javascript - 刷新页面时本地存储加载投票(损坏)

Question

当我刷新页面时，我无法显示 voteUp/Down，因为如果我执行 voteUp/Down(+1 或 -1) 并刷新页面，这会再次返回 voteUp/Down (0)。过去我使用 JSON，但社区推荐不使用 JSON。所以我在这一刻拥有了它。谢谢。

    var voteUp = document.getElementById('vote-up');
var voteDown = document.getElementById('vote-down');

var handUp = once(function() {
    var total = Number(voteUp.innerHTML);
    total += 1;
    voteUp.innerHTML = total;
    saveVote();
});

voteUp.addEventListener('click', handUp);

var handDown = once(function() {
    var total = Number(voteDown.innerHTML);
    total -= 1;
    voteDown.innerHTML = total;
    saveVote();
});

voteDown.addEventListener('click', handDown);

function once(fn, context) {
    var result;

        return function() {
            if(fn) {
                result = fn.apply(context);
                fn = null;
            }
            return result;
        };
}

function saveVote() {
    var votes = voteUp, voteDown;
        localStorage.setItem('data', votes);
            console.log('saveVote');
}

function loadVote() {
    var votes = localStorage.getItem('data');
        if(!votes){
            return;
        }
            console.log(localStorage.getItem('data'));
}

loadVote();

Answer 1

var voteUp = document.getElementById('vote-up');

function handUp() {
   var total = Number(voteUp.innerHTML);
   total += 1;
   voteUp.innerHTML = total;
}

voteUp.addEventListener('click',handUp,false);

它通过单​​击来增加每个值。