Jquery remove script bad coding

Question

What is a good way to write this script more dynamic:

I have a variable which can have a value up to 18.

If $count is 1 I want to remove everything up to 18.

If $count is 2 I want to remove everything up to 18.

And so on.

        if($count == 1) {
            $('#input-1').remove();
            // remove up to 18
        }
        if($count == 2) {
            $('#input-2').remove();
            // remove up to 18
        }
        if($count == 3) {
            $('#input-3').remove();
            // remove up to 18
        }
        ...
        if($count == 18) {
            $('#input-18').remove();
        }

I did this:

            for(var i = 0; i < 18; i++) {
                if($count== i) {
                    $('#input-'+i).remove();
                }
            }

but it does not remove the divs from i to 18

EDIT: (to be clear)

I want to check each, like $count == i then I want to remove from i to 18

Your question is a bit odd. Can you please add the HTML too? — Ionut Necula
– Ionut Necula, Commented Mar 22, 2017 at 12:22

Ousmane D. · Accepted Answer · 2017-03-22 12:46:53Z

2

This should do:

for(var i = $count; i <= 18;i++){
   $('#input-'+i).remove();
}

EDIT

I want to check each, like $count == i then I want to remove from i to 18

The outer for loop allows us to check if i is equal to $count, if the expression is satisfied (true) then we simply begin another loop starting at i. Within the second for loop, we begin removing the elements from i and beyond.

for(var i = 1; i <=18; i++) {
    if($count == i) {
       for(var j = i; j <= 18; j++){
         $('#input-'+j).remove();
       }
      return;
    }
}

Note - This will send a return value of undefined to the caller. However, you can specify a different return value if you wish to.

edited Mar 22, 2017 at 12:46

answered Mar 22, 2017 at 12:25

Ousmane D.

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8 Comments

J.Alan Over a year ago

but I want to check each like $count == i then remove from i to 18

Ousmane D. Over a year ago

@J.Alan okay i will do another solution for that if you like.

Ousmane D. Over a year ago

@J.Alan just do me a favour and update your question with the comment you've just written --> " but I want to check each like $count == i then remove from i to 18"

J.Alan Over a year ago

Seems to work so far I will test it again after a few hours then I am going to tick your answer :D

J.Alan Over a year ago

Thanks alot everything is clear to me now, have a nice week!

|

j08691 · Accepted Answer · 2017-03-22 12:26:13Z

1

Something like this $("#input-"+$count).remove()

edited Mar 22, 2017 at 12:26

j08691

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answered Mar 22, 2017 at 12:23

diavolic

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2 Comments

dgo Over a year ago

Based upon OP's stated intention - this makes the most sense.

dgo Over a year ago

Okay - now the question was edited - this no longer works

Weedoze · Accepted Answer · 2017-03-22 12:25:12Z

0

You can use a simple for loop

for(let i=$count;i<=18;i++){
    $('#input-'+i).remove();
}

This will iterate starting from $count to 18 and will remove all input with id between $count and 18

answered Mar 22, 2017 at 12:25

Weedoze

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1 Comment

J.Alan Over a year ago

but I want to check each like $count == i then remove from i to 18

dschu · Accepted Answer · 2017-03-22 12:31:29Z

The easiest method would be a loop with a variable using Template-Strings

const addElements = function() {
  // Add elements dynamically
  for (let i = 0; i < 10; i++) {
    const div = `<div id="element-${i}">I\'m ${i}</div>`;
    $('body').append(div);
  }
}

const removeElements = function() {
  // Remove elements dynamically
  for (let i = 0; i < 10; i++) {
    const $div = $(`div#element-${i}`);

    $div.remove();
  }
}


$('#addBtn').on('click', function() {
  addElements();
});

$('#removeBtn').on('click', function() {
  removeElements();
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<button id="removeBtn">Remove</button>
<button id="addBtn">Add</button>

Kakashi Hatake · Accepted Answer · 2017-03-22 12:32:43Z

0

Use the following.!

for (var i = $count; i <= 18; i++) {
  $('#input-' + i).remove();
}

edited Mar 22, 2017 at 12:32

answered Mar 22, 2017 at 12:24

Kakashi Hatake

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2 Comments

J.Alan Over a year ago

but I want to check each like $count == i then remove from i to 18

Kakashi Hatake Over a year ago

@Andreas noted. Thanks.. (y)

dgo · Accepted Answer · 2017-03-22 12:40:18Z

0

I believe you want this:

var start=$('input-'+$count);
var end=$('#input-18').index();
$('input').slice(start,end).remove();

If the only inputs you have are $('input-1) to $('#input-18'), you can skip the check and do:

  $('input').slice($count-1).remove();

answered Mar 22, 2017 at 12:40

dgo

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