Update Cart Page by Ajax on quantity changeUpdate quantity of products in cart to the max available if quantity entered is greater thanAuto update cart quantity when change quantityTrying to run an AJAX script from the admin area in magentoGet the magento cart quantity with ajaxproduct quickview not working with infinite scrollMagento 2: How to create custom ui_component form to submit data by ajaxAjax cart validation on change qtyMagento 2 cart event observers not redirecting correctlyShopping cart page, update quantity by ajaxCant Access a Value on Input or Button with type Submit
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Update Cart Page by Ajax on quantity change
Update quantity of products in cart to the max available if quantity entered is greater thanAuto update cart quantity when change quantityTrying to run an AJAX script from the admin area in magentoGet the magento cart quantity with ajaxproduct quickview not working with infinite scrollMagento 2: How to create custom ui_component form to submit data by ajaxAjax cart validation on change qtyMagento 2 cart event observers not redirecting correctlyShopping cart page, update quantity by ajaxCant Access a Value on Input or Button with type Submit
I have to update the main cart on change of quantity by Ajax without click on "Update Cart Button" and without reloading the page.
By the following code, I'm able to update cart table but m not getting a solution to update the "Cart Summary" block.
<script>
require(['jquery', 'Magento_Customer/js/customer-data',
'jquery/jquery-storageapi'], function ($)
// $("#submitbutton").hide();
var form = $('form#form-validate');
var qtyfields = $('input.qty');
$('.page.messages').each(function ()
var thismessage = $(this);
thismessage.attr('id', 'messages');
);
form.find(qtyfields).each(function (e)
var thisfield = $(this);
$(this).change(function ()
console.log('change detected');
form.submit();
);
);
form.on('submit', function (e)
e.preventDefault();
$.ajax(
url: form.attr('action'),
data: form.serialize(),
type: 'post',
success: function (res)
var parsedResponse = $.parseHTML(res);
var result = $(parsedResponse).find("#form-validate");
$("#form-validate").replaceWith(result);
//console.log(result);
//location.reload();
,
error: function ()
console.log('error');
);
console.log('form submitted');
);
);
</script>
Please give me some solution.
cart ajax magento-2.1.7
edited 14 mins ago
magefms
1,588225
asked Aug 29 '17 at 12:17
MagecodeMagecode
519421
add a comment |
I have to update the main cart on change of quantity by Ajax without click on "Update Cart Button" and without reloading the page.
By the following code, I'm able to update cart table but m not getting a solution to update the "Cart Summary" block.
<script>
require(['jquery', 'Magento_Customer/js/customer-data',
'jquery/jquery-storageapi'], function ($)
// $("#submitbutton").hide();
var form = $('form#form-validate');
var qtyfields = $('input.qty');
$('.page.messages').each(function ()
var thismessage = $(this);
thismessage.attr('id', 'messages');
);
form.find(qtyfields).each(function (e)
var thisfield = $(this);
$(this).change(function ()
console.log('change detected');
form.submit();
);
);
form.on('submit', function (e)
e.preventDefault();
$.ajax(
url: form.attr('action'),
data: form.serialize(),
type: 'post',
success: function (res)
var parsedResponse = $.parseHTML(res);
var result = $(parsedResponse).find("#form-validate");
$("#form-validate").replaceWith(result);
//console.log(result);
//location.reload();
,
error: function ()
console.log('error');
);
console.log('form submitted');
);
);
</script>
Please give me some solution.
cart ajax magento-2.1.7
edited 14 mins ago
magefms
1,588225
asked Aug 29 '17 at 12:17
MagecodeMagecode
519421
add a comment |
I have to update the main cart on change of quantity by Ajax without click on "Update Cart Button" and without reloading the page.
By the following code, I'm able to update cart table but m not getting a solution to update the "Cart Summary" block.
<script>
require(['jquery', 'Magento_Customer/js/customer-data',
'jquery/jquery-storageapi'], function ($)
// $("#submitbutton").hide();
var form = $('form#form-validate');
var qtyfields = $('input.qty');
$('.page.messages').each(function ()
var thismessage = $(this);
thismessage.attr('id', 'messages');
);
form.find(qtyfields).each(function (e)
var thisfield = $(this);
$(this).change(function ()
console.log('change detected');
form.submit();
);
);
form.on('submit', function (e)
e.preventDefault();
$.ajax(
url: form.attr('action'),
data: form.serialize(),
type: 'post',
success: function (res)
var parsedResponse = $.parseHTML(res);
var result = $(parsedResponse).find("#form-validate");
$("#form-validate").replaceWith(result);
//console.log(result);
//location.reload();
,
error: function ()
console.log('error');
);
console.log('form submitted');
);
);
</script>
Please give me some solution.
cart ajax magento-2.1.7
edited 14 mins ago
magefms
1,588225
asked Aug 29 '17 at 12:17
MagecodeMagecode
519421
I have to update the main cart on change of quantity by Ajax without click on "Update Cart Button" and without reloading the page.
By the following code, I'm able to update cart table but m not getting a solution to update the "Cart Summary" block.
<script>
require(['jquery', 'Magento_Customer/js/customer-data',
'jquery/jquery-storageapi'], function ($)
// $("#submitbutton").hide();
var form = $('form#form-validate');
var qtyfields = $('input.qty');
$('.page.messages').each(function ()
var thismessage = $(this);
thismessage.attr('id', 'messages');
);
form.find(qtyfields).each(function (e)
var thisfield = $(this);
$(this).change(function ()
console.log('change detected');
form.submit();
);
);
form.on('submit', function (e)
e.preventDefault();
$.ajax(
url: form.attr('action'),
data: form.serialize(),
type: 'post',
success: function (res)
var parsedResponse = $.parseHTML(res);
var result = $(parsedResponse).find("#form-validate");
$("#form-validate").replaceWith(result);
//console.log(result);
//location.reload();
,
error: function ()
console.log('error');
);
console.log('form submitted');
);
);
</script>
Please give me some solution.
cart ajax magento-2.1.7
cart ajax magento-2.1.7
edited 14 mins ago
magefms
1,588225
asked Aug 29 '17 at 12:17
MagecodeMagecode
519421
edited 14 mins ago
magefms
1,588225
asked Aug 29 '17 at 12:17
MagecodeMagecode
519421
edited 14 mins ago
magefms
1,588225
edited 14 mins ago
magefms
1,588225
edited 14 mins ago
magefms
1,588225
1,588225
asked Aug 29 '17 at 12:17
MagecodeMagecode
519421
asked Aug 29 '17 at 12:17
MagecodeMagecode
519421
asked Aug 29 '17 at 12:17
MagecodeMagecode
519421
519421
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Thank you for your Answer, but this code only will submit once if call the
$("#form-validate").replaceWith(result);
And Ajax returning a whole HTML page for the result, is there any way to just get back small string of a JSON data set instead?
edited Oct 20 '17 at 23:23
Aasim Goriya
3,9671938
answered Oct 20 '17 at 17:25
Harry AliveHarry Alive
487
add a comment |
use this one. its works fine for me
- var parsedResponse = jQuery.parseHTML(res);
- var result = jQuery(parsedResponse).find(".subtotal");
- var result1 = jQuery(parsedResponse).find(".message");
- jQuery(".subtotal").replaceWith(result);
- jQuery(".message").replaceWith(result1);
answered Dec 14 '17 at 6:31
HansuHansu
13
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thank you for your Answer, but this code only will submit once if call the
$("#form-validate").replaceWith(result);
And Ajax returning a whole HTML page for the result, is there any way to just get back small string of a JSON data set instead?
edited Oct 20 '17 at 23:23
Aasim Goriya
3,9671938
answered Oct 20 '17 at 17:25
Harry AliveHarry Alive
487
add a comment |
Thank you for your Answer, but this code only will submit once if call the
$("#form-validate").replaceWith(result);
And Ajax returning a whole HTML page for the result, is there any way to just get back small string of a JSON data set instead?
edited Oct 20 '17 at 23:23
Aasim Goriya
3,9671938
answered Oct 20 '17 at 17:25
Harry AliveHarry Alive
487
add a comment |
Thank you for your Answer, but this code only will submit once if call the
$("#form-validate").replaceWith(result);
And Ajax returning a whole HTML page for the result, is there any way to just get back small string of a JSON data set instead?
edited Oct 20 '17 at 23:23
Aasim Goriya
3,9671938
answered Oct 20 '17 at 17:25
Harry AliveHarry Alive
487
Thank you for your Answer, but this code only will submit once if call the
$("#form-validate").replaceWith(result);
And Ajax returning a whole HTML page for the result, is there any way to just get back small string of a JSON data set instead?
edited Oct 20 '17 at 23:23
Aasim Goriya
3,9671938
answered Oct 20 '17 at 17:25
Harry AliveHarry Alive
487
edited Oct 20 '17 at 23:23
Aasim Goriya
3,9671938
edited Oct 20 '17 at 23:23
Aasim Goriya
3,9671938
edited Oct 20 '17 at 23:23
Aasim Goriya
3,9671938
3,9671938
answered Oct 20 '17 at 17:25
Harry AliveHarry Alive
487
answered Oct 20 '17 at 17:25
Harry AliveHarry Alive
487
answered Oct 20 '17 at 17:25
Harry AliveHarry Alive
487
487
add a comment |
add a comment |
use this one. its works fine for me
- var parsedResponse = jQuery.parseHTML(res);
- var result = jQuery(parsedResponse).find(".subtotal");
- var result1 = jQuery(parsedResponse).find(".message");
- jQuery(".subtotal").replaceWith(result);
- jQuery(".message").replaceWith(result1);
answered Dec 14 '17 at 6:31
HansuHansu
13
add a comment |
use this one. its works fine for me
- var parsedResponse = jQuery.parseHTML(res);
- var result = jQuery(parsedResponse).find(".subtotal");
- var result1 = jQuery(parsedResponse).find(".message");
- jQuery(".subtotal").replaceWith(result);
- jQuery(".message").replaceWith(result1);
answered Dec 14 '17 at 6:31
HansuHansu
13
add a comment |
use this one. its works fine for me
- var parsedResponse = jQuery.parseHTML(res);
- var result = jQuery(parsedResponse).find(".subtotal");
- var result1 = jQuery(parsedResponse).find(".message");
- jQuery(".subtotal").replaceWith(result);
- jQuery(".message").replaceWith(result1);
answered Dec 14 '17 at 6:31
HansuHansu
13
use this one. its works fine for me
- var parsedResponse = jQuery.parseHTML(res);
- var result = jQuery(parsedResponse).find(".subtotal");
- var result1 = jQuery(parsedResponse).find(".message");
- jQuery(".subtotal").replaceWith(result);
- jQuery(".message").replaceWith(result1);
answered Dec 14 '17 at 6:31
HansuHansu
13
answered Dec 14 '17 at 6:31
HansuHansu
13
answered Dec 14 '17 at 6:31
HansuHansu
13
answered Dec 14 '17 at 6:31
HansuHansu
13
13
add a comment |
add a comment |
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