Example of jumping back to the original page you want to visit after logging in in php

Example of jumping back to the original page you want to visit after logging in in php_PHP tutorial

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Release： 2016-07-13 16:57:06

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In many websites, users first visit a page to log in, but they did not log in at that time and then logged in. After waiting for the user to log in successfully, they definitely want to return to the page they last visited. Now I will introduce to you how to jump back after logging in. It turns out that the easiest way to access the page instance

is to directly use php $_SERVER['HTTP_REFERER']

If I want to log in on the A.php page

Now jump to the B.php page, we only need to add the following code to b.php

The code is as follows

代码如下	复制代码
$url = $_SERVER['HTTP_REFERER']; header("location:$url");

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$url = $_SERVER['HTTP_REFERER'];

header("location:$url");

代码如下	复制代码
protected function checkLogin() { if (没有登录){ $thisurl = "http://".$_SERVER["HTTP_HOST"].$_SERVER['PHP_SELF'];//当前URL $thisurl = urlencode($thisurl);//这里要注意需要把获取到的url转码，不然后面不好传递URL redirect("http://".$_SERVER["HTTP_HOST"]."/cityosweb/default.php/Index/login?url=".$thisurl); } }

However, the above method has many shortcomings, such as parameters, etc., but under the IE browser, if you jump through the location of js, then this value cannot be obtained.

代码如下	复制代码
$this->checkLogin();

Now I will make a comprehensive one.

First create a method to determine whether you are logged in, if not logged in

if (not logged in){ $thisurl = "http://".$_SERVER["HTTP_HOST"].$_SERVER['PHP_SELF'];//Current URL $thisurl = urlencode($thisurl);//It should be noted here that the obtained url needs to be transcoded, otherwise it will be difficult to pass the URL later

The code is as follows

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代码如下	复制代码
public function login() { $url = $_GET['url']; $this->assign('url',$url); $this->assign('title','Login'); $this->display('user/reg_new.html'); }

protected function checkLogin() {

redirect("http://".$_SERVER["HTTP_HOST"]."/cityosweb/default.php/Index/login?url=".$thisurl); } }