如何构造if ... else语句以返回true

Question

Full code https://codepen.io/3noki/pen/xjyErQ

function checkForMatch() {
  if ( $(openedCardsList[0]).is(openedCardsList[1]) ) {
  $(openedCardsList[0]).removeClass('open show').addClass('match');
  $(openedCardsList[1]).toggleClass('open show');
  $(openedCardsList[1]).toggleClass('match');
  $(openedCardsList) = [];
  }

  else {unFlip()}
}

This code currently only returns false values, I would like to know how to structure this in either pure DOM or pure jquery format to be able to return a true value.

openedCardsList[0]===openedCardsList[1]

Has not returned true either

When the card at array 0 and 1 are the same I want this value to be returned true.

Answer 1

How about checking/matching against the (FontAwesome) icon class names?

So when creating the card (or generating its HTML), you'd add fa to the element's "data" (using jQuery.data() ), like this:

function createCardHTML() {
  symbol = cards.children("i");
  symbol.each(function(index, item) {
    $(item).addClass(cardsList[index])
      // If the icon's class name is `fa-diamond`, `$(item).data('fa')` would be `diamond`.
      .data('fa', cardsList[index].substring(3)); // <- here's the data
  });
  return symbol;
}

And then, in checkForMatch() :

var b = $('i.fa', openedCardsList[0]).data('fa');
var c = $('i.fa', openedCardsList[1]).data('fa');
var eq = ( b && c && b === c );

Demo