Question

有一个网站允许＆＃34;关注＆＃34;和喜欢＆＃34; Instagram的页面没有访问令牌。它有一个按钮，当你点击它时，它会打开某个人的Instagram页面的新窗口，如果你关注他/她并关闭Instagram页面，他们就会明白你已经关注了这个页面。这怎么可能？

以下是他们的示例代码：

<script>
    function func(){
        FBLL_671661 = window.open('getulr.php?i=JTIxMTglRTQlREUlMTAlOTElODQlMTF6JUUzJTg5JTIwWiUzRCUzQyVDNSVCMDclOUQ=','FB_671661',status=0,toolbar=0,width=550,height=195);
        if(FBLL_671661.closed){
            confirmSingleFB(671661, "simeumesmo", "31084ff3990fe6142306251b671661", "19", 0, "UV9W49wV%2FowWe%2BaHIkEwMsDaWfvZfnZkmRS8XOk6GlyALO%2Bpg8Q%2BwWhyXUA%3D", "10","simeumesmo");
        }
    }

    function confirmSingleFB(id,url,linkId,network,isPriority,code,cpc,link_title){
        url=encodeURIComponent(url);
        $.ajax({
            type:"POST",
            url:"./ajax.php",
            data:'act=checkFollowed&params={"id":"'+id+'","url":"'+url+'", "network":"'+network+'", "link_id":"'+linkId+'", "IXY5pZpE":"'+code+'", "cpc":"'+cpc+'", "title":"'+link_title+'"}',
            success:function(result){
                $(container).html(result);}
        });
    }
</script>


<button onclick=func()> follow </button>

simeumesmo是必须遵循的人的用户名。显然我不知道geturl.php和ajax.php背后是什么

得到Instagram＆喜欢/按下按钮响应

0 个答案: