在JavaScript中检测IE版本(v9之前)

如果我们网站的用户使用的是v9之前版本的Internet Explorer，我希望将他们弹出到一个错误页面。不值得我们花时间和金钱去支持iev9之前的版本。所有其他非ie浏览器的用户都没问题，不应该被弹出。以下是提议的代码:

if(navigator.appName.indexOf("Internet Explorer")!=-1){     //yeah, he's using IE
    var badBrowser=(
        navigator.appVersion.indexOf("MSIE 9")==-1 &&   //v9 is ok
        navigator.appVersion.indexOf("MSIE 1")==-1  //v10, 11, 12, etc. is fine too
    );

    if(badBrowser){
        // navigate to error page
    }
}

这段代码能行吗?

为了阻止一些可能会出现在我面前的评论:

Yes, I know that users can forge their useragent string. I'm not concerned. Yes, I know that programming pros prefer sniffing out feature-support instead of browser-type but I don't feel this approach makes sense in this case. I already know that all (relevant) non-IE browsers support the features that I need and that all pre-v9 IE browsers don't. Checking feature by feature throughout the site would be a waste. Yes, I know that someone trying to access the site using IE v1 (or >= 20) wouldn't get 'badBrowser' set to true and the warning page wouldn't be displayed properly. That's a risk we're willing to take. Yes, I know that Microsoft has "conditional comments" that can be used for precise browser version detection. IE no longer supports conditional comments as of IE 10, rendering this approach absolutely useless.

还有其他明显需要注意的问题吗?

当前回答

我为此做了一个方便的下划线混合。

_.isIE();        // Any version of IE?
_.isIE(9);       // IE 9?
_.isIE([7,8,9]); // IE 7, 8 or 9?

_.mixin({ isIE: function(mixed) { if (_.isUndefined(mixed)) { mixed = [7, 8, 9, 10, 11]; } else if (_.isNumber(mixed)) { mixed = [mixed]; } for (var j = 0; j < mixed.length; j++) { var re; switch (mixed[j]) { case 11: re = /Trident.*rv\:11\./g; break; case 10: re = /MSIE\s10\./g; break; case 9: re = /MSIE\s9\./g; break; case 8: re = /MSIE\s8\./g; break; case 7: re = /MSIE\s7\./g; break; } if (!!window.navigator.userAgent.match(re)) { return true; } } return false; } }); console.log(_.isIE()); console.log(_.isIE([7, 8, 9])); console.log(_.isIE(11)); <script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.8.3/underscore-min.js"></script>

2016-05-23 19:11:34

其他回答

它帮助了我

function IsIE8Browser() {
  var rv = -1;
  var ua = navigator.userAgent;
  var re = new RegExp("Trident\/([0-9]{1,}[\.0-9]{0,})");
  if (re.exec(ua) != null) {
    rv = parseFloat(RegExp.$1);
  }
  return (rv == 4);
}

2018-01-30 07:30:50

我喜欢这样:

<script>
   function isIE () {
       var myNav = navigator.userAgent.toLowerCase();
       return (myNav.indexOf('msie') != -1) ? parseInt(myNav.split('msie')[1]) : false;
   }    
   var ua = window.navigator.userAgent;
   //Internet Explorer | if | 9-11

   if (isIE () == 9) {
       alert("Shut down this junk! | IE 9");
   } else if (isIE () == 10){
       alert("Shut down this junk! | IE 10");
   } else if (ua.indexOf("Trident/7.0") > 0) {
       alert("Shut down this junk! | IE 11");
   }else{
       alert("Thank god it's not IE!");
   }

</script>

2015-08-21 18:23:47

检测IE及其版本再容易不过了，你所需要的只是一点原生/香草Javascript:

var uA = navigator.userAgent;
var browser = null;
var ieVersion = null;

if (uA.indexOf('MSIE 6') >= 0) {
    browser = 'IE';
    ieVersion = 6;
}
if (uA.indexOf('MSIE 7') >= 0) {
    browser = 'IE';
    ieVersion = 7;
}
if (document.documentMode) { // as of IE8
    browser = 'IE';
    ieVersion = document.documentMode;
}

下面是一种用法:

if (browser == 'IE' && ieVersion <= 9) 
    document.documentElement.className += ' ie9-';

适用于所有IE版本，包括更高版本的低兼容性视图/模式，documentMode是IE专有的。

2014-06-18 13:45:46

var isIE9OrBelow = function()
{
   return /MSIE\s/.test(navigator.userAgent) && parseFloat(navigator.appVersion.split("MSIE")[1]) < 10;
}

2015-04-24 02:07:41

你的代码可以做检查，但正如你所认为的，如果有人试图使用IE v1或> v19访问你的页面将不会得到错误，所以可能更安全的做检查的Regex表达式如下代码:

var userAgent = navigator.userAgent.toLowerCase();
// Test if the browser is IE and check the version number is lower than 9
if (/msie/.test(userAgent) && 
    parseFloat((userAgent.match(/.*(?:rv|ie)[\/: ](.+?)([ \);]|$)/) || [])[1]) < 9) {
  // Navigate to error page
}

2012-06-09 22:25:10

在JavaScript中检测IE版本(v9之前)

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