Modal form Submission to PHP using Jquery .ajax - php

I am trying to post a modal form to a table using php, jquery .ajax but it never works.. tried debugging using firebug and i don't see any errors. i tested the form by using form action="notes_functions.php" and it worked fine.
Profile.php
<div class="modal-body">
<form class="noteform" id="notesmodal" method="post">
<fieldset>
<label>Title</label>
<input type="text" class="form-control" name="note_title" placeholder="Enter a title for your note">
<label>Note</label>
<textarea rows="4" cols="50" class="form-control" name="note_content" placeholder="note"></textarea>
<label>Note type</label>
<div class="panel-body">
<input type="tagsinput" id="teetete" class="tagsinput" value="" />
</div>
<label for="exampleInputFile">Attach a document</label>
<input type="file" id="exampleInputFile3">
<p class="help-block">PDF, DOCX and image files are supported.</p>
<div class="checkbox">
<label>
<input type="checkbox"> Check me out
<input type="label" name="note_account" value="<?php echo $acctname ?>"/>
</label>
</div>
<input type="hidden" name="note_creator" value="<?php echo $_SESSION['username'];?>"/>
</fieldset>
<button class="btn btn-default" id="submitnote" >ADD</button>
</form>
</div>
this is my js code
$(function(){
$("button#submitnote").click(function(){
$.ajax ({
type:"POST",
url:"notes_functions.php",
data: $('form.noteform').serialize(),
success: function(msg){
$("#thanks").html(msg)
$("form.noteform").modal('hide');
},
error: function(){
alert("failure");
}
});
});
});
notes_functions.php
<?php
include_once 'dbconnect.php';
if (isset($_POST['note_title'])) {
$notetitle = strip_tags($_POST['note_title']);
$noteContent = strip_tags($_POST['note_content']);
$noteAccount = strip_tags($_POST['note_account']);
$noteCreator = strip_tags($_POST['note_creator']);
mysql_query("INSERT INTO account_notes (note_title, note_contents, note_account, note_creator)
VALUES ('$notetitle','$noteContent', '$noteAccount', '$noteCreator') ");
echo "Name = ".$notetitle;
echo $noteCreator;
}
?>

You should really be using .submit() rather than click (for submit-by-enter, etc.) and returning false to block conventional submits. You also need to make sure the code to bind the events runs after the form element is created. The easiest way to do this is to put it in the document ready handler.
jQuery(document).ready(function ($) {
$("#notesmodal").submit(function () {
$.ajax({
type: "POST",
url: "notes_functions.php",
data: $('form.noteform').serialize(),
success: function (msg) {
$("#thanks").html(msg)
$("form.noteform").modal('hide');
},
error: function () {
alert("failure");
}
});
return false;
});
});
And change the ADD button to be:
<input type="submit" name="submit" class="btn btn-default" id="submitnote" value="ADD" />

Related

How to open a Modal dialog instead when submitting data to a php script? [duplicate]

I am trying to post a modal form to a table using php, jquery .ajax but it never works.. tried debugging using firebug and i don't see any errors. i tested the form by using form action="notes_functions.php" and it worked fine.
Profile.php
<div class="modal-body">
<form class="noteform" id="notesmodal" method="post">
<fieldset>
<label>Title</label>
<input type="text" class="form-control" name="note_title" placeholder="Enter a title for your note">
<label>Note</label>
<textarea rows="4" cols="50" class="form-control" name="note_content" placeholder="note"></textarea>
<label>Note type</label>
<div class="panel-body">
<input type="tagsinput" id="teetete" class="tagsinput" value="" />
</div>
<label for="exampleInputFile">Attach a document</label>
<input type="file" id="exampleInputFile3">
<p class="help-block">PDF, DOCX and image files are supported.</p>
<div class="checkbox">
<label>
<input type="checkbox"> Check me out
<input type="label" name="note_account" value="<?php echo $acctname ?>"/>
</label>
</div>
<input type="hidden" name="note_creator" value="<?php echo $_SESSION['username'];?>"/>
</fieldset>
<button class="btn btn-default" id="submitnote" >ADD</button>
</form>
</div>
this is my js code
$(function(){
$("button#submitnote").click(function(){
$.ajax ({
type:"POST",
url:"notes_functions.php",
data: $('form.noteform').serialize(),
success: function(msg){
$("#thanks").html(msg)
$("form.noteform").modal('hide');
},
error: function(){
alert("failure");
}
});
});
});
notes_functions.php
<?php
include_once 'dbconnect.php';
if (isset($_POST['note_title'])) {
$notetitle = strip_tags($_POST['note_title']);
$noteContent = strip_tags($_POST['note_content']);
$noteAccount = strip_tags($_POST['note_account']);
$noteCreator = strip_tags($_POST['note_creator']);
mysql_query("INSERT INTO account_notes (note_title, note_contents, note_account, note_creator)
VALUES ('$notetitle','$noteContent', '$noteAccount', '$noteCreator') ");
echo "Name = ".$notetitle;
echo $noteCreator;
}
?>
You should really be using .submit() rather than click (for submit-by-enter, etc.) and returning false to block conventional submits. You also need to make sure the code to bind the events runs after the form element is created. The easiest way to do this is to put it in the document ready handler.
jQuery(document).ready(function ($) {
$("#notesmodal").submit(function () {
$.ajax({
type: "POST",
url: "notes_functions.php",
data: $('form.noteform').serialize(),
success: function (msg) {
$("#thanks").html(msg)
$("form.noteform").modal('hide');
},
error: function () {
alert("failure");
}
});
return false;
});
});
And change the ADD button to be:
<input type="submit" name="submit" class="btn btn-default" id="submitnote" value="ADD" />

Two-steps form with modal window jquery + php

I'm just learning php and js and trying for a day to make this thing work...
Here's how it's supposed to be: after filling form you push button "Show modal" it shows modal window with input field. After submitting this form all the data from the form + phone from modal are send via email. When all input fields are simply in a form it all works fine. But when I show phone field in a modal it fails...
Here's the form
<div class="form">
<form class="form-deal form-deal1" id="form-order-3" name="special-price">
<input type="text" class="sqmeters" name="sqmeters" value="" />
<label>Select option:</label>
<div class="selectWrap">
<select class="formSelect" name="filmtype">
<option>Option one</option>
<option>Š¢Option two</option>
</select>
</div>
<label>Radio buttons</label>
<input type="radio" id="rb1" class="radio" name="regulator" value="r1"><label class="radioLabel" for="rb1">r1</label>
<input type="radio" id="rb2" class="radio" name="regulator" value="rb2"><label class="radioLabel" for="rb2">rb2</label>
<input type="radio" id="rb3" class="radio" name="regulator" value="rb3"><label class="radioLabel" for="rb3">rb3</label>
<label>Options:</label>
<div class="selectWrap">
<select class="formSelect" name="mount">
<option>opt1</option>
<option>opt2</option>
<option>opt3</option>
</select>
</div>
<div class="modal">
<input class="step2 hidden" type="text" name="phone" maxlength="14" placeholder="Phone:" required="" />
<button type="submit" class="btn step2 hidden">Submit form</button>
<span class="message"></span>
</div>
</form>
<a class="btn topcalcbtn">Show modal</a>
And the JS
$("form").submit(function(e){
e.preventDefault();
});
$(document).on('click', '.form-deal .btn', function () {
if ($(this).parent('form').find('input[name="phone"]').val().length>0 &&
$(this).parent('form').find('input[name="sqmeters"]').val().length>0)
AjaxFormRequest($(this).parent('form').attr('id'),'../mail_order.php');
});
function AjaxFormRequest(form_id, url) {
jQuery.ajax({
url: url,
type: "POST",
dataType: "html",
data: jQuery("#" + form_id).serialize(),
success: function (response) {
$('#'+form_id+' .message').text("Thanks!");
$('#'+form_id+' *:not(".message")').fadeOut(500);
$('.overlay, .popup').delay(2000).fadeOut(500);
$(window).attr('location','http://lp.tsk-energy.ru/thanx.html');
},
error: function (response) {
$('#'+form_id+' .message').text("Error");
}
});
}
Any help would be appreciated. Thanks in advance!

AJAX code not updating table

Here is my form, from which i want to update the values. When i try this with simple php+html it works perfectly!! But when i try to post values through ajax call it doesn't work.Any suggestions please.
HTML
<form class="form-block" role="form" action="" method="post">
<div class="form-group">
<?php
$email=$_SESSION["login_email"];
$query=mysqli_query($con,"select * from customers where email='$email'");
while($row=mysqli_fetch_array($query,MYSQLI_ASSOC))
{
?>
<label for="name">Full Name</label>
<input type="text" name="name" class="form-control" id="name" value="<?= $row["name"]; ?>" placeholder="Full Name">
<label for="comment">Address</label>
<textarea class="form-control" name="address" id="address" rows="5" id="comment" ></textarea>
<label for="telephone">Telephone</label>
<input type="tel" class="form-control" name="phone" id="phone" placeholder="Enter Mobile Number" value="" >
<label for="city">City</label>
<input type="text" class="form-control" name="city" id="city" placeholder="Enter City" value="<?= $row["city"]; ?>" >
</div>
<?php
}?>
<input type="submit" class="btn btn-default" name="add" id="add" value="Update"/>
<span class='msg'></span>
<div id="error"></div>
</form>
AJAX
$(document).ready(function() {
$('#add').click(function()
{
$.ajax({
url: "address_update.php",
type: "POST",
async: true,
data: { Name:$("#name").val(),Address:$("#address").val(), Phone:$("#phone").val(), City:$("#city").val()}, //your form data to post goes here as a json object
dataType: "html",
success: function(data) {
if(data)
{
//$('#output').html(data);
$("body").load("index.php?page=form");//.hide();//.fadeIn(1500).delay(6000);
}
else
{
$("#error").html("<span style='color:#cc0000'>Error:</span> Invalid username and password. ");
}}});
});});
PHP
<?php
include ("db/db.php");
session_start();
$name=$_POST['Name'];
$address=$_POST['Address'];
$phone=$_POST['Phone'];
$city=$_POST['City'];
$email=$_SESSION['login_email'];
$sql=mysqli_query($con,"Update customers set name='$name',address='$address',phone='$phone',city='$city' where email='$email'");
if($sql)
{
echo "updated";
}
This selector:
$('#add')
doesn't find anything. Because there is no HTML element in your markup with id="add".
You do have an element with name="add". So either add an id to that element:
<input id="add" type="submit" class="btn btn-default" name="add" value="Update"/>
or change your selector to target the existing element:
$('input[name="add"]')
Note: The same is also true of $('#address') and $('#phone'). No such elements exist in your markup. Take a look at the jQuery documentation for selectors.
Remove the IF loop mentioned under
if(isset($_SESSION["login_email"]))
Additionally, you nee to change the button type from submit to button
<input type="submit" class="btn btn-default" name="add" value="Update"/>
Above sends the form, whatever you add to the click event is done in parralel.
<input type="button" class="btn btn-default" name="add" value="Update"/>
only executes your JavaScript, when using:
$('input [name="add"]').click(function(){
// your improved ajax call here
}
Try this:
$.ajax({
.
.
data: {
'Name': $name,
'address': $address,
.
.
}
});

form with input file not submitted when using bootstrapFileInput

I am submitting a form which contains some text field, dropdown and a file input. The form is submitted using JQuery/Ajax and the data is sent to php. If I submit the form without using bootstrapFileInput, all the data of the form is sent to php and correctly processed but as soon as I enable bootstrapFileInput, the input file is missing while all other fields are correctly submitted.
I get no error in the console.
HTML :
<div class="form-inline">
<div id="singUser">
<form role="form" name="setLimit" id="setLimit" class="form-inline" enctype="multipart/form-data">
<input type="text" id="userid" name="userid" class="form-control" placeholder="Enter User ID">
<input type="text" id="quota" name="quota" class="form-control" placeholder="Enter Quota limit in GB">
<select class="form-control" id="ltype" name="ltype">
<option value="desktop">Desktop</option>
<option value="server">Server</option>
</select>
<input type="checkbox" id="rmlimit" name="rmlimit"> Remove Limit (Set to Shared storage)
<button type="submit" id="sgUserButn" class="btn btn-default btn-primary" >Set Limit!</button>
</div>
<span id="multUser">
<p>Select a file with a list of user ids and the limit in GB you wish to assign download the example file.</p>
<img src="img/csvex.png" id="csvexp" ><br><br>
<input class="file-input-wrapper" type="file" id="lsUsers" name="lsUsers" data-filename-placement="inside" title="Select CSV File">
<button id="mulUserButn" type="submit" class="btn btn-default btn-primary" >Set Limit!</button>
</span>
</form>
</div>
JQuery
$(document).ready(function() {
$.ajaxSetup({
cache: false
});
$('input[type=file]').bootstrapFileInput();
$('.file-inputs').bootstrapFileInput();
$('#setLimit')
.submit(function(e) {
busyStatus();
$.ajax({
url: 'testProcess.php',
type: 'POST',
data: new FormData(this),
processData: false,
contentType: false,
complete: function() {
idleStatus();
},
success: function(result) {
//$(document.body).append(result);
//alert( $(result).toArray() );
$("#result").html(result);
}
});
e.preventDefault();
return false;
});
});
PHP
<?php
print_r($_POST);
echo '<br>';
print_r($_FILES);
?>
I am really out of ideas and would really appreciate some help.
Thanks in advance.

Form Cross domain POST request using PHP

I am trying to send data from a form to a php file so I can store it in a database, but its not working...
The code for the form is not on the same server as the php file, because the form will be on a mobile app.
html
<div data-role="page" id="createclub">
<div data-role="content">
<form id="cname" align="left" action="post">
<label for="name">Enter Name:</label>
<input type="text" id="name" value="" />
<input type="submit" value="Submit" data-inline="true">
</form>
<div id="result"></div>
</div>
<script type="text/javascript">
$(document).ready(function(){
$("#cname").submit( function () {
$.post(
'http://www.clubbedin.isadcharity.org/createclub.php',
$("#cname").serialize(),
function(data){
$("#result").html(data);
alert("Data " + data);
}
);
return false;
});
});
</script>
php file
$name = $_POST['name'];
THANK YOU!!!
Add this at the beginning of your PHP file:
header("access-control-allow-origin: *");
More info on cross domain policy here.
I think you need to prevent the default function of the submit button using .preventDefault() because as I look on your code you want to submit your form using ajax
$("#cname").submit(function (e) {
e.preventDefault();
$.ajax({
url: 'http://www.clubbedin.isadcharity.org/createclub.php',
crossDomain: true, //set as a cross domain requests
type: 'post',
data: $("#cname").serialize(),
success: function (data) {
$("#result").html(data);
alert("Data " + data);
},
});
});
and please use .ajax() so that you can set your ajax request to a cross-domain request
http://api.jquery.com/jQuery.ajax/
Your input element <input type="text" id="name" value="" /> must set up the name attribute as name="name".
form #cname after edited
<form id="cname" align="left" action="post">
<label for="name">Enter Name:</label>
<input type="text" id="name" name="name" value="" />
<input type="submit" value="Submit" data-inline="true">
</form>
You can gather more informations from the jQuery API Documention:
http://api.jquery.com/
http://api.jquery.com/serialize/

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