Home > Web Front-end > JS Tutorial > ExtJS method to determine IE browser type_extjs

ExtJS method to determine IE browser type_extjs

WBOY
Release: 2016-05-16 17:00:55
Original
1388 people have browsed it

The code is under srccoreext.js in ext

The latest ext3.0beat1 code is as follows:

Copy the code The code is as follows:

ua = navigator.userAgent.toLowerCase(),
check = function(r){
return r.test(ua);
},
isStrict = document.compatMode == "CSS1Compat",
isOpera = check(/opera/),
isChrome = check(/chrome/),
isWebKit = check(/webkit/),
isSafari = !isChrome && check(/safari/),
isSafari3 = isSafari && check(/version/3/),
isSafari4 = isSafari && check(/version/4/),
isIE = !isOpera && check(/msie/),
isIE7 = isIE && check(/msie 7/),
isIE8 = isIE && check(/msie 8/),
isGecko = !isWebKit && check( /gecko/),
isGecko3 = isGecko && check(/rv:1.9/),
isBorderBox = isIE && !isStrict,
isWindows = check(/windows|win32/),
isMac = check(/macintosh|mac os x/),
isAir = check(/adobeair/),
isLinux = check(/linux/),
isSecure = /^https/i.test(window. location.protocol);

And under 2.2.1 (in sourcecoreext.js) it is
Copy code The code is as follows:

var ua = navigator.userAgent.toLowerCase();
var isStrict = document.compatMode == "CSS1Compat",
isOpera = ua.indexOf ("opera") > -1,
isChrome = ua.indexOf("chrome") > -1,
isSafari = !isChrome && (/webkit|khtml/).test(ua),
isSafari3 = isSafari && ua.indexOf('webkit/5') != -1,
isIE = !isOpera && ua.indexOf("msie") > -1,
isIE7 = !isOpera && ua.indexOf("msie 7") > -1,
isIE8 = !isOpera && ua.indexOf("msie 8") > -1,
isGecko = !isSafari && !isChrome && ua.indexOf ("gecko") > -1,
isGecko3 = isGecko && ua.indexOf("rv:1.9") > -1,
isBorderBox = isIE && !isStrict,
isWindows = (ua. indexOf("windows") != -1 || ua.indexOf("win32") != -1),
isMac = (ua.indexOf("macintosh") != -1 || ua.indexOf( "mac os x") != -1),
isAir = (ua.indexOf("adobeair") != -1),
isLinux = (ua.indexOf("linux") != -1 ),
isSecure = window.location.href.toLowerCase().indexOf("https") === 0;
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