PHP determines whether the webpage is loaded by Android webview

WBOY
Release: 2016-07-28 08:26:43
Original
1514 people have browsed it

一. Idea:
1. Android uses webview to load web pages, and sets its userAgent specific characters on the android side
2. Server-side PHP determines whether the obtained user agent contains specific characters given by Android
3. If there is, load the webpage for Android webview

II. Code
Android side requires settings:
(These two lines of code: Find the userAgent on the Android side, change the Android string to APP_WEBVIEW Android, APP_WEBVIEW is the sign for php)

<code>String userAgent = webView<span>.getSettings</span>()<span>.getUserAgentString</span>()<span>;//找到webview的useragent</span>
webView<span>.getSettings</span>()<span>.setUserAgentString</span>(userAgent<span>.replace</span>(<span>"Android"</span>, <span>"APP_WEBVIEW Android"</span>))<span>;//在useragent上添加APP_WEBVIEW 标识符,服务器会获取该标识符进行判断</span></code>
Copy after login

php server side:

<code><span><span><?php</span><span>if</span>(strpos(<span>$_SERVER</span>[<span>'HTTP_USER_AGENT'</span>],<span>'APP_WEBVIEW'</span>) !== <span>false</span>){
    <span>echo</span><span>"<script>alert('当前网页是Android webview 加载');</script>"</span>;  
}<span>else</span>{
    <span>echo</span><span>"<script>alert('当前网页不是Android webview 加载');</script>"</span>; 
}
<span>?></span></span></code>
Copy after login

Three. Description
If it is iOS, the code judgment on the PHP side is the same. You only need to add the "APP_WEBVIEW" string to the userAgent on the iOS side

').addClass('pre-numbering').hide(); $(this).addClass('has-numbering').parent().append($numbering); for (i = 1; i ').text(i)); }; $numbering.fadeIn(1700); }); });

The above introduces how PHP determines whether a web page is loaded by Android webview, including the content. I hope it will be helpful to friends who are interested in PHP tutorials.

Related labels:
source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template