所以我有这个拖放脚本。我想做的是保存拖动项目的位置。所以如果有人刷新页面。拖动的图像应保留在同一位置。我只想在 javascript 中执行此操作
.fill {
background-image: url('https://source.unsplash.com/random/150x150');
height: 150px;
width: 150px;
cursor: pointer;
}
.hold {
border: solid 5px #ccc;
}
.empty {
height: 56rem;
width: 36rem;
margin: 10px;
border: solid 3px salmon;
background: white;
}
.hovered {
background: #f4f4f4;
border-style: dashed;
}<!DOCTYPE html>
<html lang="en" dir="ltr">
<head>
<link rel="stylesheet" href="style.css" />
<link rel="stylesheet" href="https://stackpath.bootstrapcdn.com/bootstrap/4.1.3/css/bootstrap.min.css" integrity="sha384-MCw98/SFnGE8fJT3GXwEOngsV7Zt27NXFoaoApmYm81iuXoPkFOJwJ8ERdknLPMO" crossorigin="anonymous">
<meta charset="utf-8">
<title></title>
</head>
<body>
<div class="container">
<div class="row">
<div class="col-md-6">
<div class="empty">
<div class="fill" draggable="true"> </div>
</div>
</div>
<div class="col-md-6">
<div …我从开放的API获取用户数据，将其存储在本地存储中，可以显示所有数据，但是只有第一个值被保存在本地存储中。如何存储所有数据？
我的代码：
String.prototype.capitalize = function() {
return this.charAt(0).toUpperCase() + this.slice(1);
}
function createNode(element) {
return document.createElement(element);
}
function append(parent, element ) {
return parent.appendChild(element);
}
const ul = document.getElementById('authors');
const url = 'https://randomuser.me/api/?results=10';
fetch(url)
.then((resp) => resp.json())
.then(function(data) {
let authors = data.results;
return authors.map(function(author) {
const myObj = {
name: `${author.name.first}`,
lastname : `${author.name.last}`,
email : `${author.email}`,
location : `${author.location.city}, ${author.location.street}`,
phone : `${author.phone}`
}
let li = createNode('li'),
img = createNode('img'),
span = createNode('span');
let myObj_serialized = JSON.stringify(myObj); …