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

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");
}

其他回答

如果没有人添加addeventlister -方法,而你使用的是正确的浏览器模式,那么你可以检查ie8或更低版本

if (window.attachEvent && !window.addEventListener) {
    // "bad" IE
}

传统Internet Explorer和attachEvent (MDN)

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
}

or

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

or

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

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 >');}

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
}

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>