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

ExtJS method to determine IE browser type_extjs

WBOY
WBOYOriginal
2016-05-16 17:00:551360browse

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;
Statement:
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