检测用户是否已从Chrome扩展程序登录 [英] Detecting whether user is logged in or not from a Chrome extension

问题描述

是否有一种简单的方法可以从Chrome扩展程序中检测到用户是否已登录他的Google帐户。

我想有一种讨厌的方式这通过加载需要验证的html / css / js资源来完成。
但我想以清晰的方式做到这一点。

thx和最好的，
Viktor

解决方案

在铬扩展邮件列表中已经有一些讨论：
http://groups.google.com/a/chromium.org/group/chromium-extensions/ browse_thread / thread / 6e46a3a6e46d9110 /

基本上，根据用户的建议，您向Google.com发送Xml Http请求，并通过正则表达式搜索，如果当前用户已登录：

来源：Guillaume Boudreau（位于铬扩展邮件列表）

  var currentUser; 
 var xhr = new XMLHttpRequest（）; 
 xhr.onreadystatechange = function（data）{
 if（xhr.readyState == 4）{
 currentUser = null; 
 if（xhr.status == 200）{
 var re = new RegExp（/< b class =？gb4？> [\s] *（[^ <+] @ [^<] +）< \ / b个/ I）; 
 var m = re.exec（xhr.responseText）; 
 if（m& m.length == 2）{
 currentUser = m [1]; 
 
 
 console.log（当前记录的用户（在google.com上）：+ 
 currentUser）; 
} 
}; 
 xhr.open（'GET'，'https://www.google.com/'，false）; 
 xhr.send（）; 
  

is there an easy way to detect from a Chrome extension that the user is logged into his google account or not.

I guess there is a nasty way to figure this out by loading a html/css/js resource that requires authentication. But i'd like to do this in a "clear" way.

thx and best, Viktor

解决方案

there has been some discussions on that within the chromium-extensions mailing list: http://groups.google.com/a/chromium.org/group/chromium-extensions/browse_thread/thread/6e46a3a6e46d9110/

Basically, as a user suggested, you send an Xml Http Request to google.com, and search through regex, if a current user is signed in:

Source: Guillaume Boudreau (on chromium-extensions mailing list)

var currentUser;
var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function(data) {
       if (xhr.readyState == 4) {
               currentUser = null;
               if (xhr.status == 200) {
                       var re = new RegExp(/<b class="?gb4"?>[\s]*([^<]+@[^<]+)<\/b>/i);
                       var m = re.exec(xhr.responseText);
                       if (m && m.length == 2) {
                               currentUser = m[1];
                       }
               }
               console.log("Currently logged user (on google.com): " +
currentUser);
       }
};
xhr.open('GET', 'https://www.google.com/', false);
xhr.send();

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