我在使用LocalStorage的计算机上使用JSON.parse作为简单数据库。 它工作顺利,直到我正在检查这个“数据库”; 下面是用于向LocalStorage输入信息的代码:
var users = JSON.parse(localStorage.registeredUsers);
users.push({username:name, password:userpass, connected:false});
localStorage.registeredUsers = JSON.stringify(users);
当我'检查那些注册用户时,我收到错误“Uncaught SyntaxError:Unexpected token u”:
var users = JSON.parse(localStorage.registeredUsers);
if(users[userindex].connected)
{.........}
错误指向JSON.parse行。 我试图通过一些类似的主题弄清楚但却无法找到方法。
我推入localstorage数组的代码:
function regBtn(event)
{
event.preventDefault();
name=document.forms["regform"]["username"].value;
userpass=document.forms["regform"]["password"].value;
localStorage.username=name;
localStorage.password=userpass;
if(!(localStorage.registeredUsers))
{
localStorage.registeredUsers = '[]';
}
var users = JSON.parse(localStorage.registeredUsers);
users.push({username:name, password:userpass, connected:false});
localStorage.registeredUsers = JSON.stringify(users);
$('#mainContent').load('HomePage.html');
}
答案 0 :(得分:7)
尝试
var users = localStorage.registeredUsers? JSON.parse(localStorage.registeredUsers) : [];
或者如果您不喜欢三元运算符,
var users=[];
if(localStorage.registeredUsers){
users=JSON.parse(localStorage.registeredUsers);
}
可能有帮助
答案 1 :(得分:0)
function setCookies(){
var cookiesObjects = {};
cookiesObjects.name = document.getElementById('usr').value;
cookiesObjects.email = document.getElementById('email').value;
cookiesObjects.age = document.getElementById('age').value;
var actualString = JSON.stringify(cookiesObjects);
document.cookie = "user="+ actualString;
}
function getCookies(){
var nameValueArray = document.cookie.split("=");
var cookieObject = JSON.parse(nameValueArray[1]);
document.getElementById('usr').value=cookieObject.name;
document.getElementById('email').value=cookieObject.email;
document.getElementById('age').value=cookieObject.age;
}
function clearCookies(){
document.getElementById('usr').value= "";
document.getElementById('email').value= "";
document.getElementById('age').value= "";
}
and saying Error : Uncaught SyntaxError: Unexpected token u
我无法确定错误的原因。