I am trying to prevent a button to be clicked multiple times to avoid resending requests. The disabling works but my data to be sent or updated is not executed.
<form class="detail_form" method="POST" action="{{ url('update', $id) }}" enctype="multipart/form-data">
#csrf
<button class="btn btn-update accntfrm_btn" type="submit" id="btn-update">Update</button>
</form>
$("#btn-update").on('click', function (event) {
event.preventDefault();
$(this).prop('disabled', true);
setTimeout(function(){el.prop('disabled', false); }, 3000);
});
How can I execute my updates and disallow the multiple clicks at the same time?
Use like this in action attribute of Form,
{{ route('update', ['id'=>$id]) }}
I guess it is your route,
Route::post('/update/{id}','YourController#your_function')->name('update');
and in your controller,
public function your_function(Request $request, $id){ // your code }
and if you want to go pure laravel,
use Form class
{!! Form::open(['route' => ['update', 'id'=>$id], 'files' => true, 'class' => 'detail_form']) !!}
event.preventDefault();
prevents default action of the form, this means that your form is not going to submit to the server. what you can do is use ajax or maybe axios if you have it installed to send your information to the server. Since you obviously have jquery, you can make an ajax request to your server to update your information like so
`const route = "{{ route('update', $id)}}";`
or
const route = "/route/to/your/server";
`$.post(route,
{//add a body if you need to send some information to the server
//it is optional},
function(data, status){// in this callback you can create a feedback
//from a successful trip to and from the server
//for your users
})`
.fail(function(error){
//in this callback you can handle errors from a failed trip
//to and from the server
});
Related
I've been searching for similar question in a while but I couldn't find what would actually help my issue.
Using Laravel 5.4.
So I have a resource controller and its index method that returns a view with some data attached to it.
Then I want to make an ajax request from the view returned which is a search request.
e.preventDefault();
let q = $('#inputserver').val();
$.ajaxSetup({
headers: {
'X-CSRF-TOKEN': $('meta[name="csrf-token"]').attr('content')
}
});
$.ajax({
url: "/servers",
type: 'GET',
data: {'data': q},
success: function(response){
console.log('Successo');
}
})
That, for how a resource controller's methods are structured should invoke the index method, in which I want to identify if I have an Ajax request incoming.
If I do, I'll search with a query in an Eloquent Model for the data retrieved by the search form and of course I want to show only the matching results.
This is my controller code:
if(!$request->ajax()){
$colonna = 'id';
$servers = Server::orderBy($colonna, 'desc')->paginate(10);
return view('servers.index', array('servers' => $servers));
}
else{
$servers= Server::where('name', '=', $request->data)->paginate(10);
return view('servers.index', array('servers' => $servers));
}
The issue is that nothing is happening, so the ajax request isn't even considered, can someone help me with this? I'm almost sure the issue is some obvious things I forgot or didn't consider.
Thank you in advance, I'll edit if you would need some more info about it.
EDIT:
This is the route I have Route::resource('servers', 'ServerController');
EDIT2:
I'm sorry ids are in Italian, but I of course select them correctly when using jQuery.
<div class="input-group mb-2 mr-sm-2 mb-sm-0">
<div class="input-group-addon">
<span>
<i class="fa fa-search"></i>
</span>
</div>
{{Form::text('search', null, array('class' => 'form-control', 'id' => 'inputserver' , 'placeholder' => 'Cerca..'))}}
<span class="input-group-btn">
<button class="btn btn-default" type="button" id="cercaserver">Go!</button>
</span>
The blade file is messy.Try to create a form open and form close and make the button submit of type. And try to change your ajax to this:
$(document).ready(function() {
$('#cercaserver').on('submit', function (e) {
e.preventDefault();
var input = $('#inputserver').val();
$.ajax({
type: "GET",
url: './servers',
data: {input: input},
});
});
});
make sure you are loading jquery.
What do you mean by nothing's happening? What was shown in the console when the ajax request was fired?
Also, you're returning a view, you might want to return a json array of your results?
return $servers;
Laravel will automagically convert it into a JSON response
https://laravel.com/docs/5.4/responses#creating-responses
Or if you want to be specific:
return response()->json($servers);
https://laravel.com/docs/5.4/responses#json-responses
Edit:
I think I already know the problem, in your resource controller function index, is there a parameter called $request? It might be non existing and for sure will throw a 500 internal server error because you used it in your condition.
Hi there once again SO community. I've been developing a site and so far it's going pretty well. But today after a long day searching for a solution I can't understand nor find what the right path is...
I want to click on a button and a profile page where you can edit the fields appear. I can redirect to the page I want but I don't know how to send the user data so I can populate the fields.
Here is my button code on my view
<button class="btn btn-xs btn-warning dropdown-toggle" type="button" data-toggle="dropdown" aria-expanded="false" style="border-color: black" id="dados_{{ $user->username }}"> Alterar Dados Pessoais
<i class="glyphicon glyphicon-cog"></i>
</button>
Here is the button AJAX request handler
if((this.id).indexOf("dados") != -1){
var content = this.id.replace("dados_", "");
$.get('callPermissions', {usernameSend:content, '_token': $('meta[name=csrf-token]').attr('content'),}, function(data){
window.location.replace('settings');
});
And here is my callPermission Controller
public function callPermissions(Request $request)
{
if($request->ajax()){
$usernames = Input::get('usernameSend');
if(isset($usernames)){
$user = User::Where('username', '=', $usernames)->first();
$returnHTML = view('userOptions.settings')->render();
return view('userOptions.settings');
}else{
Log::warning("Username não existe na base de dados.");
}
}
}
and here my Settings Controller
public function settings(Request $request)
{
return view('userOptions.settings');
}
And here is the route
Route::get('/callPermissions', 'SidebarController#callPermissions');
I know the controller is wrong and from what I've read I should verify if the AJAX is successful and if it is handle i on the AJAX request. But from what I've understand I'm not using the Controller at all (even though it goes there). How should I send the user information (in this case the username, then I can get everything from the database) and then send it to the view? I've been searching and trying out stuff that doesn't work...since the return view("your_view") on the Controller doesn't work.
Sorry if I've been confusing and if you need additional information feel free to ask!
Thanks for your help!!
Edit: If I return this on the controller
return view('userOptions.settings', compact('user'));
and do a replace with the Ajax request as show above and add this to the settings view
<p> {{ $user->name }} </p>
I get the following error Undefined variable: user (View: C:\wamp64\www\siteXL\ideiasxl\resources\views\userOptions\settings.blade.php)
Is there anyway to send the parameters with a compact alike or I need to send it through the link? Was avoiding to show the username on the url.
Edit2: For further clarification, this works as intended
<button onclick="window.location='{{url('/settings/' . $user->username)}}'" type="button" id="dadosPessoais" class="btn btn-default">Alterar Dados Pessoais
<i class="glyphicon glyphicon-wrench"></i>
</button>
but I was trying not to send id's and usernames through the URL.
If this is not achievable it's ok, but if there's a way I can't find it, that's why I'm asking
I think you have to add a parameter in the Route and receive the data in the controller function. I'd do something like this:
Route:
Route::get('/callPermissions/{user}', 'SidebarController#callPermissions');
Controller:
public function callPermissions(Request $request, $user)
{
//get data related to $user
}
Ajax call:
$.get('callPermissions/'+userIdVariable, {usernameSend:content, '_token': $('meta[name=csrf-token]').attr('content'),}, function(data){
window.location.replace('settings');
});
This would send the user id through the route.
To get the user id with JavaScript, you can make a hidden field in the Blade file and set the user id as the value. For example, if you using Form helper:
{{ Form::hidden('user_id', $user->id, array('id' => 'js-user-id')) }}
And then, in the JavaScript, you can get the value using something like this:
var userIdVariable = $('#js-user-id')->val();
I have a global search form that submits to search action of a controller:
<?=Html::beginForm(['/feqh/search'], 'get', ['class' => 'navbar-form navbar-left', 'role' => 'search', 'id' => 'searchForm']);?>
<div class="form-group has-feedback Right">
<input id="q" type="text" class="form-control" placeholder="<?=yii::t('app','Search');?>" name="q" value="<?= Html::encode(\Yii::$app->getRequest()->getQueryParam('q',""));?>" />
<i class="form-control-feedback glyphicon glyphicon-search"></i>
</div>
<button type="submit" class="btn btn-default"><?=yii::t('app','Submit');?> <i class="glyphicon glyphicon-ok"></i></button>
</form>
I decided to make pretty URL for search through rules as following:
'search/<q:\w+>' => 'feqh/search',
However, submitting the form always generate the following URL:
example.com/feqh/search?q=anySearchString
However, example.com/search/anySearchString is accessible. Here the problem with submitting using the form.
I tried to change the form action URL:
<?=Html::beginForm(['feqh/search'] i.e removing the initial / but It does not make any difference.
By the way, the following rule is working too:
'search' => 'feqh/search', it makes example.com/search?q=anySearchString. However, the applying of this rule preventexample.com/search/anySearchString`
This has nothing to do with your pretty URL configuration (and not even Yii)... It's a browser thing. It only knows how to submit a form is posted as either a GET or a POST.
So since you are posting in GET mode it will simply add the inputs as query parameters to your URL.
If you want the URL in the address bar to represent your pretty URL you'll have to take control over the submit and perhaps issue a redirect instead?
$('#searchForm').submit(function() {
window.location = $(this).attr("action") + '/' + $('#q').val();
return false;
});
It's the only way I can think of right now.
You can try something like:
'search/<q:w>' => 'feqh/search/variable_name/<q>'
Then in your
actionSearch()
Do something like
$query = isset($_REQUEST['variable_name']) ? $_REQUEST['variable_name'] : '';
Or u can try to make redirect action next to your search action and change the search form so it leads to redirect.
Put this into common/main.php rules(advanced app):
'controller/action/<param:[\w-]+>/<page:[\d]+>' => 'controller/action',
'controller/action/<param>' => 'controller/action',
You need to change "controller","action", and "param" into your controller action and parameter.
This is mainly for search problem i encountered so i posted it here in hope it helps someone.
I have a subscribe box :
My goal is when the user enter the email, I want to save it to my database.
I already know how to achieve this using form post via Laravel.
Form POST via Laravel
public function postSubscribe() {
// Validation
$validator = Validator::make( Input::only('subscribe_email'),
array(
'subscribe_email' => 'email|unique:subscribes,email',
)
);
if ($validator->fails()) {
return Redirect::to('/#footer')
->with('subscribe_error','This email is already subscribed to us.')
->withErrors($validator)->withInput();
}else{
$subscribe = new Subscribe;
$subscribe->email = Input::get('subscribe_email');
$subscribe->save();
return Redirect::to('/thank-you');
}
}
Now, I want to achieve this using Ajax call to avoid the page load and also to learn more about Ajax. Here are what I've tried :
Form
{!! Form::open(array('url' => '/subscribe', 'class' => 'subscribe-form', 'role' =>'form')) !!}
<div class="form-group col-lg-7 col-md-8 col-sm-8 col-lg-offset-1">
<label class="sr-only" for="mce-EMAIL">Email address</label>
<input type="email" name="subscribe_email" class="form-control" id="mce-EMAIL" placeholder="Enter email" required>
<!-- real people should not fill this in and expect good things - do not remove this or risk form bot signups-->
<div style="position: absolute; left: -5000px;"><input type="text" name="b_168a366a98d3248fbc35c0b67_73d49e0d23" value=""></div>
</div>
<div class="form-group col-lg-3 col-md-4 col-sm-4"><input type="submit" value="Subscribe" name="subscribe" id="subscribe" class="btn btn-primary btn-block"></div>
{!! Form::close() !!}
Ajax Call
<script type="text/javascript">
$(document).ready(function(){
$('#subscribe').click(function(){
$.ajax({
url: '/subscribe',
type: "post",
data: {'subscribe_email':$('input[name=subscribe_email]').val(), '_token': $('input[name=_token]').val()},
dataType: 'JSON',
success: function (data) {
console.log(data);
});
});
});
</script>
Controller
public function postSubscribeAjax() {
// Getting all post data
if(Request::ajax()) {
$data = Input::all();
die;
}
dd($data);
}
Route
Route::post('/subscribe','SubscribeController#postSubscribeAjax');
Result
I keep getting:
Undefined variable: data
Which mean my Request::ajax() is not containing anything? Why is that?
How can I achieve this using an Ajax call?
This was resolved in Chat
We had to fix a couple syntax errors and change the input button to a html button so that it would prevent the form from submitting. There are other ways we could have done this, but this worked.
So I'm going to address a few things here. Starting from bottom moving on up!!
Route
You can do this without the slash.
Route::post('subscribe','SubscribeController#postSubscribeAjax');
Controller
Your Controller is good enough, but you need to actually display the result from the ajax. We will handle that.
Remove the die in your check if there is an Ajax request.
Ajax
<script type="text/javascript">
$(document).ready(function(){
$('#subscribe').click(function(e){
e.preventDefault(); // this prevents the form from submitting
$.ajax({
url: '/subscribe',
type: "post",
data: {'subscribe_email':$('input[name=subscribe_email]').val(), '_token': $('input[name=_token]').val()},
dataType: 'JSON',
success: function (data) {
console.log(data); // this is good
}
});
});
});
</script>
Form
Pretty sure the form is good.
Assuming the "Undefined variable" error is happening in your postSubscribeAjax method, it looks like you have a problem with scope. Specifically, when the PHP interpreter gets to the line dd($data), the variable $data has not been defined yet.
This is not surprising, since $data is assigned inside the block of code following an if statement - it is not available outside that block.
You can solve this problem by adding $data = null; before the if statement, BUT... this is only going to show you that you have a different problem. Here's why:
You're getting an "undefined variable" error. The means that your code is skipping the if block (how do I know this? because if it didn't skip it, it would execute the die statement and you'd never see the error). This means that your code is not correctly identifying the request as an AJAX request.
What I would do is this: skip the if block entirely. Simply get your data and process it.
It's been a while since I looked at this stuff, but IIRC, there is no reliable way to detect AJAX requests. To the server, an AJAX request is just like any other request. Some Javascript libraries (jQuery included, if I'm not mistaken) set a header on the request to identify it as AJAX, but it is neither universally required nor universally detected.
The only benefit to this AJAX detection is to prevent your users from accessing the AJAX URL directly from a browser. And, ultimately, if someone really wants to do that, you can't stop them since they can (with a little work) spoof any request.
When I've made multistep forms in the past I would generally store the form data in the session before returning it to the view, that way the data persists if the user refreshes the page or clicks the browser's native back buttons.
Transferring my past logic to Laravel I built the following form consisting of three stages:
[Input -> Confirm -> Success]
Routes.php
Route::group(array('prefix' => 'account'), function(){
Route::get('register', array(
'before' => 'guest',
'as' => 'account-create',
'uses' => 'AccountController#getCreate'
));
Route::post('register', array(
'before' => 'guest|csrf',
'as' => 'account-create-post',
'uses' => 'AccountController#postCreate'
));
Route::get('register/confirm', array(
'before' => 'guest',
'as' => 'account-create-confirm',
'uses' => 'AccountController#getCreateConfirm'
));
Route::post('register/confirm', array(
'before' => 'guest|csrf',
'as' => 'account-create-confirm-post',
'uses' => 'AccountController#postCreateConfirm'
));
Route::get('register/complete', array(
'before' => 'guest',
'as' => 'account-create-complete',
'uses' => 'AccountController#getCreateComplete'
));
});
AccountController.php
<?php
class AccountController extends BaseController {
private $form_session = 'register_form';
public function getCreate()
{
if(Session::has($this->form_session))
{
// get forms session data
$data = Session::get($this->form_session);
// clear forms session data
Session::forget($this->form_session);
// load the form view /w the session data as input
return View::make('account.create')->with('input',$data);
}
return View::make('account.create');
}
public function postCreate()
{
// set the form input to the session
Session::set($this->form_session, Input::all());
$validation_rules = array(
'email' => 'required|max:50|email|unique:users',
'password' => 'required|max:60|min:6',
'password_conf' => 'required|max:60|same:password'
);
$validator = Validator::make(Input::all(), $validation_rules);
// get forms session data
$data = Session::get($this->form_session);
// Return back to form w/ validation errors & session data as input
if($validator->fails()) {
return Redirect::back()->withErrors($validator);
}
// redirect to the confirm step
return Redirect::route('account-create-confirm');
}
public function getCreateConfirm()
{
// prevent access without filling out step1
if(!Session::has($this->form_session)) {
return Redirect::route('account-create');
}
// get forms session data
$data = Session::get($this->form_session);
// retun the confirm view w/ session data as input
return View::make('account.create-confirm')->with('input', $data);
}
public function postCreateConfirm()
{
$data = Session::get($this->form_session);
// insert into DB
// send emails
// etc.
// clear forms session data
Session::forget($this->form_session);
// redirect to the complete/success step
return Redirect::route('account-create-complete');
}
public function getCreateComplete() {
return View::make('account.create-complete');
}
}
create.blade.php
<form action="{{ URL::route('account-create-post') }}" method="post">
Email: <input type="text" name="email" value="{{ (isset($input['email'])) ? e($input['email']) : '' }}">
#if($errors->has('email'))
{{ $errors->first('email') }}
#endif
<br />
Password: <input type="text" name="password" value="">
#if($errors->has('password'))
{{ $errors->first('password') }}
#endif
<br />
Password Confirm: <input type="text" name="password_conf" value="">
#if($errors->has('password_conf'))
{{ $errors->first('password_conf') }}
#endif
<br />
{{ Form::token() }}
<input type="submit" value="Confirm">
</form>
create-confirm.blade.php
Email: {{ $input['email']; }}
Password: {{ $input['password']; }}
<form action="{{ URL::route('account-create-confirm-post') }}" method="post">
{{ Form::token() }}
return
<input type="submit" name="submit_forward" value="Submit">
</form>
The above works fine, however I am wondering if this is the best way to approach multi-step forms in Laravel?
When I have created multi-part forms, I have always done it in a way so that the user can always come back and finish the form later, by making each form persist what it has to the database.
For instance
Step 1 - Account Creation
I would have the user create their authentication details at this step, create the user account (with password) here and also log the user in, redirecting to the dashboard. There I can do a check to see if the user has a profile and if they don't, redirect them to the profile creation form.
Step 2 - Profile Creation
Because we have an authenticated user, the profile creation form can save its data to the currently logged in user. Subsequent sections follow the same process but check the existence of the previous step.
Your question seems to be about confirming whether a user wishes to create an account. What I would do in your situation would be, on the form you created to confirm the user account, I would keep the user's data in hidden input fields.
Email: {{ $input['email'] }}
Password: {{ $input['password'] }}
<form action="{{ URL::route('account-create-confirm-post') }}" method="post">
<input type="hidden" name="email" value="{{ $input['email'] }}">
<input type="hidden" name="password" value="{{ $input['password'] }}">
{{ Form::token() }}
return
<input type="submit" name="submit_forward" value="Submit">
</form>
Although displaying the user's chosen password back to them on this page seems to be a bit superfluous when you ask them to confirm their password on the previous page, plus some users might question why their password is being shown in plaintext on the screen, especially if they are accessing the site from a public computer.
The third option I would suggest would be to create the user account and soft-delete it (Laravel 4.2 Docs / Laravel 5 Docs), returning the user's account number to the new form:
Email: {{ $input['email'] }}
Password: {{ $input['password'] }}
<form action="{{ URL::route('account-create-confirm-post') }}" method="post">
<input type="hidden" name="id" value="{{ $user_id }}">
{{ Form::token() }}
return
<input type="submit" name="submit_forward" value="Submit">
</form>
then undo the soft-delete when the user confirms their account. This has the added bonus that you could track people trying to sign up multiple times for an account and not completing the process and see if there's a problem with your UX.
Conclusion
Of course, you could also still do it the way you always have with a session, all I have tried to do here is show you some other ways you can approach it, as with everything to do with the best way of doing something, this is a highly opinionated subject and is likely to get many opposing views on how it should be done. The best way to do it is the way that works best for you and your users... mainly your users.
There are two ways to do it (that i can think of). I prefer second one.
Client side - everything can be handled by javascript. Basic validation (if field is email, if field has enough characters etc.) would be checked with javascript. After confirmation, AJAX request would go through server side validation and if anything went wrong you could highlight invalid inputs. "check if email is available" button (via AJAX) would be great too.
Server side - pretty much what you did but I would move it to service - it would make it much cleaner.
public function getCreate() {
if ($this->formRememberService->hasData()) {
return View::make('account.create')
->with('input', $this->formRememberService->getData());
}
return View::make('account.create');
}
public function postCreate() {
$this->formRememberService->saveData(Input::all());
// ...
}
public function postCreateConfirm() {
// ...
$this->formRememberService->clear();
return Redirect::route('account-create-complete');
}
Adding "forget me" action would be nice (especially if form requires more private data).
Why getCreate() has Session::forget()? If someone goes back to change something and accidently leaves your site his data will be lost.
1st) Create a custom hidden field in the form containing a random md5 character set to submit it with the form... (it can be the timestamp, the user ip address, and country concatenated together as 3 md5 strings separated by whatever character , or #, so it can be working as a token of the form)
2nd) pass the hidden field into your controller and validate it after getting the user input from the form by generating the same values in your controller, encrypting these values as md5 too, then concatenate them all together, and compare the values that is coming from the user input form with the values you are generating in your controller.
3rd) Put the values of the form in your controller in a session then regenerate the session id every visit to every view the user is going to visit.
4th) update the timestamp in your session according the timestamp the user is visiting every page.
Just because you know Laravel, does not mean you have to do everything in Laravel.
Multi-step forms should never involve server-side magic. The best and easiest you can do is to hide certain steps with display:none; and switch to the next step using javascript toggling visibilities only.