我做了一个简单的Android应用程序与本地菜单和我的网站内容的webview。我需要检测我的网站是否包含在一个webview中，以隐藏菜单栏。
经过长时间的研究，我找到了这样的方法:
if($_SERVER['HTTP_X_REQUESTED_WITH'] == "myAppPackage"){
//the site is included in webview
}
这个解决方案适用于很多设备，但是对于Galaxy S4 Mini(Android 4.2.2)这个变量是空白的!
其他http头变量:
- 路径 /usr/地方/bin:/usr/bin:/bin
- UNIQUE_ID U6LV8AAAEAABFtKXcAAADY
- HTTP_HOST 主机名
- HTTP_ACCEPT text/html, application/xhtml + xml应用程序/xml; q = 0.9 /; q = 0.8
- HTTP_USER_AGENT Mozilla/5.0 (Windows NT 6.1;WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/27.0.1453.116Safari/537.36
- HTTP_REFERER http://webx225.aruba.it/CP/index.php
- HTTP_ACCEPT_ENCODING gzip、缩小sdch
- HTTP_ACCEPT_LANGUAGE葡萄酒,,q = 0.8, en - us; q = 0.6, en; q = 0.4
- HTTP_COOKIE PHPSESSID = 5 deb3527097d0f767aba45d0aa042;acopendivids =没有什么结果;acgroupswithpersist =没有什么结果;wooTracker = lzp5HpPQNDCd;__utma = 250126209.968543423.1402584011.1403102882.1403106109.27;__utmc = 250126209;__utmz = 250126209.1402752662.8.2.utmcsr =谷歌| utmccn =(有机)| utmcmd =有机| utmctr =(不提供% 20)
- HTTP_CACHE_CONTROL max-stale = 0
- HTTP_CONNECTION维生
- HTTP_X_BLUECOAT_VIA b1ae316f3a2874e7
- SERVER_SIGNATURE无值
- SERVER_SOFTWARE Apache/2.2
- SERVER_NAME 主机名
- SERVER_ADDR 62.149.140.235
- SERVER_PORT 80
- REMOTE_ADDR 62.249.32.77
- DOCUMENT_ROOT /web/根/主机名/home/
- SERVER_ADMIN postmaster@hostname
- SCRIPT_FILENAME/web/根/主机名/home/ aruba__php__test 。php
- REMOTE_PORT 43924
- GATEWAY_INTERFACE CGI/1.1
- SERVER_PROTOCOL HTTP/1.1
- REQUEST_METHOD得到
- QUERY_STRING无值
- REQUEST_URI / aruba__php__test 。php
- SCRIPT_NAME / aruba__php__test 。php
- PHPRC no value
多亏了greenapps的想法，这是最终的解决方案。
Android APP MainActivity:
public View onCreateView
....
WebView webView = (WebView) rootView.findViewById(R.id.my_webview);
String agentModified = webView.getSettings().getUserAgentString().concat(" MobileApplication(mypackage)");
webView.getSettings().setUserAgentString(agentModified);
:
if(strpos($_SERVER['HTTP_USER_AGENT'], 'com.vivicrema.android') !== false)
$isMobileApp = true;
真是太神奇了!