submit form at each step using Jquery form wizard - model-view-controller

I am using Jquery Form wizard plugin in a MVC application. http://thecodemine.org/
I have a form with 4 steps. in one of step i have upload functionality.
I want submit functionality at each step and also back and next steps. later steps are optional.
I was able to add a submit button in navigation. On clicking it, form data related to active step is only submitted and other data is null.
For more clarity of my issue:
View :
<form id="myform" method="post" action="/Controller/Action">
<div id="fieldWrapper">
<fieldset class="step fieldset" id="first">
<legend class="legend">First step</legend>
Some input controls
</fieldset>
<fieldset class="step fieldset" id="second">
<legend class="legend">Second step</legend>
some more input controls (optional)
</fieldset>
<fieldset class="step fieldset" id="third">
<legend class="legend">Third step</legend>
some more input controls with filu upload
(optional)
</fieldset>
<fieldset class="step fieldset" id="fourth">
<legend class="legend">Fourth step</legend>
some more input controls (optional)
</fieldset>
</div>
<div id="demoNavigation">
<input class="navigation_button" id="back" value="Back" type="reset" />
<button type="submit" id="submitBtn">Submit and Finish</button>
<input class="navigation_button" id="next" value="Next" type="submit" />
</div>
</form>
Script:
<script type="text/javascript">
$(function () {
$("#myform").formwizard({
validationEnabled: true,
focusFirstInput: false,
disableUIStyles: true,
textSubmit: 'Submit and Finish',
textNext: 'Continue to next step',
next: "input:submit"
}
);
});
</script>
Updated the Jquery.form.wizard.js to so that the submit button hides at last step.
Now on each step i have submit button and navigation button.
When i submit form on second step, form data in second step is only posted and rest is not posted.
I went through the samples but could not find the appropriate one.
Can anyone suggest how to achieve this?

I went through the documentation and jquery.form.wizard.js file to get beeter understanding of what is done.
I just have to write some script as follows:
<script type="text/javascript">
$(function () {
$("#myform").formwizard({
validationEnabled: true,
focusFirstInput: false,
disableUIStyles: true,
textSubmit: 'Submit and Finish',
textNext: 'Continue to next step',
next: "input:submit"
}
);
});
$('#submitBtn').click(function () {
var stepInfo = $('#myform').formwizard('state');
for (var i = 0; i < stepInfo.activatedSteps.length; i++) {
stepInfo.steps.filter("#" + stepInfo.activatedSteps[i]).find(":input").not(".wizard-ignore").removeAttr("disabled");
}
});
</script>

Related

Autosave content of TinyMce editor on change event (with laravel 7)?

I want to autosave the content of tinyMce editor
I'm using tinmce version 5.2.2 with laravel V7
tinymce configuration in app.js
require('./bootstrap');
require('tinymce/themes/silver');
require('tinymce/plugins/image');
require('tinymce/plugins/code');
require('tinymce/plugins/save');
import tinymce from 'tinymce';
tinymce.init({
selector:'textarea#inputQuestionTitle',
height:400,
setup: function(editor) {
editor.on('Change Keyup', function () {
editor.save();
//tinyMCE.triggerSave() <-- also tried
});
}
With following View
#extends('dashboard.layout')
#section('content')
<form id='addQ' action="{{route('questions.store')}}" method="post" enctype="multipart/form-data" >
#csrf
<div class="form-row align-items-center " >
<div class="col-md-12">
<label class="sr-only" for="inputQuestionTitle">Title</label>
<textarea name="title" class="form-control mb-2 " id="inputQuestionTitle" placeholder="Question Title"> </textarea>
</div>
</div>
<div class="form-row">
<button type="submit" class="btn btn-primary mb-2">Add new category </button>
</div>
</form>
<script type="text/javascript" src="{{asset('js/app.js')}}"></script>
<script type="text/javascript">
$(function(){
$('#addQ').submit(function(e){
e.preventDefault();
var frmdata = $(this).serialize();
console.log(frmdata);
});
});
</script>
#endsection
Outputs
I had using jquery to show input request with console.log
As showed in output title field does not returns any value
Please tell me how i autosave the content of tinymce editor and send it with input request ..
I had also tried to tinymce.triggerSave() before serialize() method of ajax so that input request will take tinymce content but not works
Instead of trying to update the textarea via the triggerSave() or save() methods while someone is editing the content I would just place tinymce.triggerSave() at the very beginning of the function that submits the page. This will ensure that the triggerSave() is called directly before you grab the content from the textarea.

how to validate a form without displaying any error messages

I do not understand how I can validate a form with jquery validate to display the Submit button or not.
My javascript is as follows:
ko.bindingHandlers.jqValidate = {
init: function (element, valueAccessor, option) {
var element = element;
var validateOptions = {
ignore: [":hidden:not(required)"],
focusCleanup: true,
onsubmit: true
};
function displaySubmitButton() { // Remove/Add class to submit button
if ($('form').valid()) $('.toSave').show();
else $('.toSave').hide();
}
$('form').bind('onchange keyup onpaste', function (event, element) {
if ($('form').valid() === true && $(element).not('.resetForm')) $('.toSave').show();
else if ($(element).not('.resetForm')) $('.toSave').hide();
});
}
};
My form :
<ol class="MutColor13 mouseOver" data-bind="jqValidate: {}">
<li class="rightCol">
<strong>
<label for="iEmailReply"><asp:Literal runat="server" ID="lEmailReply" /></label>
</strong>
<input id="iEmailReply" name="EmailReply" type="text" tabindex="2"
class="MutColor13 email"
data-bind="value: communication.Message.ReplyEmail, valueUpdate: 'afterkeydown'"
required="required" />
</li>
<li class="leftCol">
<strong>
<label for="iEmailFrom"><asp:Literal runat="server" ID="lEmailFrom" /></label>
</strong>
<input id="iEmailFrom" name="EmailFrom" type="text" tabindex="1"
class="MutColor13 email"
data-bind="value: communication.Message.SenderEmail, valueUpdate: 'afterkeydown'"
required="required" />
</li>
<!-- and more form input -->
</ol>
My submit button :
<div class="buttonBlock rightButton" data-bind="fadeVisible: validateMessage()">
<a class="MutButton3 toSave" data-bind="click: saveMessage"><span class="MutButton3">submit</span></a>
</div>
when I type "$ ('form'). valid ()" in the firebug console, all error messages appear. I do not think the submit button is the problem because I have not clicked at this point
How do I enable the display of the error message from the input short change while allowing the display of the submit button if fields (and any other form fields in the page) if all fields are valid?
I was inspired by this question: jquery validate: IF form valid then show submit button
a working demo : http://jquery.bassistance.de/validate/demo/
but the button is displayed continuously
ok I think this could work:
http://jsfiddle.net/Ay972/10/
what I did :
$('form').bind('change oninput', function (event, element) {
console.log('formLive :: ', $(event));
if ($('form').valid() === true && $(element).not('.resetForm')) $('.toSave').show();
else if ($(element).not('.resetForm')) $('.toSave').hide();
});
the 'change oninput' seems to work.

Input Button as Input image will not work

On my website i have a button which when clicked takes you to one of two random youtube videos. However i would like to change this to a image in stead of a button.I have tried to change it to a INPUT type="image" but this doesn't work. Here is the code i am using.
<SCRIPT language="JavaScript">
<!--
function get_random()
{
var ranNum= Math.floor(Math.random()*2);
return ranNum;
}
function getaGame()
{
var whichGame=get_random();
var game=new Array(2)
game[0]= "https://www.youtube.com/watch?feature=player_detailpage&v=NcFQF3PZFRk#t=722s";
game[1]= "https://www.youtube.com/watch?v=klBAW4MQffU";
location.href = game[whichGame];
}
//-->
</SCRIPT>
<FORM name="form1">
<center>
<INPUT type="button" onClick="getaGame()" >
</center>
</FORM>
Thanks for any help
An onclick event can be fired from any element. Here are some examples!

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

Submitting form into a div

Can anyone tell me why this is not working or give me another way into doing what I want.
I have a form on a page when click submit I want it to process into add.php and for it to open up in a DIV called right.
Form page
<script>
$("#Submit").click(function() {
var url = "add.php"; // the script where you handle the form input.
$.ajax({
type: "POST",
url: url,
data: $("#myForm").serialize(), // serializes the form's elements.
success: function(html){ $("#right").html(html); }
});
return false; // avoid to execute the actual submit of the form.
});
</script>
</head>
<body>
<form action="add.php" method="post" enctype="multipart/form-data" id ="myForm">
Name: <input type="text" name="name"><br>
E-mail: <input type="text" name = "email"><br>
Phone: <input type="text" name = "phone"><br>
Photo: <input type="file" name="photo"><br>
<input type="submit" value="Upload">
</form>
</body>
Now if I add action to form to direct it to add.php all works fine so other script is ok yet when i do it this way,nothing happens it does not load add.php into the 'right' div like I want it to.
Anyone got any suggestions?
Your $("#Submit") selector does not match anything (the submit button has no id and is defined after that bind attempt), so the function is not bound to any event, thus never executed.
Instead the form acts the way it should: upon submit it posts its content to the url specified in the 'action' attribute. This is what happens, that div is never touched.
you have to go back to understand how jquery selectors work. How to bind a function to an event.
Is this your exact code? There are a few issues:
$("#Submit").click should be wrapped in a document ready
handler so it doesnt run before the page has actually loaded.
There is no button that matches #Submit
There is no div that matches #right
Try
<script type="text/javascript">
jQuery(document).ready(function($) {
$("#Submit").click(function() {
var url = "add.php"; // the script where you handle the form input.
$.ajax({
type: "POST",
url: url,
data: $("#myForm").serialize(), // serializes the form's elements.
success: function(html){ $("#right").html(html); }
});
return false; // avoid to execute the actual submit of the form.
});
});
</script>
</head>
<body>
<div id="right">
</div>
<form action="add.php" method="post" enctype="multipart/form-data" id ="myForm">
Name: <input type="text" name="name"><br>
E-mail: <input type="text" name = "email"><br>
Phone: <input type="text" name = "phone"><br>
Photo: <input type="file" name="photo"><br>
<input type="submit" value="Upload" id="Submit">
</form>
you have to create on div in body tag which id will be right and
<input id="submit" type="submit" name="upload" >
add new div in body tag like this
<div id="right"></div>

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