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

根据微软的说法,以下是最好的解决方案,它也很简单:

function getInternetExplorerVersion()
// Returns the version of Internet Explorer or a -1
// (indicating the use of another browser).
{
    var rv = -1; // Return value assumes failure.
    if (navigator.appName == 'Microsoft Internet Explorer')
    {
        var ua = navigator.userAgent;
        var re  = new RegExp("MSIE ([0-9]{1,}[\.0-9]{0,})");
        if (re.exec(ua) != null)
            rv = parseFloat( RegExp.$1 );
    }
    return rv;
}

function checkVersion()
{
    var msg = "You're not using Internet Explorer.";
    var ver = getInternetExplorerVersion();

    if ( ver > -1 )
    {
        if ( ver >= 8.0 ) 
            msg = "You're using a recent copy of Internet Explorer."
        else
            msg = "You should upgrade your copy of Internet Explorer.";
      }
    alert( msg );
}

其他回答

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>

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

简单的解决方案,停止思考浏览器和使用年。

var year = eval(today.getYear());
if(year < 1900 )
 {alert('Good to go: All browsers and IE 9 & >');}
else
 {alert('Get with it and upgrade your IE to 9 or >');}

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");
}