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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 thanTrying to run an AJAX script from the admin area in magentoRemoving Items from the cart doesn't update the ajax cartGet the magento cart quantity with ajaxUpdate cart using Ajax Magento not working?product 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 correctlyCant Access a Value on Input or Button with type Submit

1

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

share|improve this question

edited 3 mins ago

magefms

1,588225

asked Aug 29 '17 at 12:17

MagecodeMagecode

519421

add a comment | 

1

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

share|improve this question

edited 3 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

share|improve this question

edited 3 mins ago

magefms

1,588225

asked Aug 29 '17 at 12:17

MagecodeMagecode

519421

<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>

share|improve this question

edited 3 mins ago

magefms

1,588225

asked Aug 29 '17 at 12:17

MagecodeMagecode

519421

share|improve this question

edited 3 mins ago

magefms

1,588225

asked Aug 29 '17 at 12:17

MagecodeMagecode

519421

edited 3 mins ago

magefms

1,588225

edited 3 mins ago

magefms

1,588225

asked Aug 29 '17 at 12:17

MagecodeMagecode

519421

asked Aug 29 '17 at 12:17

MagecodeMagecode

519421

2 Answers
2

active

oldest

votes

0

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?

share|improve this answer

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 | 

0

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);

share|improve this answer

answered Dec 14 '17 at 6:31

HansuHansu

13

add a comment | 

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0

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?

share|improve this answer

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 | 

0

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?

share|improve this answer

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?

share|improve this answer

edited Oct 20 '17 at 23:23

Aasim Goriya

3,9671938

answered Oct 20 '17 at 17:25

Harry AliveHarry Alive

487

$("#form-validate").replaceWith(result);

share|improve this answer

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

answered Oct 20 '17 at 17:25

Harry AliveHarry Alive

487

answered Oct 20 '17 at 17:25

Harry AliveHarry Alive

487

0

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);

share|improve this answer

answered Dec 14 '17 at 6:31

HansuHansu

13

add a comment | 

0

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);

share|improve this answer

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);

share|improve this answer

answered Dec 14 '17 at 6:31

HansuHansu

13

share|improve this answer

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