JQuery Remove input from variable

Question

I have specific div that I want to copy with jQuery. After copying, I want to modify it a little bit and the remove an Input with the ID CustomerDeliveryaddressID. Finally I want to append it somewhere. Unfortunately, the deletion of the part is not working for me. How can I do that? Here is my Code.

  $("#addDeliveryAddress").click(function(){
    var content = $('.deliveryAddressBox').html();
    content = "<div class='deliveryAddressBox'>"+content+"</div>";
    content = $(content).remove("#CustomerDeliveryaddressId");
    $(".deliveryAddressesContainer").append(content);
  });

Double check that remove() is not returning the part that is removed. If it is, you are setting the content with what was removed. — Taplar
– Taplar, Commented May 8, 2019 at 14:09

Rahul · Accepted Answer · 2019-05-08 14:26:13Z

1

Instead of removing and saving into content try replacing your remove line with

$("#addDeliveryAddress").click(function(){
    var $content = $('.deliveryAddressBox:last').clone();
    $content.wrapAll( "<div class='deliveryAddressBox'/>");
    $content.find("#CustomerDeliveryaddressId").remove();
    $(".deliveryAddressesContainer").append($content);
});

edited May 8, 2019 at 14:26

answered May 8, 2019 at 14:12

Rahul

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4 Comments

DS87 Over a year ago

I replaced your code with the line. But it's still not working. I checkout out with only the "find", if it finds the right supposed input. It is finding it. But removing it does somehow not work.

Rahul Over a year ago

Check now, I took clone instead of html

DS87 Over a year ago

That's looking good. There is just one thing for the finetuning. I have several divs with .deliveraAddressBox. How is it possible to just take the last one with clone? Otherwise it duplicates all (so if I have two divs with this class, it creates 4 divs instead of just one).

Rahul Over a year ago

See I updated my answer, you need to append :last to element which wants you to fetch in the index of DOM

Taplar · Accepted Answer · 2019-05-08 14:16:43Z

0

Make sure to not store the response from remove() in a variable that overwrites the value you want.

$("#addDeliveryAddress").click(function(){
    var $content = $( $('.deliveryAddressBox').html() );

    // wrap the contents with the div
    $content.wrapAll("<div class='deliveryAddressBox'>");
    // remove the element you do not want
    $content.find("#CustomerDeliveryaddressId").remove();
    // append it to the DOM
    $(".deliveryAddressesContainer").append($content);
});

answered May 8, 2019 at 14:16

Taplar

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Comments

Selvakumar · Accepted Answer · 2019-05-08 14:33:40Z

0

$("#addDeliveryAddress").click(function(){
    var content = $('.deliveryAddressBox').html();
    content = "<div class='deliveryAddressBox'>"+content+"</div>";
    content = $(content).remove("#CustomerDeliveryaddressId");
    $(".deliveryAddressesContainer").append(content);
  });

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

Hi , Replace $(content).remove("#CustomerDeliveryaddressId"); with $(content).remove("#CustomerDeliveryaddressId");

answered May 8, 2019 at 14:33

Selvakumar

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