홈>2020-02-01 01:19

javascript - 반복되는 If 문을 개선하려면 어떻게해야합니까?

스크롤 효과와 성공을 작성하려고했지만 더 잘 될 것이라고 생각하고 이러한 코드를 단순화하기 위해 조언이 필요합니다.

var nowIndex = 0;
var pcMenuSelect = $('body').find('.carFormatMenu li');
var mbMenuSelect = $('body').find('.menu-cartype .item');
var dropText = mbMenuSelect.parents('.dropdown').find('.text');
var dropValue = mbMenuSelect.parents('.dropdown').find('input')
if (nowIndex == 0){
    dropText.text(mbMenuSelect.eq(0).text());
    pcMenuSelect.removeClass('current');
    pcMenuSelect.eq(0).addClass('current');
    mbMenuSelect.removeClass('active selected');
    mbMenuSelect.eq(0).addClass('active selected');
    dropValue.eq(0).val(mbMenuSelect.eq(0).text());
}
if (nowIndex == 1){
    dropText.text(mbMenuSelect.eq(1).text());
    pcMenuSelect.removeClass('current');
    pcMenuSelect.eq(1).addClass('current');
    mbMenuSelect.removeClass('active selected');
    mbMenuSelect.eq(1).addClass('active selected');
    dropValue.eq(1).val(mbMenuSelect.eq(1).text());
}
if (nowIndex == 2){
    dropText.text(mbMenuSelect.eq(2).text());
    pcMenuSelect.removeClass('current');
    pcMenuSelect.eq(2).addClass('current');
    mbMenuSelect.removeClass('active selected');
    mbMenuSelect.eq(2).addClass('active selected');
    dropValue.eq(2).val(mbMenuSelect.eq(2).text());
}
if (nowIndex == 3){
    dropText.text(mbMenuSelect.eq(3).text());
    pcMenuSelect.removeClass('current');
    pcMenuSelect.eq(3).addClass('current');
    mbMenuSelect.removeClass('active selected');
    mbMenuSelect.eq(3).addClass('active selected');
    dropValue.eq(3).val(mbMenuSelect.eq(3).text());
}
if (nowIndex == 4){
    dropText.text(mbMenuSelect.eq(4).text());
    pcMenuSelect.removeClass('current');
    pcMenuSelect.eq(4).addClass('current');
    mbMenuSelect.removeClass('active selected');
    mbMenuSelect.eq(4).addClass('active selected');
    dropValue.eq(4).val(mbMenuSelect.eq(4).text());
}
if (nowIndex == 5){
    dropText.text(mbMenuSelect.eq(5).text());
    pcMenuSelect.removeClass('current');
    pcMenuSelect.eq(5).addClass('current');
    mbMenuSelect.removeClass('active selected');
    mbMenuSelect.eq(5).addClass('active selected');
    dropValue.eq(5).val(mbMenuSelect.eq(5).text());
}
if (nowIndex == 6){
    dropText.text(mbMenuSelect.eq(6).text());
    pcMenuSelect.removeClass('current');
    pcMenuSelect.eq(6).addClass('current');
    mbMenuSelect.removeClass('active selected');
    mbMenuSelect.eq(6).addClass('active selected');
    dropValue.eq(6).val(mbMenuSelect.eq(6).text());
}
if (nowIndex == 7){
    dropText.text(mbMenuSelect.eq(7).text());
    pcMenuSelect.removeClass('current');
    pcMenuSelect.eq(7).addClass('current');
    mbMenuSelect.removeClass('active selected');
    mbMenuSelect.eq(7).addClass('active selected');
    dropValue.eq(7).val(mbMenuSelect.eq(7).text());
}

javascript

답변 # 1

LalithKumar의 답변 사용 체인에 추가.

var nowIndex = 0;
    var pcMenuSelect = $('body').find('.carFormatMenu li');
    var mbMenuSelect = $('body').find('.menu-cartype .item');
    var dropText = mbMenuSelect.parents('.dropdown').find('.text');
    var dropValue = mbMenuSelect.parents('.dropdown').find('input');
if (nowIndex>0){
    dropText.text(mbMenuSelect.eq(nowIndex).text());
    pcMenuSelect.removeClass('current').eq(nowIndex).addClass('current');
    mbMenuSelect.removeClass('active selected').eq(nowIndex).addClass('active selected');
    dropValue.eq(nowIndex).val(mbMenuSelect.eq(nowIndex).text());
}

답변 # 2

다음과 같이하십시오 :

var nowIndex = 0;
    var pcMenuSelect = $('body').find('.carFormatMenu li');
    var mbMenuSelect = $('body').find('.menu-cartype .item');
    var dropText = mbMenuSelect.parents('.dropdown').find('.text');
    var dropValue = mbMenuSelect.parents('.dropdown').find('input');
if (nowIndex>0){
    dropText.text(mbMenuSelect.eq(nowIndex).text());
    pcMenuSelect.removeClass('current');
    pcMenuSelect.eq(nowIndex).addClass('current');
    mbMenuSelect.removeClass('active selected');
    mbMenuSelect.eq(nowIndex).addClass('active selected');
    dropValue.eq(nowIndex).val(mbMenuSelect.eq(nowIndex).text());
}

답변 # 3

Dinesh의 답변에 추가 :

var nowIndex = 0;
    var pcMenuSelect = $('body').find('.carFormatMenu li');
    var mbMenuSelect = $('body').find('.menu-cartype .item');
    var dropMenu=mbMenuSelect.parents('.dropdown'); 
    if (nowIndex> 0){
        dropMenu.find('.text').text(mbMenuSelect.eq(nowIndex).text());
        pcMenuSelect.removeClass('current').eq(nowIndex).addClass('current');
        mbMenuSelect.removeClass('active selected').eq(nowIndex).addClass('active selected');
        dropMenu.find('input').eq(nowIndex).val(mbMenuSelect.eq(nowIndex).text());
    }

mbMenuSelect의 캐싱을 완료했습니다. 그러나$( 'body')은 아닙니다. 찾기 기능을 다시 실행하면 실행 속도가 느려지기 때문입니다.

답변 # 4

공통 기능을 캡슐화하는 기능을 만드십시오.

var pcMenuSelect = $('body').find('.carFormatMenu li')
var mbMenuSelect = $('body').find('.menu-cartype .item')
var dropText = mbMenuSelect.parents('.dropdown').find('.text')
var dropValue = mbMenuSelect.parents('.dropdown').find('input')
function updateSelect(index) {
  dropText.text(mbMenuSelect.eq(index).text())
  pcMenuSelect
    .removeClass('current')
    .eq(index)
    .addClass('current')
  mbMenuSelect
    .removeClass('active selected')
    .eq(index)
    .addClass('active selected')
  dropValue.eq(index).val(mbMenuSelect.eq(index).text())
}
// call the function with the index
updateSelect(nowIndex)

Python2.net

javascript - 반복되는 If 문을 개선하려면 어떻게해야합니까?

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