I have one form with a few submit buttons. I want to submit the form via POST to itself to process the filled out form fields... does the jquery override the submit?
$('#myButton').click('myAction') <!-- not actual code, just for the idea -->
<input type="button" type="submit" id="myButton" value="do something">
you can use something like this:
html form:
<form id="myform" method="post" name="form" action="">
<input type="text" name="name" value="">
<input type="text" name="other" value="">
<input type="submit" type="submit" name="myButton" value="do something">
</form>
js:
$('#myform').on('submit',function(e){
e.preventDefault();
var addnew='&addnew=1234';//additional data
var _data = $('#myform').serialize();
_data = _data+addnew;
console.log(_data);
});
sample jsfiddle
Related
I am using $_SESSION to dynamically create forms for my web store. These forms hold the custom info for the product that the customer wants. This is the layout:
Page1
Customer fills out form that looks something like this:
<form action="page2" method="post">
<input type="text" name="size">
<input type="text" name="color">
<input type="submit" name="submit" value="Review Order">
</form>
Page2
Customer reviews order details and has the option of adding more products. Customer goes back to page1 to order another one. All of the customer's orders will show on page2 in their respective form.
Looks like this:
Size: 1
Color: blue
Click Here To Checkout
Size: 2
Color:green
Click Here To Checkout
Size:3
color:red
Click Here To Checkout
What I want is one button that will add ALL orders to the PayPal cart. Sure they can add every order individually by clicking on Click Here To Checkout, but then they will have to go through a big loop to add multiple products.
I want the customer to be able to add as many products as possible and then click one button that adds all of the orders to the shopping cart.
This is what I tried but it obviously didn't work:
<script>
$(document).ready(function(){
$('#clickAll').on('click', function() {
$('input[type="submit"]').trigger('click');
});
});
</script>
<form action="" method="post">
<input type="text" name="name">
<input type="submit" name="submit" value="submit">
</form>
<form action="" method="post">
<input type="text" name="name">
<input type="submit" name="submit" value="submit">
</form>
<form action="" method="post">
<input type="text" name="name">
<input type="submit" name="submit" value="submit">
</form>
<button id="clickAll">Submit All</button>
Here is the php script that generates the dynamic forms using $_SESSION:
<?php
if(isset($_POST['submit'])) :
$test = array(
'size' => $_POST['size'],
'color' => $_POST['color'],
'submit' => $_POST['submit']
);
$_SESSION['testing'][] = $test;
endif;
if(isset($_SESSION['testing'])) :
foreach($_SESSION['testing'] as $sav) {
?>
<form action="paypal.com/..." method="post">
<input type="text" name="size" value="<?php echo $sav['size']; ?>">
<input type="text" name="color" value="<?php echo $sav['color']; ?>">
<input type="submit" name="submit" value="Click Here to Checkout">
</form>
<?php } endif; ?>
So the question is, how can I submit all of the forms with ONE button?
Have you tried to do it with $.ajax? You can add an foreach, or call another form on the Onsucces function. Another approach is changing all to one form with an array that points to the right "abstract" form:
<form action="" method="post">
<input type="text" name="name[]">
<input type="text" name="example[]">
<input type="text" name="name[]">
<input type="text" name="example[]">
<input type="text" name="name[]">
<input type="text" name="example[]">
<button id="clickAll">Submit All</button>
</form>
And in php:
foreach ($_POST['name'] as $key => $value) {
$_POST['name'][$key]; // make something with it
$_POST['example'][$key]; // it will get the same index $key
}
Here is a working jsFiddle: http://jsfiddle.net/SqF6Z/3/
Basically, add a class to each form and trigger() a submit on that class. Like so:
HTML (example only):
<form action="http://www.google.com" method="get" class="myForms" id="1stform">
<input type="text" value="1st Form" name="q1" />
</form>
<form action="http://www.google.com" method="get" class="myForms" id="2ndform">
<input type="text" value="2nd Form" name="q2" />
</form>
<form action="http://www.google.com" method="get" class="myForms" id="3rdform">
<input type="text" value="3rd Form" name="q3" />
</form>
<input type="button" id="clickMe" value="Submit ALL" />
jQuery:
$('.myForms').submit(function () {
console.log("");
return true;
})
$("#clickMe").click(function () {
$(".myForms").trigger('submit'); // should show 3 alerts (one for each form submission)
});
FWIW, I do this by creating an iframe, making that the target for the second form then submit both like this
//create the iframe
$('<iframe id="phantom" name="phantom">').appendTo('#yourContainer');
and create the dual submit like this:
function dualSubmit() {
document.secondForm.target = 'phantom';
document.secondForm.submit();
document.firstForm.submit();
}
works!
first create loop get all forms id and send them to ajax.
<script name="ajax fonksiyonları" type="text/javascript">
function validate(form){
//get form id
var formID = form.id;
var formDetails = $('#'+formID);
$.ajax({
type: "POST",
url: 'ajax.php',
data: formDetails.serialize(),
success: function (data) {
// log result
console.log(data);
//for closing popup
location.reload();
window.close()
},
error: function(jqXHR, text, error){
// Displaying if there are any errors
console.log(error);
}
});
return false;
}
//this function will create loop for all forms in page
function submitAll(){
for(var i=0, n=document.forms.length; i<n; i++){
validate(document.forms[i]);
}
}
create button for submit in order
<a class="btn" id="btn" onclick="submitAll();" href="">Save & Close</a>
then stop ajax call after success.also dont forget to log to console.
this code works in popup and closing popup after all ajax completed.
I am using $_SESSION to dynamically create forms for my web store. These forms hold the custom info for the product that the customer wants. This is the layout:
Page1
Customer fills out form that looks something like this:
<form action="page2" method="post">
<input type="text" name="size">
<input type="text" name="color">
<input type="submit" name="submit" value="Review Order">
</form>
Page2
Customer reviews order details and has the option of adding more products. Customer goes back to page1 to order another one. All of the customer's orders will show on page2 in their respective form.
Looks like this:
Size: 1
Color: blue
Click Here To Checkout
Size: 2
Color:green
Click Here To Checkout
Size:3
color:red
Click Here To Checkout
What I want is one button that will add ALL orders to the PayPal cart. Sure they can add every order individually by clicking on Click Here To Checkout, but then they will have to go through a big loop to add multiple products.
I want the customer to be able to add as many products as possible and then click one button that adds all of the orders to the shopping cart.
This is what I tried but it obviously didn't work:
<script>
$(document).ready(function(){
$('#clickAll').on('click', function() {
$('input[type="submit"]').trigger('click');
});
});
</script>
<form action="" method="post">
<input type="text" name="name">
<input type="submit" name="submit" value="submit">
</form>
<form action="" method="post">
<input type="text" name="name">
<input type="submit" name="submit" value="submit">
</form>
<form action="" method="post">
<input type="text" name="name">
<input type="submit" name="submit" value="submit">
</form>
<button id="clickAll">Submit All</button>
Here is the php script that generates the dynamic forms using $_SESSION:
<?php
if(isset($_POST['submit'])) :
$test = array(
'size' => $_POST['size'],
'color' => $_POST['color'],
'submit' => $_POST['submit']
);
$_SESSION['testing'][] = $test;
endif;
if(isset($_SESSION['testing'])) :
foreach($_SESSION['testing'] as $sav) {
?>
<form action="paypal.com/..." method="post">
<input type="text" name="size" value="<?php echo $sav['size']; ?>">
<input type="text" name="color" value="<?php echo $sav['color']; ?>">
<input type="submit" name="submit" value="Click Here to Checkout">
</form>
<?php } endif; ?>
So the question is, how can I submit all of the forms with ONE button?
Have you tried to do it with $.ajax? You can add an foreach, or call another form on the Onsucces function. Another approach is changing all to one form with an array that points to the right "abstract" form:
<form action="" method="post">
<input type="text" name="name[]">
<input type="text" name="example[]">
<input type="text" name="name[]">
<input type="text" name="example[]">
<input type="text" name="name[]">
<input type="text" name="example[]">
<button id="clickAll">Submit All</button>
</form>
And in php:
foreach ($_POST['name'] as $key => $value) {
$_POST['name'][$key]; // make something with it
$_POST['example'][$key]; // it will get the same index $key
}
Here is a working jsFiddle: http://jsfiddle.net/SqF6Z/3/
Basically, add a class to each form and trigger() a submit on that class. Like so:
HTML (example only):
<form action="http://www.google.com" method="get" class="myForms" id="1stform">
<input type="text" value="1st Form" name="q1" />
</form>
<form action="http://www.google.com" method="get" class="myForms" id="2ndform">
<input type="text" value="2nd Form" name="q2" />
</form>
<form action="http://www.google.com" method="get" class="myForms" id="3rdform">
<input type="text" value="3rd Form" name="q3" />
</form>
<input type="button" id="clickMe" value="Submit ALL" />
jQuery:
$('.myForms').submit(function () {
console.log("");
return true;
})
$("#clickMe").click(function () {
$(".myForms").trigger('submit'); // should show 3 alerts (one for each form submission)
});
FWIW, I do this by creating an iframe, making that the target for the second form then submit both like this
//create the iframe
$('<iframe id="phantom" name="phantom">').appendTo('#yourContainer');
and create the dual submit like this:
function dualSubmit() {
document.secondForm.target = 'phantom';
document.secondForm.submit();
document.firstForm.submit();
}
works!
first create loop get all forms id and send them to ajax.
<script name="ajax fonksiyonları" type="text/javascript">
function validate(form){
//get form id
var formID = form.id;
var formDetails = $('#'+formID);
$.ajax({
type: "POST",
url: 'ajax.php',
data: formDetails.serialize(),
success: function (data) {
// log result
console.log(data);
//for closing popup
location.reload();
window.close()
},
error: function(jqXHR, text, error){
// Displaying if there are any errors
console.log(error);
}
});
return false;
}
//this function will create loop for all forms in page
function submitAll(){
for(var i=0, n=document.forms.length; i<n; i++){
validate(document.forms[i]);
}
}
create button for submit in order
<a class="btn" id="btn" onclick="submitAll();" href="">Save & Close</a>
then stop ajax call after success.also dont forget to log to console.
this code works in popup and closing popup after all ajax completed.
<?php
echo $_POST['textvalue'];
echo $_post['radiovalue'];
?>
<div id="hidethis">
<form method="POST" action="">
<label>Tekst Value</label>
<input type="text" name="textvalue">
<label>Radio Value</label>
<input type="radio" name="radiovalue" value="autogivevalue">
<input type="submit" id="submit" name="submit" value="submit">
</div>
http://jsfiddle.net/Bjk89/2/ here is it with the jQuery.
What i try to do is to hide the <div id="hidethis"> when it's clicking submit.
I know i can make another page where i can recieve the values without the <form> section, but i want to put both in one page, make the <div id="hidethis"> hidden after submit.
So i'll be able to get echo $_POST['textvalue']; and echo $_post['radiovalue']; as results
RESULT MUST BE LIKE
A Text // This is the value you input into Tekst Value
autogivevalue // This is the value from the radio button
----- INVISIBLE -----
<form is hidden because we set it in jQuery so>
</form>
Try this. No need to use jQuery here.
<?php
if($_POST) {
echo $_POST['textvalue'];
echo $_post['radiovalue'];
} else {
?>
<form method="POST" action="">
<label>Tekst Value</label>
<input type="text" name="textvalue">
<label>Radio Value</label>
<input type="radio" name="radiovalue" value="autogivevalue">
<input type="submit" id="submit" name="submit" value="submit">
</form>
<?php
}
?>
Try adding '#' in your jquery code. Your version does not have # next to submit. Also your form is missing a closing tag here and in your JSFiddle code.
Try this:
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script>
$(document).ready(function () {
$('#submit').click(function () {
$('form').submit();
$('#hidethis').hide();
});
});
</script>
<form method='post' id="hidethis" name='form'>
<input type="text" name="textvalue">
<input type="radio" name="radiovalue" value="1">
<input type="button" id="submit" name="submit" value="submit">
</form>
I'm using multiple html on my page.
<form id="form1">
<input type="text" name="foo" id="foo">
<input type="submit" value="submit">
</form>
<form id="form2">
<input type="text" name="foo" id="foo">
<input type="submit" value="submit">
</form>
When the form is submitted I am access the data as follows:bar
var bar=$(this).find('input[name=foo]').val();
That works fine, but let's say I want to change the form data back on the html page after I have processed the submitted data. How do I reference the form which was submitted?
Something like:
$(this).$('#foo').val('Hello');
Is it possible?
Thanks!
If this is an ajax call, you can store a reference to the form before submitting:
var form = $(this);
$.ajax (...).done(function (){
form.find(...).val (...)
});
i have input type and button, i need when insert into
<input type="text" name="bills_ID" id="bills_ID" value="2">
i will get id for this item and put it into here
BillsPrint.php?bills_ID=id
this is full code
<form id="wrapper">
<input type="text" name="bills_ID" id="bills_ID" value="2">
<input name="print" type="submit" id="print" value="print" class="css3buttonblue" onclick="window.open('BillsPrint.php?bills_ID='this.id, '_blank')" />
</form>
how can i put the id on this link BillsPrint.php?bills_ID=2 without refresh page
You could do this:
<form id="wrapper">
<input type="text" name="bills_ID" id="bills_ID" value="2">
<input name="print" type="submit" id="print" value="print" class="css3buttonblue" />
</form>
<script>
$(function(){
$('#print').click(function(){
window.open('BillsPrint.php?bills_ID='+$('#bills_ID').val(), '_blank')
});
});
</script>
<form id="wrapper">
<input type="text" name="bills_ID" id="bills_ID" value="2">
<input name="print" type="submit" id="print" value="print" class="css3buttonblue"
onclick="window.open('BillsPrint.php?bills_ID=' + $("#bills_ID").val(), '_blank')" />
</form>
Remove the onclick etc. And use get method. Use this simple code and the variables will come in URL as query string.
<form action="BillsPrint.php" method="get" id="wrapper">
<input type="text" name="bills_ID" id="bills_ID" value="2">
<input name="print" type="submit" id="print" value="print" class="css3buttonblue">
</form>
You forget to specify action of form. Use this code it will work. :)
Hope it will help.
Try this:
$(document).ready(function(){
$('form').submit(function(){
var id = $('input[name="bills_ID"]').val();
window.open('BillsPrint.php?bills_ID='+id, '_blank');
return false;
});
});
$("#print").click(function(){
var id = $("#bills_ID").val();
$.ajax({
url: 'BillsPrint.php?bills_ID=' + id,
success: function(data) {
$('.result').html(data);
alert(data); //alerts output from the BillsPrint.php page
}
});
});