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

如微软参考页所述，IE版本10不再支持条件注释。

然后使用:

if((ieDetector.isIE) && (ieDetector.Version <= 9))
{

}

2014-10-07 03:20:59

其他回答

轻松检测MSIE (v6 - v7 - v8 - v9 - v10 - v11):

if (navigator.userAgent.indexOf('MSIE') !== -1 || navigator.appVersion.indexOf('Trident/') > 0) {
   // MSIE
}

2013-12-05 21:58:05

返回IE版本，否则返回false

function isIE () {
  var myNav = navigator.userAgent.toLowerCase();
  return (myNav.indexOf('msie') != -1) ? parseInt(myNav.split('msie')[1]) : false;
}

例子:

if (isIE () == 8) {
 // IE8 code
} else {
 // Other versions IE or not IE
}

if (isIE () && isIE () < 9) {
 // is IE version less than 9
} else {
 // is IE 9 and later or not IE
}

if (isIE()) {
 // is IE
} else {
 // Other browser
}

2013-04-13 01:01:53

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

这是我喜欢的做法。它提供了最大限度的控制。(注意:条件语句仅在IE5 - 9中支持。)

首先正确地设置ie类

<!DOCTYPE html>
<!--[if lt IE 7]> <html class="lt-ie9 lt-ie8 lt-ie7"> <![endif]-->
<!--[if IE 7]>    <html class="lt-ie9 lt-ie8"> <![endif]-->
<!--[if IE 8]>    <html class="lt-ie9"> <![endif]-->
<!--[if gt IE 8]><!--> <html> <!--<![endif]-->    
<head>

然后你可以使用CSS来创建样式异常，或者，如果你需要，你可以添加一些简单的JavaScript:

(function ($) {
    "use strict";

    // Detecting IE
    var oldIE;
    if ($('html').is('.lt-ie7, .lt-ie8, .lt-ie9')) {
        oldIE = true;
    }

    if (oldIE) {
        // Here's your JS for IE..
    } else {
        // ..And here's the full-fat code for everyone else
    }

}(jQuery));

感谢Paul Irish。

2012-06-09 22:38:20

在JavaScript中检测IE版本(v9之前)

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