jQuery not hiding divs (mvc 4) - javascript

I have 2 divs:
<div class="lobsummary">
<input type="submit" name ="sbmtbtn">
<input type="submit" name ="sbmtbtn">
</div>
<div class="applicationsummary">
</div>
I want to show applicationsummary when sbmtbtn is submitted so I used the below script:
$('.lobsummary input:submit').submit(function (event) {
$('.applicationsummary').show();
$('.lobsummary').hide();
});
But its not working.
EDIT: The divs are enclosed in <% Html.BeginForm("Index", "Home"); %> which I am using to pass values to my controller from the view.

Use event.preventDefault() it stops the default action
$('.lobsummary input[type=submit]').click(function (event) {
event.preventDefault();
alert($(this).index() + 1);
$('.applicationsummary').show();
$('.lobsummary').hide();
});

#Rajaprabhu Aravindasamy is right,use submit() over a form:
$("form").submit( function () {
return false;
} );
HTML on the client side will be like:
<div class="lobsummary">
<form action="/"method="post">
<input type="submit" name ="sbmtbtn">
</form>
<form action="/"method="post">
<input type="submit" name ="sbmtbtn">
</form>
</div>
<div class="applicationsummary">
</div>
Just use click() to test your jquery selection:
$('.applicationsummary').hide();
$(".lobsummary form:eq(0)").click(function () {
alert(1)
$('.applicationsummary').show();
});
$(".lobsummary form:eq(1)").click(function () {
alert(2)
$('.applicationsummary').hide();
});
Then replace the click to submit.

When the form was submitted, it reloaded the whole page and that's why it was not hiding because it was going back to its default view. So I used if statements in my view to hide the forms.

Related

Submit form on change of dropdown list in Semantic UI?

I would like to submit form when an element is selected - skipping pressing the submit button.
I tried using onchange="this.form.submit()", but it's not working here.
This is the code:
<form action="" method="get">
<div class="ui floating dropdown labeled search icon button dd">
<input type="hidden" name="nutr_code">
<span class="text">Select nutrient</span>
<div class="menu">
<div class="item" data-value="ca">Calcium</div>
<div class="item" data-value="fe">Iron</div>
<div class="item" data-value="mg">Magnesium</div>
<div class="item" data-value="zn">Zinc</div>
</div>
</div>
<br><br>
<input type="submit" value="Show results">
</form>
Because you are using GET method you can redirect with javascript by constructing the url yourself.
$('.ui.dropdown').dropdown({
'onChange': function (value, text, $choice) {
location.href = 'http://example.com/?nutr_code=' + value;
}});
Second option is changing the input field 'nutr_code' with the value from the callback as shown above
$('input[name="nutr_code"]').val(value);
and submit the <FORM/> from js.
$('form').submit();
EDIT:
Example of second option.
$('.ui.dropdown').dropdown({
'onChange': function (value, text, $choice) {
// Uncomment if semantic is not updating the input before submit.
//$('input[name="nutr_code"]').val(value);
$('form').submit();
}});
You need to define action in form tag or add some method to submit button
<input type="submit" value="Show results" onclick="someFunction()">
You need to define someFunction() too if you follow this way. Thanks
You Can Use Jquery
$(document).ready(function(){
$("input[name='nutr_code']").on('input',function(e){
$("input[type='submit']").trigger( "click" );
});
});

Send button value by post request via jquery

My code
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<form>
<button id="test" value="123" name="test" >ok</button>
</form>
<script>
$(document).ready(function () {
$("#test").click(function() {
var combinedData = $("#test").serialize();
$.post(
"element_submit.php",
combinedData
).done(function(data) {
//alert("Successfully submitted!");
$("#result").html(data);
}).fail(function () {
//alert("Error submitting forms!");
})
});
});
</script>
<div id="result" ></div>
The element_submit.php file
<?php
//just to test it should output in the #result div
echo $_POST['test'];
?>
What I am trying to do is submit the with the value="attribute" so the data is serialized and send the post request, it's not like a submit when user insert a value and submit,What I need is to get the value attribute and submit to the php, this code is only for To simplify and illustrate what I am trying to do, because in this page I have the following buttons with ids #follow #unfollow so I need a way to get the button value to make the user follow and unfollow.
you need to serialize the form - not the elements within it .You can also have the triggering button outside the form which will prevent hte form from submitting on the button click.
<form id="testForm">
<input type="hidden" name="testInput" value="123"/>
</form>
<button name="test" id="testButton">submit</button>
...
<script>
$(document).ready(function () {
$("#testButton").click(function() {
var combinedData = $("#testForm").serialize();...
$(document).ready(function () {
$("#testButton").click(function() {
var combinedData = $("#testForm").serialize();
console.log(combinedData);
})
})
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<form id="testForm">
<input type="hidden" value="123" name="testinput"/>
</form>
<button id="testButton">Click</button>
Straight JS might help you out. Include a function that sends the id and get the value of that id. Then just send a regular post of the value without serialize... easier.
<script>
function fetchButtonValue(theId){
var p = document.getElementById(theId).value;
alert (p);
}
</script>
<button id="myFormBtn" value ="woo"
onclick="fetchButtonValue(this.id)">My Button</button>
this works...
You could also put a class on the button let's say class="followBTN" then on a click you could just snag the value by $(this).val() I'd use this method if I had more than one button per page.

how to bind bootstrap loading button success / error on form submit

I'm trying to bind a bootstrap button with loading state to a form.
I would like the button to keep the data-loading state until the form is submitted. If the form errors out the data-loading should ideally stop or display a different message.
How can I achieve this?
My code for the button:
http://www.bootply.com/128762
what would be the best approach for a form that is not ajax?
Are You looking Something Like this Below
$('#loading-example-btn').click(function () {
if($("#txt").val().length > 0)
{
var btn = $(this)
$(this).attr('value','Loading');
}
else
{
$(this).attr('value','error On Submit');
}
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<form method="post">
<input type="text" name="test" id="txt">
<input type="submit" id="loading-example-btn" value="Submit" data-loading-text="Loading...">
</form>

jquery - hide not working

By clicking on "empty cart?", I am unsetting the cart array which is working fine. Now I wanted another form to be hidden if this "empty cart" button is clicked.
html:
<form id="f1" action="checkout.php" method="post">
<input name="cust_login" type="submit" value="Continue" />
</form>
<br>
<button>empty cart?</button>
javascript:
<script>
$(document).ready(function(){
$("button").click(function(){
$("#f1").hide();
});
});
</script>
php code for unsetting array:
if (isset($_GET['cmd']) && $_GET['cmd'] == "emptycart") {
unset($_SESSION["cart_array"]);
But hide is not working here.
My question is, can we use button and link together as I have written ? If not, then how to implement it?
Try to use this:
$(document).ready(function(){
var cmd = '<?php echo $_GET["cmd"] ;?>' ;
if(cmd == "emptycart")
{
$("#f1").hide();
}
});
You can try
$(document).ready(function(){
$("button").click(function(){
$("#f1").attr('style','display: block !important')
});
});
or to add a class to style file that do the same and then just add this class when clicking on the button.
<form id="f1"
action="checkout.php"
style="<?= (isset($_SESSION['cart_array']) && count($_SESSION['cart_array']))? "display:none;": ""; ?>"
method="post">
<input name="cust_login" type="submit" value="Continue" />
</form>
<br>
<button>empty cart?</button>
On rendering form - check if cart is empty - render it on hidden state
But in this case your js code had no sense. So you can add
<script>
$(document).ready(function(){
$(document).on('click', "a[href='cart.php?cmd=emptycart']", function(ev){
ev.preventDefault();
$.get($(this).attr('href', function() {
$("#f1").hide();
});
});
});
</script>
Istead of your js code - this one prevent page reload ( which happens when you click on link ) and send request via AJAX
You can try this solution
HTML
<form id="f1" action="checkout.php" method="post">
<input name="cust_login" type="submit" value="Continue" />
</form>
<br>
<button>empty cart?</button>
JS
$(document).ready(function(){
$("button").click(function(){
//$("#f1").hide();
window.location.href = "cart.php?cmd=emptycart";
});
});

Multiple show/hide forms in jquery

Hello I want to have list of files in directory and a form below each of them that allows my users to name them.
That's all clear - I made it in php, but now I want to have this list and hidden forms, and when I'm clicking on one of my file's name, the form shows under the clicked name.
Something like here: http://papermashup.com/demos/jquery-sliding-div/#
Here is the code: http://papermashup.com/simple-jquery-showhide-div/
But it works in a way, that when i click on one of files, all forms shows or all hides. How to fix it to work only for clicked file?
JSFIDDLE EXAMPLE: http://jsfiddle.net/qbNrR/
#UPDATE - SIMILAR PROBLEM
Hey, I've got similar problem with submitting ajax forms - using this tutorial: http://net.tutsplus.com/tutorials/javascript-ajax/submit-a-form-without-page-refresh-using-jquery/
my forms are in div id=#upform and when i'm trying to submit any of them via $.ajax it submits only the first one, here's the code:
<script>
$(function() {
$(".button").click(function() {
var txt = $(".tekst#test").val();
var dataString = 'tekst=' + tekscior;
$.ajax({
type: "POST",
url: "upload/base",
data: dataString,
success: function() {
$('#upform').html("<div id='message'></div>");
$('#message')
.html("<h2>described!</h2>")
.append("<p>thanks!</p>")
.hide()
.fadeIn(1500, function() {
$('#message')
.append("<img id='checkmark' src='http://artivia-dev2/i/check.png' />");
});
}
});
return false;
});
});​
</script>
AND Here are my forms:
// ONLY THIS ONE IS SUBMITTED, EVEN WHEN I'M SUBMITTING THE SECOND ONE!
<div class="slidingDiv">
<div id="upform">
<form name="contact" action="">
<input type="text" value="TESTFORM" class="tekst" id="test">
<input type="submit" name="submit" class="button" id="submit" value="Send" />
<form>
</div>
<div class="slidingDiv">
<div id="upform">
<form name="contact" action="">
<input type="text" value="TESTFORM" class="tekst" id="test">
<input type="submit" name="submit" class="button" id="submit" value="Send" />
<form>
</div>
You can use the next() jQuery method. Then your code will look something like:
$(document).ready(function() {
$(".slidingDiv").hide();
$(".show_hide").show();
$('.show_hide').click(function(e) {
$(e.target).next(".slidingDiv").slideToggle();
});
});​
Try event.currentTarget to get the form that triggered the click event
on click event use jquery like
$(this).show(); // or hide();
as in example
$('.show_hide').click(function(){
//$(".slidingDiv").slideToggle();
$(this).slideToggle();
});

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