php中判断解决方案
php中判断
有一个a页面,页面有如下内容。
通过meta跳转到另一个页面b.php
1,打开a页面后进入b.php
2,直接打开b.php
这两种方式都是打开了b.php
问题
在b.php页面中怎样判断是第一种方式还是第二种方式打开的
------解决方案--------------------
可以使用jquery cookie判断
一个插件:https://github.com/carhartl/jquery-cookie
A页面
$(document).ready(function(){
$.cookie('自定义cookie名', '');//A页面加载时,清空cookie,方便重复使用。
}
$('a').click(function(){
$.cookie('自定义cookie名', '自定义cookie值');//点击超链,设置cockie
}
B页面,
if($_COOKIE['自定义cookie名']=='自定义cookie值'){
//条件2,直接打开b.php
}else{
//条件1,meta跳转
}
------解决方案--------------------
试了很多浏览器看
发现 chrome,pera,afari可以请求头的Referer来判断来源,ie ff都不可以...
如果是点击a连接过去
测试了一下 Referer都有记录
------解决方案--------------------
php header是获取不了referer的
但是js和meta跳转是可以获取的,可以使用server.referer来确定
如楼上说的有的browser不行,我就不确定了。
------解决方案--------------------
可通过 $_SERVER['HTTP_REFERER'] 进行判断
------解决方案--------------------
用js判断下来源不就可以了
------解决方案--------------------
一向不用Referer做关键判断,只做参考数据
如果A和B都是自己可控的话,用session做关键判断
如果A不可控或者不可信的话,不要以这个思路做重要的后续工作判断
------解决方案--------------------
有个笨方法

------解决方案--------------------
同意4楼。。。
------解决方案--------------------
a连到b,但b php header到c,
在c中,获取不到b,获到的是a
------解决方案--------------------
给个参数标识下

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