use jQuery to send javascript array to php - php

I'm trying to pass javascript array to php using ajax and jQuery.
I have simple page that contains a series of numbers that I've made selectable via jQuery UI (see below) When I select a group of numbers, I use array.push to add those numbers to an array called "shift". Here's the question: What's the simplest way to send this array to PHP? Will it remain an array after it comes over? I'm new to coding and would appreciate any help I can get. After a lot of research, here's what I've tried. Oh, I've managed to figure out how to submit the form to PHP, it's the jQuery UI array that i'm stuck on.
here's my main.php
<!DOCTYPE html>
<html>
<head>
<title>test</title>
<link rel="stylesheet" type="text/css" href="main.css">
</head>
<body>
<div class = "container">
<form action="post.php" method="post" id="add">
<input type="text" class="leader" name="name" placeholder="Leader">
<input type="text" class="date" name="date" placeholder="date">
<input type="text" class="time" name="time" placeholder="time">
<input type="text" class="score" name="score" placeholder="score">
<input type="submit" id="btn" value="send" />
</form>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.0/jquery.min.js"></script>
<script type="text/javascript" src="jquery-ui-1.10.4.custom.js"></script>
<script src="globe.js"></script>
<ul id ="hours_zoned">
<li class="nine">9</li>
<li class="ten">10</li>
<li class="eleven">11</li>
<li class="twelve">12</li>
<li class="one">1</li>
<li class="two">2</li>
<li class="three">3</li>
<li class="four">4</li>
<li class="five">5</li>
<li class="six">6</li>
<li class="seven">7</li>
<li class="eight">8</li>
<li class="nine">9</li>
</ul>
</div>
</body>
</html>
here's my post.php
<html>
<body>
/* I'm using all of these echoes to verify that the data is coming
over. Only "shift" fails to come over */
NAME <?php echo $_POST["name"]; ?><br>
DATE <?php echo $_POST["date"]; ?><br>
TIME <?php echo $_POST["time"]; ?><br>
SCORE: <?php echo $_POST["score"]; ?><br>
SHIFT: <?php echo $_POST["shift"]; ?>
<?php
include("db.php");
$name = $_POST["name"];
$date = $_POST["date"];
$time = $_POST["time"];
$query = "INSERT INTO leaders (name, shift_date, shift_time) VALUES ('$name', '$date', '$time')";
$result = $db->query($query);
?>
</body>
</html>
here is my globe.js
var shift = []; //create array to store zoned hours
$(function() {
$( "#hours_zoned" ).selectable();
$( "#hours_zoned" ).on('click', 'li', function() {
$(this).addClass("clicked ui-selected");
$( this ).css( "background-color", "#e74c3c" );
$( this ).selectable({ disabled: true });
var floorTime = $(this).text(); // get the value of the hour that was clicked
shift.push(floorTime); // add that hour to the floorTime
});
/*$('#add').on('submit', function() {
var name = $('.leader').val(); */
$('#btn').on('submit', function() {
$.ajax({
type: 'POST',
url: 'post.php',
data: shift,
success: function(msg) {
alert(msg);
}
});
});
return false;
});

You are only sending the key
data: shift,
according to the docs - Object must be Key/Value pairs. (https://api.jquery.com/jQuery.ajax/)
try
data: {shift: shift},
so it is now
$('#btn').on('submit', function() {
$.ajax({
type: 'POST',
url: 'post.php',
data: {shift: shift},
success: function(msg) {
alert(msg);
}
});
return false;
});

EDIT: Your ajax function was a bit jacked up. Fixed that for you also. See new code.
Your ajax submit isn't running. Add return false; to prevent the main form from submitting. I'm thinking the easiest way to add the array would be to insert it into a hidden input immediately after being pushed in your jQuery function. Then you would just serialize the form data and send it all in your ajax function. See below:
New form:
<form action="post.php" method="post" id="add">
<input type="text" class="leader" name="name" placeholder="Leader">
<input type="text" class="date" name="date" placeholder="date">
<input type="text" class="time" name="time" placeholder="time">
<input type="text" class="score" name="score" placeholder="score">
<input type="hidden" class="shift" name="shift">
<input type="submit" id="btn" value="send" />
</form>
New function:
var shift = []; //create array to store zoned hours
$(function() {
$( "#hours_zoned" ).selectable();
$( "#hours_zoned" ).on('click', 'li', function() {
$(this).addClass("clicked ui-selected");
$( this ).css( "background-color", "#e74c3c" );
$( this ).selectable({ disabled: true });
var floorTime = $(this).text(); // get the value of the hour that was clicked
shift.push(floorTime); // add that hour to the floorTime
$("#add .shift").val(shift); // add shift array to hidden input in form
});
/*$('#add').on('submit', function() {
var name = $('.leader').val(); */
$('#add').submit(function() {
$.ajax({
type: 'POST',
url: 'post.php',
data: $("#add").serialize(),
success: function(msg) {
alert(msg);
}
});
return false; // you need this to prevent the other form submission
});
return false;
});
Untested, but that should get you going in the right direction.

Just wanna add to the answers given here..
You also need to check if the items selected are in array already (prevent duplicate).. so to do that you can do it like this
*code is taken from Matt answer
if($.inArray(floorTime, shift) === -1) {
shift.push(floorTime);
$("#add .shift").val(shift);
}

Try parse_str().
$array = parse_str( $_POST['data'] );

Related

how can i insert two different form using one controller method by one trigger in codeigniter?

I have try but, i want to do it with ajax on click one button, want to send two form value to table using one method bu controller in codeigniter.
Suppose these are your forms
<form name="frm1" id="frm1">
<input type="text" name="txt1" >
<input type="text" name="txt2">
</form>
<form name="frm2" id="frm2">
<input type="text" name="txt3" >
<input type="text" name="txt4">
</form>
<button type="button" id="btn">Submit</button>
Ajax Script
<script type="text/javascript">
var BASE_URL = "<?php echo base_url(); ?>";
$( document ).ready(function() {
$( "#btn" ).click(function() {
$.ajax({
type: 'POST',
url:BASE_URL + "Controller/save",
data: $('#frm1, #frm2').serialize(),
success: function (data) {
data = JSON.parse(data);
console.log(data);
}
});
});
});
</script>
By this in the save method you can get the values of the forms in controller
function save()
{
echo $this->input->post('txt1');
echo $this->input->post('txt2');
echo $this->input->post('txt3');
echo $this->input->post('txt4');
}

PHP Jquery Ajax POST call, not work

As the title says, i have try many times to get it working, but without success... the alert window show always the entire source of html part.
Where am i wrong? Please, help me.
Thanks.
PHP:
<?php
if (isset($_POST['send'])) {
$file = $_POST['fblink'];
$contents = file_get_contents($file);
echo $_POST['fblink'];
exit;
}
?>
HTML:
<html>
<head>
<script type="text/javascript" src="https://code.jquery.com/jquery-1.8.3.min.js"></script>
<script>
$(document).ready(function() {
$("input#invia").click(function(e) {
if( !confirm('Are you sure?')) {
return false;
}
var fbvideo = $("#videolink").val();
$.ajax({
type: 'POST',
data: fbvideo ,
cache: false,
//dataType: "html",
success: function(test){
alert(test);
}
});
e.preventDefault();
});
});
</script>
</head>
<div style="position:relative; margin-top:2000px;">
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input id="videolink" type="text" name="fblink" style="width:500px;">
<br>
<input id="invia" type="submit" name="send" value="Get Link!">
</form>
</div>
</html>
Your Problem is that you think, that your form fields are automatic send with ajax. But you must define each one into it.
Try this code:
<script>
$(document).ready(function() {
$("input#invia").click(function(e) {
if( !confirm('Are you sure?')) {
return false;
}
var fbvideo = $("#videolink").val();
$.ajax({
type: 'POST',
data: {
send: 1,
fblink: fbvideo
},
cache: false,
//dataType: "html",
success: function(test){
alert(test);
}
});
e.preventDefault();
});
});
</script>
Instead of define each input for itself, jQuery has the method .serialize(), with this method you can easily read all input of your form.
Look at the docs.
And maybe You use .submit() instead of click the submit button. Because the user have multiple ways the submit the form.
$("input#invia").closest('form').submit(function(e) {
You must specify the url to where you're going to send the data.
It can be manual or you can get the action attribute of your form tag.
If you need some additional as the send value, that's not as input in the form you can add it to the serialized form values with formSerializedValues += "&item" + value;' where formSerializedValues is already defined previously as formSerializedValues = <form>.serialize() (<form> is your current form).
<html>
<head>
<script type="text/javascript" src="https://code.jquery.com/jquery-1.8.3.min.js"></script>
<script>
$(document).ready(function() {
$("#invia").click(function(e) {
e.preventDefault();
if (!confirm('Are you sure?')) {
return false;
}
// Now you're getting the data in the form to send as object
let fbvideo = $("#videolink").parent().serialize();
// Better if you give it an id or a class to identify it
let formAction = $("#videolink").parent().attr('action');
// If you need any additional value that's not as input in the form
// fbvideo += '&item' + value;
$.ajax({
type: 'POST',
data: fbvideo ,
cache: false,
// dataType: "html",
// url optional in this case
// url: formAction,
success: function(test){
alert(test);
}
});
});
});
</script>
</head>
<body>
<div style="position:relative; margin-top:2000px;">
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input id="videolink" type="text" name="fblink" style="width:500px;">
<br>
<input id="invia" type="submit" name="send" value="Get Link!">
</form>
</div>
</body>

AJAX JQUERY Php form not working

issues are still there...pls help
I am unable to load external file while using AJAX jquery. I want to use Jquery ajax to pop up form then validate, enter data in mysql. but starting from a simple ajax function. kindly let me know where i am going wrong
<link rel="stylesheet" type="text/css" media="all" href="test_style.css">
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script>
$(document).ready(function(){
$("#ajax-contact-form").submit(function(){
var str = $(this).serialize();
$.ajax(
{
type: "POST",
url:"contact.php",
data: str,
success:function(result)
{
$("#div1").html(result);
}
});
});
});
</script>
</head>
<body>
<div id="contact_form">
<form id="ajax-contact-form" name="contact" action="">
<fieldset>
<label for="name" id="name_label">Name</label>
<input type="text" name="name" id="name" size="30" value="" class="text-input" />
<label class="error" for="name" id="name_error">This field is required.</label>
<INPUT class="button" type="submit" name="submit" value="Send Message">
</fieldset>
</form>
</div>
</body>
</html>
and contact.php file is
<?php
echo "Hello";
?>
You need to return false; to prevent the form from submitting and refreshing the page and check if your $("#div1") is missing.
$(document).ready(function(){
$("#ajax-contact-form").submit(function(){
var str = $(this).serialize();
$.ajax(
{
type: "POST",
url:"contact.php",
data: str,
success:function(result)
{
$("#div1").html(result);
}
});
return false;
});
});
Simple. Since you are posting your form through ajax, you must prevent the default form submit by returning a false inside the submit method. Below is the correct version:
<script>
$(document).ready(function(){
$("#ajax-contact-form").submit(function(){
var str = $(this).serialize();
$.ajax({
type: "POST",
url:"contact.php",
data: str,
success:function(result) {
$("#div1").html(result);
}
});
return false;
});
});
</script>
You can use a more simple form of post request as follows:
$.post("url",{var1: value1, var2: value2},function(data,status){
if(status=='success')
alert(data);
});
the second argument you can pass as many using this post request. The first argument url, if ofcourse relative to the document in which this js is loaded or you can give the exact url on the server.
According to your php file, data=='Hello'.
Similar is the procedure for any GET request also.
Make sure you are missing the div1
please use
<div id="div1"><div>

Submit an AJAX FORM from within an AJAX Success

I need to perform another AJAX Form Post from within the first forms success function.
Example, this does 2 AJAX requests.
Search Movie => Pick Movie Wanted => View Specific Movie Details
I am able to load the results into a div <div id="results"></div> just fine but once I select a movie title it isnt performing another AJAX Request, the request goes to the main window.
Here is the initial search page that handles the results.
<script type="text/javascript">
$(document).ready(function(){
$("#searchtitle").submit(function() {
var id = $(this).children('input[name="thetitle"]').attr('value');
$.ajax({
type: "POST",
url: "s.php",
data: $('#searchtitle').serialize(),
cache: false,
success: function(data){
$('#status').html(data);
}
});
return false;
});
});
</script>
<form id="searchtitle">
<input type="text" name="thetitle" />
<input type="submit" name="submit" class="button expand postfix" value="Search" />
</form>
<div id="status"></div>
s.php which returns results within #results
<?php
if(empty($_POST['thetitle'])) {
?>
<div class="alert-box error">
<div class="alert-error"></div>
Error: Nothing Found
</div>
<?php
}
if(!empty($_POST['thetitle'])) {
$myid = strtoupper($_POST['thetitle']);
$searchReults = $tmdb_V3->searchMovie($myid,'en');
?>
<?php
foreach($searchReults['results'] as $result) {
?>
<form class="sform">
<input type="hidden" name="mid" value="<?php echo $result['id']); ?>" />
<h5><?php echo $result['title']; ?></h5>
<span class="mreleased">Year: <?php echo $result['year']; ?></span>
<input type="submit" class="button" value="Select">
</form>
<?php
}
}
?>
This is the code that will post the results from s.php
<script type="text/javascript">
$(".sform").submit(function () {
$.ajax({
type: 'POST',
url: 'sx.php',
data: $(this).closest("form").serialize();
success: function (response) {
$('#status').load(response);
$('#status').find('script').each(function (i) {
eval($(this).text());
});
}
});
return false;
}
</script>
I have tried putting this within s.php, within the bottom of the initial search page, in the head of the initial page and no luck, it submits fine just not the sx.php where it should.
In s.php the statement:
<input type="hidden" name="mid" value="<?php echo $result['id']); ?>" />
Should be:
<input type="hidden" name="mid" value="<?php echo $result['id']; ?>" /> //remove extra bracket
In your javascript code in s.php there are some typos:
data: $(this).closest("form").serialize(); // here should be comma not semicolon
After return false you should close the script properly } should be });.
And since you are trying to submit the dynamic content $(".sform").submit(function () will not work. You should use on for dynamic contents. So the correct script would be:
<script type="text/javascript">
$(document).on('submit', '.sform', function () {
$.ajax({
type: 'POST',
url: 'sx.php',
data: $(this).closest("form").serialize(),
success: function (response) {
$('#status').load(response);
$('#status').find('script').each(function (i) {
eval($(this).text());
});
}
});
return false;
});
</script>
I have checked and verified in my localhost (with a simple setup). It is making both ajax request. Hope this helps!

onclick form send via ajax no page refresh

I've been racking my brains for days looking at examples and trying out different things to try and get my form to submit with Ajax without a page refresh. And Its not even sending the data now.. I don't know what I'm doing wrong..Can someone run through my ajax and form please.
Toid is the users id and newmsg is the text in which the user submits. The two values get sent to the insert.php page.
I would really appreate the help. I'm new to Ajax, and I look at some of it and don't have a clue. If I finally got it working, It may help me realize what I've done wrong. I am looking up tutorials and watching videos..but it can be very time consuming for something that would be simple to someone in the know on here. It maybe that I've got the wrong idea on the ajax and it makes no sense at all, sorry about that.
<script type="text/javascript">
$(document).ready(function(){
$("form#myform").submit(function() {
homestatus()
event.preventDefault();
var toid = $("#toid").attr("toid");
var content = $("#newmsg").attr("content");
$.ajax({
type: "POST",
url: "insert.php",
data: "toid="+content+"&newmsg="+ newmsg,
success: function(){
}
});
});
return false;
});
</script>
<form id="myform" method="POST" class="form_statusinput">
<input type="hidden" name="toid" id="toid" value="<?php echo $user1_id ?>">
<input class="input" name="newmsg" id="newmsg" placeholder="Say something" autocomplete="off">
<div id="button_block">
<input type="submit" id="button" value="Feed" onsubmit="homestatus(); return false" >
</div>
</form>
INSERT.PHP
$user1_id=$_SESSION['id'];
if(isset($_POST['toid'])){
if($_POST['toid']==""){$_POST['toid']=$_SESSION['id'];}
if(isset($_POST['newmsg'])&isset($_POST['toid'])){
if($_POST['toid']==$_SESSION['id']){
rawfeeds_user_core::create_streamitem("1",$_SESSION['id'],$_POST['newmsg'],"1",$_POST['toid']);
}else{
rawfeeds_user_core::create_streamitem("3",$_SESSION['id'],$_POST['newmsg'],"1",$_POST['toid']);
Try using firebug to identify bugs in your code. It's a really good companion for developing javascript. Nearly all of your bugs led to error messages in the firebug console.
You had several errors in your code, here is the corrected version:
$(document).ready(function(){
$("form#myform").submit(function(event) {
event.preventDefault();
var toid = $("#toid").val();
var newmsg = $("#newmsg").val();
$.ajax({
type: "POST",
url: "insert.php",
data: "toid=" + content + "&newmsg=" + newmsg,
success: function(){alert('success');}
});
});
});
And here the corrected html:
<form id="myform" method="POST" class="form_statusinput">
<input type="hidden" name="toid" id="toid" value="<?php echo $user1_id; ?>">
<input class="input" name="newmsg" id="newmsg" placeholder="Say something" autocomplete="off">
<div id="button_block">
<input type="submit" id="button" value="Feed">
</div>
</form>
Actually onsubmit event has to be used with form so instead of
<input type="submit" id="button" value="Feed" onsubmit="homestatus(); return false" >
it could be
<form id="myform" method="POST" class="form_statusinput" onsubmit="homestatus();">
and return the true or false from the function/handler, i.e.
function homestatus()
{
//...
if(condition==true) return true;
else return false;
}
Since you are using jQuery it's better to use as follows
$("form#myform").on('submit', function(event){
event.preventDefault();
var toid = $("#toid").val(); // get value
var content = $("#newmsg").val(); // get value
$.ajax({
type: "POST",
url: "insert.php",
data: "toid=" + toid + "&newmsg=" + content,
success: function(data){
// do something with data
}
});
});
In this case your form should be as follows
<form id="myform" method="POST" class="form_statusinput">
...
</form>
and input fields should have a valid type and value attribute, Html form and Input.
I think you should read more about jQuery.
Reference : jQuery val and jQuery Ajax.
change the form to this
<form id="myform" ... onsubmit="homestatus(); return false">
you don't need the onsubmit attribute on the submit button, but on the form element instead
homestatus might be out of scope
function homestatus () {
var toid = $("#toid").attr("toid");
var content = $("#newmsg").attr("content");
$.ajax({
type: "POST",
url: "insert.php",
data: "toid="+content+"&newmsg="+ newmsg,
success: function(){
}
});
}
This isn't tested, but try this (I annotated some stuff using comments)
<script type="text/javascript">
$(document).ready(function(){
$("form#myform").submit(function(event) {
// not sure what this does, so let's take it out of the equation for now, it may be causing errors
//homestatus()
// needed to declare event as a param to the callback function
event.preventDefault();
// I think you want the value of these fields
var toid = $("#toid").val();
var content = $("#newmsg").val();
$.ajax({
type: "POST",
url: "insert.php",
data: "toid="+toid +"&newmsg="+ content,
success: function(){
}
});
return false;
});
});
</script>
<form id="myform" method="POST" class="form_statusinput">
<input type="hidden" name="toid" id="toid" value="<?php echo $user1_id ?>">
<input class="input" name="newmsg" id="newmsg" placeholder="Say something" autocomplete="off">
<div id="button_block">
<input type="submit" id="button" value="Feed" / >
</div>
</form>
It's a lot easier to let .serialize() do the work of serializing the form data.
The submit handler also needs event as a formal parameter, otherwise an error will be thrown (event will be undefined).
With a few other changes, here is the whole thing:
<script type="text/javascript">
$(document).ready(function(){
$("form#myform").submit(function(event) {
event.preventDefault();
homestatus();
var formData = $(this).serialize();
$.ajax({
type: "POST",
url: "insert.php",
data: formData,
success: function(data) {
//...
}
});
});
});
</script>
<form id="myform" class="form_statusinput">
<input type="hidden" name="toid" id="toid" value="<?php echo $user1_id ?>">
<input class="input" name="newmsg" id="newmsg" placeholder="Say something" autocomplete="off">
<div id="button_block">
<input type="submit" id="button" value="Feed" >
</div>
</form>
Unless you are omitting some of your code, the problem is this line:
homestatus()
You never defined this function, so the submit throws an error.
You may want to take a look at jQuery (www.jquery.com) or another js framework.
Such frameworks do most of the stuff you normally have to do by hand.
There are also a bunch of nice helper functions for sending form data or modifying html elements

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