label for="firstname" class="contact-input-text">First Name</label> <br/>
<input id="first-name" name="firstname" type="text" maxlength="30" autofocus /></p><p><label for="lastName" class="contact-input-text">Last Name</label> <br/>
<input id="last-name" name="lastname" type="text" maxlength="30" autofocus /></p>
<p><input type="submit" id="submit-button" class="contact-input-text" value="submit" /></p>
</form>
</div>
</div>
<script>
$('form.ajax').on('submit', function(){
var jsondata = {};
$(this).find('[name]').each(function(i, data){
console.log(data);
var that = $(this);
var key = that.attr('name');
var value = that.val();
jsondata[key] = value;
});
console.log(jsondata);
$.ajax({
type: 'POST',
url: 'ajax.php',
dataType: 'json',
data: jsondata,
success: function(response){
console.log(response);
},
error: function(xhr){
console.log(xhr);
}
});
return false;
</script>
Here is the ajax.php file....
<?php
if(isset($_POST['submit'])) {
$file = "data.json";
$json_string = json_encode($_POST,JSON_PRETTY_PRINT);
file_put_contents($file,$json_string,FILE_APPEND);
}
?>
This is the directory structure :
index.html (contains the form input fields and the ajax request)
ajax.php
/styles
/images
have you ensure with correct url in ajax?
maybe not thi:
url: 'ajax.php'
but this:
url: 'api/ajax.php'