How to validate Laravel 4 form (html) with js framework - laravel-4

I am developed a laravel web application with forms. My client is not satisfied with laravel form validation because its validating from server side and reload the page, its taking more time. I want to validate form from client side.. which js framework is most suitable for laravel 4. I need the laravel form validation methords like min,max,numeric,alphanumeric,alpha. With custom validation error messages.
Also please give the best tutorial link for your suggestion. ?

Have you considered making an AJAX call to the backend, making use of Laravel's form validators, as the user tabs to the next input field? That way the form is validated prior to submit.

I manage the validation from you can use any framework and any CMS try it.
p{color:red;}
<script src="http://ajax.aspnetcdn.com/ajax/jquery.validate/1.9/jquery.validate.min.js"></script>
<script type="text/javascript">
(function($,W,D)
{
var JQUERY4U = {};
JQUERY4U.UTIL =
{
setupFormValidation: function()
{
//form validation rules
$("#register-form").validate({
rules: {
roll: {required : true, minlength : 2,digits: true},
lastname: {required : true},
agree: "required"
},
messages: {
roll: {required :"<p>Please enter roll</p>", minlength : "<p>Must be at least 2 characters</p>",digits :"<p>Please enter integer Numper Only</p>" },
lastname: {required :"<p>Please enter lastname</p>" },
agree: "Please accept our policy"
}
});
}
}
//when the dom has loaded setup form validation rules
$(D).ready(function($) {
JQUERY4U.UTIL.setupFormValidation();
});
})(jQuery, window, document);
</script>
<form method="post" action="<?php echo base_url(); ?>classname/functionname/" id="register-form" novalidate="novalidate" />
<input class="form-control" type="text" name="roll" value="">
<input class="form-control" type="text" name="lastname" value="">
</form>

Related

Play framework write Action with Ok(...) that doesn't load new page

Play framework 2.4.x. A button is pressed on my home page that executes some code via Ajax, and returns its results beneath the button without loading a new page. The results wait for a user to input some text in a field and press "submit". Those results Look like this:
<li class="item">
<div>
<h3>Email: </h3>
<a>#email.tail.init</a>
<h3>Name: </h3>
<a>#name</a>
</div>
<div>
<h3>Linkedin: </h3>
<form class="linkedinForm" action="#routes.Application.createLinkedin" method="POST">
<input type="number" class="id" name="id" value="#id" readonly>
<input type="text" class="email" name="email" value="#email" />
<input type="text" class="emailsecondary" name="emailsecondary" value="" />
<input type="text" class="name" name="email" value="#name" />
<input type="text" class="linkedin" name="linkedin" value="" />
<input type="submit" value="submit" class="hideme"/>
</form>
</div>
<div>
<form action="#routes.Application.delete(id)" method="POST">
<input type="submit" value="delete" />
</form>
</div>
</li>
Along with some jquery that slides up a li after submission:
$(document).ready(function(){
$(".hideme").click(function(){
$(this).closest('li.item').slideUp();
});
});
However, since a form POST goes inside an Action that must a return an Ok(...) or Redirect(...) I can't get the page to not reload or redirect. Right now my Action looks like this (which doesn't compile):
newLinkedinForm.bindFromRequest.fold(
errors => {
Ok("didnt work" +errors)
},
linkedin => {
addLinkedin(linkedin.id, linkedin.url, linkedin.email, linkedin.emailsecondary, linkedin.name)
if (checkURL(linkedin.url)) {
linkedinParse ! Linkedin(linkedin.id, linkedin.url, linkedin.email, linkedin.emailsecondary, linkedin.name)
Ok(views.html.index)
}else{
Ok(views.html.index)
}
}
)
Is it possible to return Ok(...) without redirecting or reloading? If not how would you do a form POST while staying on the same page?
EDIT: Here is my attempt at handling form submission with jquery so far:
$(document).ready(function(){
$(".linkedinForm").submit(function( event ) {
var formData = {
'id' : $('input[name=id]').val(),
'name' : $('input[name=name]').val(),
'email' : $('input[name=email']).val(),
'emailsecondary' : $('input[name=emailsecondary]').val(),
'url' : $('input[name=url]').val()
};
jsRoutes.controllers.Application.createLinkedin.ajax({
type :'POST',
data : formData
})
.done(function(data) {
console.log(data);
});
.fail(function(data) {
console.log(data);
});
event.preventDefault();
};
});
This is an issue with the browser's behavior on form submission, not any of Play's doing. You can get around it by changing the behavior of the form when the user clicks submit.
You will first want to attach a listener to the form's submission. You can use jQuery for this. Then, in that handler, post the data yourself and call .preventDefault() on the event. Since your javascript is now in charge of the POST, you can process the data yourself and update your page's HTML rather than reloading the page.
What you need is use ajax to submit a form, check this: Submitting HTML form using Jquery AJAX
In your case, you can get the form object via var form = $(this), and then start a ajax with data from the form by form.serialize()
$.ajax({
type: form.attr('method'),
url: form.attr('action'),
data: form.serialize(),
success: function (data) {
alert('ok');
}
});
In order to accomplish this task, i had to use play's javascriptRouting
This question's answer helped a lot.
I'm not experienced with jquery so writing that correctly was difficult. For those that find this, here is my final jquery that worked:
$(document).ready(function(){
$("div#results").on("click", ".hideme", function(event) {
var $form = $(this).closest("form");
var id = $form.find("input[name='id']").val();
var name = $form.find("input[name='name']").val();
var email = $form.find("input[name='email']").val();
var emailsecondary = $form.find("input[name='emailsecondary']").val();
var url = $form.find("input[name='url']").val();
$.ajax(jsRoutes.controllers.Application.createLinkedin(id, name, email, emailsecondary, url))
.done(function(data) {
console.log(data);
$form.closest('li.item').slideUp()
})
.fail(function(data) {
console.log(data);
});
});
});
Note that my submit button was class="hideme", the div that gets filled with results from the DB was div#results and the forms were contained within li's that were class="item". So what this jquery is doing is attaching a listener to the static div that is always there:
<div id="results">
It waits for an element with class="hideme" to get clicked. When it gets clicked it grabs the data from the closest form element then sends that data to my controller via ajax. If the send is successful, it takes that form, looks for the closest li and does a .slideUp()
Hope this helps

Easiest way of converting a regular from to perform submissions via ajax?

I'm trying to convert simple forms such as:
<form action="/api.php", method="get">
... radios, checkboxes etc
<input type="submit" value="Submit">
</form>
To perform ajax submission instead of reloading the page when hitting submit using angular.
I plan on removing the submit button and replacing it with a regular button with ng-click="submit()". My submit function would look something like this:
$scope.submit = function() {
$http.get('/api', { params: ??? })
.success(...));
}
However the difficulty I have here is attaching the get params from my form inputs. I'm not sure how to reference them. Would I have to add ng-model to every single input element?
I have a lot of forms that require "converting" and I was wondering what would be the least intrusive way (least changes to markup) to do this? The reason is because a previous developer has left me with a soup of ugly html changing things will be costly.
Yes, its EASY.
html
<form action="/api.php" ng-submit="submit()">
<input type="text" name="name" ng-model="user.name">
<input type="submit" value="Submit">
</form>
js
$scope.submit = function() {
$scope.user = {};
$http({
method: 'GET',
url: '/api.php?name=' + $scope.user.name
}).
success(function(data, status, headers, config) {
console.log(JSON.stringify(data));
});
};

Automate AJAXed forms with Jquery

I want to improve my website and figured out a good way to do it was by submitting forms via AJAX. But, I have so many forms that it would be inpractical to do $('#formx').submit(). I was wondering if there was a way to do this automatically by making an universal markup like;
<form class="ajax_form" meta-submit="ajax/pagename.php">
<input type="text" name="inputx" value="x_value">
<input type="text" name="inputy" value="y_value">
</form>
And have this submit to ajax/pagename.php, where it automatically includes inputx and inputy?
This would not only save me a lot of time but also a lot of lines of code to be written.
First question so I hope it's not a stupid one :)
Something like this should work for all forms. It uses jQuery - is this available in your project? This specific code chunk hasn't been tested per say, but I use this method all the time. It is a wonderful time saver. Notice I changed meta-submit to data-submit so that its value can be fetched using $('.elemenet_class').data('submit');
HTML
<!-- NOTE: All Form items must have a unique 'name' attribute -->
<form action="javascript:void(0);" class="ajax_form" data-submit="ajax/pagename.php">
<input type="text" name="inputx" value="x_value">
<input type="text" name="inputy" value="y_value">
<input type="submit" value="go" />
</form>
JavaScript
$('.ajax_form').submit(function(e){
var path = $(this).attr('data-submit'); //Path to Action
var data = $(this).serialize(); //Form Data
$.post(path, {data:data}, function(obj){
});
return false;
})
PHP
//DEBUGGING CODE
//var_dump($_POST);
//die(null);
$data = $_POST['data'];
$inputx = $data['inputx'];
$inputy = $data['inputy'];
you can create ajax fot text boxes so that it can update to database whenever change the focus from it.
<form id="ajax_form1">
<fieldset>
<input type="text" id="inputx" value="x_value" />
<input type="text" id="inputy" value="y_value" />
</fieldset>
</form>
<script>
$(document).ready(function()
{
$("form#ajax_form1").find(":input").change(function()
{
var field_name=$(this).attr("id");
var field_val=$(this).val();
var params ={ param1:field_name, param2:field_val };
$.ajax({ url: "ajax/pagename.php",
dataType: "json",
data: params,
success: setResult
});
});
});
</script>

jQuery Mobile avoid changePage() in ajax form submit

I have a post form, I want the submit to be done in an AJAX way but I don't want the page URL to change.
For example:
<form action="/countries/change_country" data-ajax="true" method="post">
<select name="country_code">
<option value="FR" selected="selected">France</option>
<option value="DE">Germany</option>
</select>
<input type="submit" value="change country" />
</form>
When I click in the change country button the form is sent through AJAX what is nice but the page URL is changed to /countries/change_country what is not nice because this URL doesn't exist in my server which is very picky with HTTP verbs.
I know it is possible to change this default behavior for the whole application but I would like to deactivate the changePage() only for this form.
Submit the form through Javascript/jQuery instead.
First: Disable the default submit-behaviour
$('#form').on('submit', function (e) {
if (e.preventDefault) e.preventDefault();
return false;
});
Second: Serialize the form data with jQuery
var serializedFormData = $('#form').serialize();
Third: Post your form with $.ajax();
$.ajax({
url: '/countries/change_country',
type: 'POST',
data: serializedFormData
});

Ajax form perfect until inside another page

I have a page which uses ajax to submit a comment form, add it to a db, then redisplay the page, hopefully without reloading the page its on.
If I access the script on it's own it works great, yet when I load it into another page it doesn't add the data and also refreshes the page on submit, which I want to avoid, which is the whole point of doing things this way.
Anyway, here's how I load the page:
<div id="wall_comments" class="msgs_holder"></div>
<script type="text/javascript">
$('#wall_comments').load('/pages/comment.php', { wl_id:"<?=$wl_id?>" });
</script>
and then the page itself with jquery code:
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.0/jquery.min.js" type="text/javascript"></script>
<div style="width:100%; overflow:auto;">
<form method=post>
<input type="text" class="inp" name="comment" id="comment">
<input type=submit value="do it" name="action" onclick="update()">
<input type=hidden name="wl_id" value="<?=$_REQUEST[wl_id]?>" id="wl_id">
<input type=hidden name="user_id" value="<?=$userfromcookie?>" id="user_id">
</form>
</div>
<script type="text/javascript">
function update(){
var wl_idVal = $("#wl_id").val();
var commentVal = $("#comment").val();
var user_idVal = $("#user_id").val();
$.ajax({
type: "POST",
url: "/pages/comment.php",
cache: false,
data: { submit: "", wl_id: wi_idVal, comment: commentVal, user_id: user_idVal }
});
}
</script>
And finally enter info into db (I know this should be mysqli and it will be)
if(isset($_POST['action'])){
$wl_id = mysql_real_escape_string($_POST['wl_id']);
$comment = mysql_real_escape_string($_POST['comment']);
$user_id = mysql_real_escape_string($_POST['user_id']);
$addcomment = mysql_query("insert into list_wall (
event_id,
user_id,
comment
) VALUES (
'$wl_id',
'$user_id',
'$comment'
) ",$db);
if(!$addcomment) { echo 'result error add comment'; echo mysql_error(); exit; } // debug
}
The problem is when you click the submit button, the page is submitted and the function update couldn't work. You have to cancel the default submit mechanism by using return false;
<input type=submit value="do it" name="action" onclick="update() return false;">
Another thing.
The onclick on the submitbutton will not work as excpected if the submit is caused without clicking the button.
For example mobile safari on iPhone can submit forms directly without triggering the button.
If you add UmairP's version of the onclick to the form element as an onsubmit method you should get the same result on every platform as far as I know.
You can see more details on iPhone forms in my own question on another issue.
How can I prevent the Go button on iPad/iPhone from posting the form

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