Ajax Request on jQuery Mobile using with Codeginiter - php

Currently Im trying to do ajax request using jQuery Mobile. I`m trying to use example on http://www.giantflyingsaucer.com/blog/?p=2574. Yes, it works perfectly. But problems was coming when i tried to use the code inside codeigniter.
Here is my controller :
<?php
class Ajax extends CI_Controller {
function __construct() {
}
function index() {
?>
<!DOCTYPE html>
<html>
<head>
<title>Submit a form via AJAX</title>
<link rel="stylesheet" href="http://code.jquery.com/mobile/1.0a4/jquery.mobile-1.0a4.min.css" />
<script src="http://code.jquery.com/jquery-1.5.2.min.js"></script>
<script src="http://code.jquery.com/mobile/1.0a4/jquery.mobile-1.0a4.min.js"></script>
</head>
<body>
<script>
function onSuccess(data, status)
{
data = $.trim(data);
$("#notification").text(data);
}
function onError(data, status)
{
// handle an error
}
$(document).ready(function() {
$("#submit").click(function(){
var formData = $("#callAjaxForm").serialize();
$.ajax({
type: "POST",
url: "callajax",
cache: false,
data: formData,
success: onSuccess,
error: onError
});
return false;
});
});
</script>
<!-- call ajax page -->
<div data-role="page" id="callAjaxPage">
<div data-role="header">
<h1>Call Ajax</h1>
</div>
<div data-role="content">
<form id="callAjaxForm">
<div data-role="fieldcontain">
<label for="firstName">First Name</label>
<input type="text" name="firstName" id="firstName" value="" />
<label for="lastName">Last Name</label>
<input type="text" name="lastName" id="lastName" value="" />
<h3 id="notification"></h3>
<button data-theme="b" id="submit" type="submit">Submit</button>
</div>
</form>
</div>
<div data-role="footer">
<h1>GiantFlyingSaucer</h1>
</div>
</div>
</body>
</html>
<?php
}
function callajax() {
echo "ok";
}
}
I want to display 'ok' in notification. Im trying to call function callajax instead of call from callajax.php is this possible or do I need to create callajax.php?
Ive tried to create callajax.php but it was failed.
Thank you,

You probably want the url to be:
url: "ajax/callajax",
As you are calling the callajax function within the controller ajax.

Related

Posting a Laravel AJAX request passes 0 arguments

I'm having trouble passing arguments through a Laravel AJAX request. There are a lot of similar questions asked, but none of them seem to address the problem I have.
Here is are my Laravel routes:
Route::get('workerAjax', function(){
return View::make('workerAjax');
});
Route::post('/workerAjax/post', 'WorkerController#storeWorker');
WorkerController.php
<?php
class WorkerController extends \BaseController {
public function storeWorker(Request $request){
return 'I am in';
}
}
workerAjax.blade.php
<!doctype html>
<html lang="{{ app()->getLocale() }}">
<head>
<meta charset="utf-8"/>
<meta http-equiv="X-UA-Compatible" content="IE=edge"/>
<meta name="viewport" content="width=device-width, initial-scale=1"/>
<meta name="_token" content="{{ csrf_token() }}" />
<title>Register worker</title>
</head>
<body>
<div class="container">
<div class="alert alert-success" style="display:none"></div>
<form id="myForm">
<!-- don't think I need this, but it doesn't work with or without -->
<input type="hidden" name="_token" value="{{ csrf_token() }}">
<div class="form-group">
<label for="firstName">First Name:</label>
<input type="text" id="firstName">
</div>
<div class="form-group">
<label for="lastName">Last Name:</label>
<input type="text" id="lastName">
</div>
<div class="form-group">
<label for="email">Email:</label>
<input type="text" id="email">
</div>
<div class="form-group">
<label for="password">Password:</label>
<input type="password" id="password">
</div>
<button class="btn btn-primary" id='ajaxSubmit'>Submit</button>
</form>
</div>
<script src="https://code.jquery.com/jquery-3.3.1.min.js"
integrity="sha256-FgpCb/KJQlLNfOu91ta32o/NMZxltwRo8QtmkMRdAu8="
crossorigin="anonymous">
</script>
<script>
jQuery(document).ready(function(){
jQuery('#ajaxSubmit').click(function(e){
e.preventDefault();
$.ajaxSetup({
headers: {
'X-CSRF-TOKEN': $('meta[name="_token"]').attr('content')
}
});
jQuery.ajax({
url: "{{ url('/workerAjax/post') }}",
method: 'post',
data: {
firstName: jQuery('#firstName').val(),
lastName: jQuery('#lastName').val(),
email: jQuery('#email').val(),
password: jQuery('#password').val(),
},
success: function(result){
jQuery('.alert').show();
jQuery('.alert').html(result.success);
}});
});
});
</script>
</body>
</html>
</html>
After filling in the form in a browser and clicking submit, I get a POST error 500 in the console.
The preview of the network error shows this message
error: {type: "Symfony\Component\Debug\Exception\FatalThrowableError",…}
file: "/path/to/folder/app/controllers/WorkerController.php"
line: 64 ( public function storeWorker(Request $request){ )
message: "Type error: Too few arguments to function WorkerController::storeWorker(), 0 passed and exactly 1 expected"
type: "Symfony\Component\Debug\Exception\FatalThrowableError"
I feel like I've done everything right, yet it's still not working. Could someone more experienced with the framework please give me a hand?
Edit:
Here is data which is sent through the request:
XHR Request
Edit 2:
My folder structure may be a piece of the puzzle, here it is:
App folder structure
Change method key to type in ajax object.
<script>
jQuery(document).ready(function(){
jQuery('#ajaxSubmit').click(function(e){
e.preventDefault();
$.ajaxSetup({
headers: {
'X-CSRF-TOKEN': $('meta[name="_token"]').attr('content')
}
});
jQuery.ajax({
url: "{{ url('/workerAjax/post') }}",
type: 'POST',
data: {
firstName: jQuery('#firstName').val(),
lastName: jQuery('#lastName').val(),
email: jQuery('#email').val(),
password: jQuery('#password').val(),
},
success: function(result){
jQuery('.alert').show();
jQuery('.alert').html(result.success);
}});
});
});
</script>
You are missing to import the request class after the namespace declaration:
use Illuminate\Http\Request;
Without it, Laravel don't know how to inject that parameters, and don't inject anything.
So using the command ssh php artisan -V, I found out I've been using Laravel 4.2 rather than Laravel 6.15 as my plesk installation lead me to believe.
I solved this problem by calling the controller function from the route directly:
Route::post('/workerAjax/{post}', function(){
return WorkerController::storeWorker(Input::all());
});
and WorkerController.php
<?php
class WorkerController extends \BaseController {
public function storeWorker($request){
$val = json_encode($request);
// use for accessing json variables
$firstName = $request['firstName'];
// alert.(result.success) returns:
// {"firstName":"John","lastName":"Doe","email":"johndoe#email.com","password":"password1"}
return Response::json(['success'=>$val],200);
}
?>
Thanks for everyone's input and I hope this helps someone in the future!

execute jquery .submit AFTER ajax success return new form

I believe my problem has something to do with the fact that my first form RETURNS a new form via ajax success .html(result) AFTER DOM has executed. My jquery within DOM isn't being recognized because elements aren't visible until after the submit of first form. HOW to get my $("#fullFormMA").on(submit,(function(e){ to execute is eluding me. Here is my html
<?php
session_start();
require_once('functions.php');
include('header.htm');?>
<title>Membership Application</title>
<meta name="description" content="">
</head>
<body>
<div id="container">
<div id="loginBanner">
<?php include ("loginMenu.php"); ?>
<?php include ("bannerIcons.php"); ?>
</div> <!--end loginBanner-->
<div id="header" class="clear">
</div> <!--end header-->
<div id="content"><div class="content">
<div id="colLt">
<?php include('tabContent.php');?>
<?php include('leftSidebar.php');?>
</div>
<div id="colRt"><div class="content">
<h1>New Member Application</h1>
<ul><li>submitting an application</li><li>submitting payment</li></ul><h6>Step #1—the application</h6>Please enter an email which will ultimately be used as your website username. This email will remain as your private email.</p><br><br>
<form method="post" name="checkUserMA" id="checkUserMA">
<label class="clear" style="width:120px">Username/Email<br><span class="small"></span></label>
<input type="text" name="usernameMA" id="usernameMA" class="green" style="width:300px;"/><br><br>
<input type="submit" id="checkUserMA" class="submit" value="Submit" />
</form>
<div class="clear"></div>
<div id="errorMA" style="background:yellow;width:200px;height:100px"></div>
<div id="resultMA"></div>
</div></div>
<div class="clear"></div>
</div></div><!--end content-->
<div id="footer">
<?php include("footer.htm") ?>
<!--<?php include("disclaimer.htm") ?>-->
</div><!--end footer-->
<div class="clear"></div>
</div><!--end container-->
<div class="clear"></div>
</body>
</html>
Here is my jquery:
$(document).ready(function() {
$('#resultMA').hide();
$('#errorMA').hide();
$("#checkUserMA").submit(function(event){
event.preventDefault();
$("#resultMA").html('');
var values = $(this).serialize();
$.ajax({
url: "checkMA.php",
type: "post",
data: values,
success: function(result){
$("#resultMA").html(result).fadeIn();
$('.error').hide();
},
error:function(){
// alert("failure");
$("#resultMA").html('There was an error. Please try again.').fadeIn();
}
});//end ajax
});
$("#fullFormMA").on(submit,(function(e){
e.preventDefault();
$("#errorMA").html('');
var values = $(this).serialize();
$.ajax({
url: "validMA.php",
type: "post",
data: values,
success: function(result){
},
error:function(){
// alert("failure");
$("#errorMA").html('There was an error. Please try again.').fadeIn();
}
});//end ajax
});
});//end dom
Here is checkMA.php...
<?php
session_start();
include('functions.php');
connect();
$username = urldecode(protect($_POST['usernameMA']));
$_SESSION['guestUser'] = $username;
$sql2 = mysql_query("SELECT username FROM members WHERE username = '$username'");
$checkNumRows = mysql_num_rows($sql2);
if (!$username){
echo "<p class='red'>Enter an email to be used as your username...</p>";
} else if ($checkNumRows == 1){
echo "<span style='font-weight:bold'>The username: ".$username." is already in use.</span>";
} else if ($checkNumRows == 0){
echo "<hr><p class='green'>This username is available.</p><p>Please continue with the registration process...</p><br>";?>
<form method="post" name="fullFormMA" action="memberAppProcess.php">
<h6>Public Information - this information will be displayed to website visitors</h6>
<label class="clear" style="width:75px">Name</label>
<label class="error" id="name_error">This field is required.</label>
<input type="text" name="firstName" id="firstName" class="left inputCheck" style="width:150px" placeholder="first name"/>
<input type="text" name="lastName" id="lastName" class="inputCheck" style="margin-left:10px" placeholder="last name"/><br><br>
<input type="submit" name="fullFormMA" id="fullFormMA" class='submit right' onClick='submitFullForm();' value="Submit application">
</form>
<?php
}?>
My #checkUserMA works but my #fullFormMA doesn't work. I would love to understand why (DOM already loaded?) and how I might fix my code to allow for a form added "after the fact" via ajax .html(result). Thank you.
The DOM is ready before your ajax success so you can write this JQuery full code
$(document).ready(function() {
$('#resultMA').hide();
$('#errorMA').hide();
$("#checkUserMA").submit(function(event){
event.preventDefault();
$("#resultMA").html('');
var values = $(this).serialize();
$.ajax({
url: "checkMA.php",
type: "post",
data: values,
success: function(result){
$("#resultMA").html(result).fadeIn();
$('.error').hide();
RunAfterAjax();
},
error:function(){
// alert("failure");
$("#resultMA").html('There was an error. Please try again.').fadeIn();
}
});//end ajax
});
function RunAfterAjax(){
$("#fullFormMA").on(submit,(function(e){
e.preventDefault();
$("#errorMA").html('');
var values = $(this).serialize();
$.ajax({
url: "validMA.php",
type: "post",
data: values,
success: function(result){
},
error:function(){
// alert("failure");
$("#errorMA").html('There was an error. Please try again.').fadeIn();
}
});//end ajax
});
}
});//end dom
It's executing, you just aren't waiting long enough for it to exist. Move the event binding for the new form to the line right after you add the new form to the document.
$("#resultMA").html(result).fadeIn();
$("#fullFormMA").on(submit,(function(e){...
$("#fullFormMA").on(submit,(function(e){ /* ... */ });
fullFormMA is an <input>, you should bind click instead of submit, and use quotes around the event name.
When you use $('#something').on('event', ...), it only works if the #something element already exists.
You could fix your code by delegating the listener to an upper existing element :
$('#content').on('click', '#fullFormMA', function() { /* ... */ });
This code will detect the click event on #fullFormMA event if it is added after an ajax response.

Ajax/PHP login form - developing an android app(Jquery.mobile) using phonegap

My Index.html file code as follows:
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>Mindcorpus - Placement</title>
<link href="style.css" rel="stylesheet" type="text/css"/>
<script src="jquery.js" type="text/javascript"></script>
<script src="jquery-mobile.js" type="text/javascript"></script>
<script src="/cordova.js" type="text/javascript"></script>
<script>
$(function()
{
$('#frm').submit(function()
{
var username = $('#textinput').val();
var username = $.trim(username);
var password = $('#passwordinput').val();
var password = $.trim(password);
if(username=='')
{
$('.error').html('Please enter username');
return false;
}
else if(password =='')
{
$('.error').html('Please enter password');
return false;
}
else
{
var user = $('[name=username]').val();
var pass = $('[name=password]').val();
$.ajax({
type: 'POST',
url: 'http://localhost/mc-new/admin/mobile1/process.php',
data: { username: user, password: pass},
success: function(data){
alert(data.success);
},
error: function(){
alert('error!');
}
});
return false;
}
});
});
</script>
</head>
<body>
<div data-role="page" id="page">
<div data-role="header">
<h1><img src="images/logo.png" /></h1>
</div>
<div data-role="content">
<div class="error"></div>
<form action="" method="post" id="frm">
<div data-role="fieldcontain">
<input type="text" name="username" placeholder="Username" id="textinput" value="" />
</div>
<div data-role="fieldcontain">
<input type="password" name="password" placeholder="******" id="passwordinput" value="" />
</div>
<button data-icon="arrow-r" type="submit">Submit</button>
</form>
</div>
<div data-role="footer">
<h4 style="font-weight:normal">Powered By: Mind Processors</h4>
</div>
</div>
</body>
</html>
And my process.php file is:
<?php
if(isset($_POST['username']) || isset($_POST['password']))
{
$data['success'] = 'Done';
echo json_encode($data);
}
?>
This functions is properly working when I use localhost on my browser. BUt when I use Dreamweaver and build & Emulate this site. It always alert error. The Ajax is not functioning in emulator. The request is not passed to process file. Please help me & sort out this problem.
You need to specify in the ajax that the expected return data type is JSON and also cross domain has to be enabled.
Try the below code for the ajax,
$.ajax({
type: 'POST',
url: 'http://localhost/mc-new/admin/mobile1/process.php',
crossDomain: true,
data: { username: user, password: pass},
dataType: 'json',
success: function(data){
alert(data.success);
},
error: function(){
alert('error!');
}
And also add the following in your php file,
<?php
header('Access-Control-Allow-Origin: *');
header('Access-Control-Allow-Methods: POST');
header('Access-Control-Max-Age: 1000');
header('Content-Type: application/json');
if(isset($_POST['username']) || isset($_POST['password']))
{
$data['success'] = 'Done';
echo json_encode($data);
}
?>
Hope this helps to someone who falls onto this question, looking for an answer.

jQuery Mobile: How to correctly submit form data

This is a jQuery Mobile question, but it also relates to pure jQuery.
How can I post form data without page transition to the page set into form action attribute. I am building phonegap application and I don't want to directly access server side page.
I have tried few examples but each time form forwards me to the destination php file.
Intro
This example was created using jQuery Mobile 1.2. If you want to see recent example then take a look at this article or this more complex one. You will find 2 working examples explained in great details. If you have more questions ask them in the article comments section.
Form submitting is a constant jQuery Mobile problem.
There are few ways this can be achieved. I will list few of them.
Example 1 :
This is the best possible solution in case you are using phonegap application and you don't want to directly access a server side php. This is an correct solution if you want to create an phonegap iOS app.
index.html
<!DOCTYPE html>
<html>
<head>
<title>jQM Complex Demo</title>
<meta name="viewport" content="width=device-width, height=device-height, initial-scale=1.0"/>
<link rel="stylesheet" href="http://code.jquery.com/mobile/1.2.0/jquery.mobile-1.2.0.min.css" />
<style>
#login-button {
margin-top: 30px;
}
</style>
<script src="http://www.dragan-gaic.info/js/jquery-1.8.2.min.js"></script>
<script src="http://code.jquery.com/mobile/1.2.0/jquery.mobile-1.2.0.min.js"></script>
<script src="js/index.js"></script>
</head>
<body>
<div data-role="page" id="login" data-theme="b">
<div data-role="header" data-theme="a">
<h3>Login Page</h3>
</div>
<div data-role="content">
<form id="check-user" class="ui-body ui-body-a ui-corner-all" data-ajax="false">
<fieldset>
<div data-role="fieldcontain">
<label for="username">Enter your username:</label>
<input type="text" value="" name="username" id="username"/>
</div>
<div data-role="fieldcontain">
<label for="password">Enter your password:</label>
<input type="password" value="" name="password" id="password"/>
</div>
<input type="button" data-theme="b" name="submit" id="submit" value="Submit">
</fieldset>
</form>
</div>
<div data-theme="a" data-role="footer" data-position="fixed">
</div>
</div>
<div data-role="page" id="second">
<div data-theme="a" data-role="header">
<h3></h3>
</div>
<div data-role="content">
</div>
<div data-theme="a" data-role="footer" data-position="fixed">
<h3>Page footer</h3>
</div>
</div>
</body>
</html>
check.php :
<?php
//$action = $_REQUEST['action']; // We dont need action for this tutorial, but in a complex code you need a way to determine ajax action nature
//$formData = json_decode($_REQUEST['formData']); // Decode JSON object into readable PHP object
//$username = $formData->{'username'}; // Get username from object
//$password = $formData->{'password'}; // Get password from object
// Lets say everything is in order
echo "Username = ";
?>
index.js :
$(document).on('pagebeforeshow', '#login', function(){
$(document).on('click', '#submit', function() { // catch the form's submit event
if($('#username').val().length > 0 && $('#password').val().length > 0){
// Send data to server through ajax call
// action is functionality we want to call and outputJSON is our data
$.ajax({url: 'check.php',
data: {action : 'login', formData : $('#check-user').serialize()}, // Convert a form to a JSON string representation
type: 'post',
async: true,
beforeSend: function() {
// This callback function will trigger before data is sent
$.mobile.showPageLoadingMsg(true); // This will show ajax spinner
},
complete: function() {
// This callback function will trigger on data sent/received complete
$.mobile.hidePageLoadingMsg(); // This will hide ajax spinner
},
success: function (result) {
resultObject.formSubmitionResult = result;
$.mobile.changePage("#second");
},
error: function (request,error) {
// This callback function will trigger on unsuccessful action
alert('Network error has occurred please try again!');
}
});
} else {
alert('Please fill all nececery fields');
}
return false; // cancel original event to prevent form submitting
});
});
$(document).on('pagebeforeshow', '#second', function(){
$('#second [data-role="content"]').append('This is a result of form submition: ' + resultObject.formSubmitionResult);
});
var resultObject = {
formSubmitionResult : null
}
I have run into same issue where I am calling another .php page from my index.html.
The .php page was saving and retrieving data and drawing a piechart. However I found that when piechart drawing logic was added, the page will not load at all.
The culprit was the line that calls the .php page from my index.html:
<form action="store.php" method="post">
If I change this to:
<form action="store.php" method="post" data-ajax="false">
, it will work fine.
On using PHP and posting data
Use
data-ajax = "false" is the best option on <form> tag.
Problem is that JQuery Mobile uses ajax to submit the form. The simple solution to this is to disable the ajax and submit form as a normal form.
Simple solution: form action="" method="post" data-ajax="false"

Send JSON to server side using Ajax

I have a html form and I am making the json object as:
var JSobObject=
'{"name":"'+personObject.GetPersonName()+
'","about":"'+personObject.GetAbout()+
'","contact":"'+personObject.GetPersonContact()+'"}';
(Here personObject holds the form data)
trying to post it to Server.php as:
if (window.XMLHttpRequest) {
xmlhttp=new XMLHttpRequest();
}
else {
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
//var url = "organizePeople.php?people=" + escape(people.toJSONString());
xmlhttp.open("POST","ServerJSON.php?person="+JSobObject,true);
xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xmlhttp.send(JSobObject);
xmlhttp.onreadystatechange=function() {
if(xmlhttp.readyState==4) {
alert(xmlhttp.responseText);
}
}
I am not getting any response from the server.php In my server i am doing this
$person = $_POST['person'];
$objArray = json_decode($person);
print_r($objArray);
Can anyone help me that what I am doing wrong? I am in the learning stage. Just using JS/AJAX I want to prepare JSON and send it to server and get the response of the object datas.
Thanks in advance
I prefer that, you do this with jQuery. It is JavaScript based library to do things (like this) easier.
$.post({
url: "Server.php",
data: JSobObject,
dataType: "json"
}).done(function(msg) {
alert(msg);
});
You need to download jQuery - of course - to get code to working!
2 thoughts I'm having.
First of all, I'd sugest you move the xmlhttp.onreadystatechange assigment to above the send() call.
Secondly, have you checked with firebug if you're getting any reponse?
Have you tried calling your php page directly, rather than through ajax?
I'm not 100% sure, but probably you'll want something like
$data = json_decode($_POST, true);
$person = $data['person'];
The reason being you're post data is entirely a JSON encoded text string, so you'll need to decode the entire post before you can access the data separately.
You should check the rawpost from php first,
$_POST maybe fill slashes Automatedly
Use this code will help you
<!DOCTYPE html>
<html>
<head>
<title>Servlet Example</title>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js"></script>
<script src="js/jquery.serializeJSON.min.js"></script>
</head>
<body>
<div class="jumbotron text-center">
<h1>Submit form Data to Database in JSON</h1>
</div>
<div class="container">
<div class="row">
<div class="col-sm-3"></div>
<div class="col-sm-5">
<h3>Enter the Details : </h3>
<form name="myform" id="myform">
<div class="form-group">
<label for="fullName">Name:</label>
<input type="text" name="fullName" class="form-control">
</div>
<div class="form-group">
<label for="email">Email:</label>
<input type="email" name="email" class="form-control">
</div>
<div class="form-group">
<label for="subject">Subject:</label>
<input type="text" name="subject" class="form-control">
</div>
<div class="form-group">
<label for="mark">Mark:</label>
<input type="number" name="mark" class="form-control">
</form>
</div>
<button type="submit" class="btn btn-success " id="submitform">Submit</button>
</div>
<div class="col-sm-3"></div>
</div>
<script>
$(document).ready(function(){
$("#submitform").click(function(e)
{
var MyForm = JSON.stringify($("#myform").serializeJSON());
console.log(MyForm);
$.ajax(
{
url : "<your url>",
type: "POST",
data : MyForm,
});
e.preventDefault(); //STOP default action
});
});
</script>
</body>
</html>

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