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

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

其他回答

使用条件注释。您正在尝试检测IE < 9的用户，条件注释将在这些浏览器中工作;在其他浏览器(IE >= 10和非IE)中，注释将被视为正常的HTML注释，这就是它们的本质。

示例HTML:

<!--[if lt IE 9]>
WE DON'T LIKE YOUR BROWSER
<![endif]-->

如果你需要，你也可以用脚本来完成:

var div = document.createElement("div");
div.innerHTML = "<!--[if lt IE 9]><i></i><![endif]-->";
var isIeLessThan9 = (div.getElementsByTagName("i").length == 1);
if (isIeLessThan9) {
    alert("WE DON'T LIKE YOUR BROWSER");
}

2012-06-09 23:03:19

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

要检测Internet Explorer 10|11，你可以在body标签后立即使用这个小脚本:

在我的情况下，我使用jQuery库加载头部。

<!DOCTYPE HTML>
<html>
<head>
    <script src="//code.jquery.com/jquery-1.11.0.min.js"></script>
</head>
<body>
    <script>if (navigator.appVersion.indexOf('Trident/') != -1) $("body").addClass("ie10");</script>
</body>
</html>

2014-08-08 21:28:17

这种检测IE的方法结合了jKey使用条件注释的答案和Owen使用用户代理的答案的优点，并避免了它们的缺点。

jKey's approach works up to version 9 and immune to user agent spoofing in IE 8 & 9. Owen's approach can fail on IE 5 & 6 (reporting 7) and is susceptible to UA spoofing, but it can detect IE versions >= 10 (now also including 12, which postdates Owen's answer). // ---------------------------------------------------------- // A short snippet for detecting versions of IE // ---------------------------------------------------------- // If you're not in IE (or IE version is less than 5) then: // ie === undefined // Thus, to detect IE: // if (ie) {} // And to detect the version: // ie === 6 // IE6 // ie > 7 // IE8, IE9 ... // ---------------------------------------------------------- var ie = (function(){ var v = 3, div = document.createElement('div'), all = div.getElementsByTagName('i'); while ( div.innerHTML = '', all[0] ); if (v <= 4) { // Check for IE>9 using user agent var match = navigator.userAgent.match(/(?:MSIE |Trident\/.*; rv:|Edge\/)(\d+)/); v = match ? parseInt(match[1]) : undefined; } return v; }());

这可以用来为包含IE版本的文档设置有用的类:

    if (ie) {
        document.documentElement.className += ' ie' + ie;
        if (ie < 9)
            document.documentElement.className += ' ieLT9';
    }

注意，如果IE处于兼容模式，它会检测正在使用的兼容模式。还要注意的是，IE版本主要适用于旧版本(<10);更高的版本更符合标准，使用modernizr.js之类的东西检查功能可能会更好。

2015-08-31 14:17:42

下面是AngularJS检查IE的方法

/**
 * documentMode is an IE-only property
 * http://msdn.microsoft.com/en-us/library/ie/cc196988(v=vs.85).aspx
 */
var msie = document.documentMode;

if (msie < 9) {
    // code for IE < 9
}

2015-07-08 12:54:12

在JavaScript中检测IE版本(v9之前)

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