Javascript方式列出IE的可用插件

Question

有没有一种快速的方法来获取可用的Active X插件的JavaScript列表？

我需要做一个测试，看看在我真正尝试运行它之前是否安装了插件。

实际上，我想创建一个页面，上面写着“已安装插件并正常工作”，或者让它优雅地失败。

如果插件不可用，我不确定如何优雅地失败。

Answer 1

try吧。

try {
  var plugin = new ActiveXObject('SomeActiveX');
} catch (e) {
  alert("Error"); // Or some other error code
}

Answer 2

如果对象无法实例化， object标签将显示其中的内容：

<object ...>
 <p>
 So sorry, you need to install the object.  Get it <a href="...">here</a>.
 </p>
</object>

因此，内置优雅故障，您根本不需要使用脚本。

Answer 3

对于Internet Explorer 11，您可以使用navigator.plugins JS API，但是您需要添加适当的registrey密钥以便IE11检测它：

HKLM\SOFTWARE\Microsoft\Internet Explorer\NavigatorPluginsList

或64位

HKLM\SOFTWARE\Wow6432\Microsoft\Internet Explorer\NavigatorPluginsList

例如，对于名称为“ABC”且mime类型为“application / abc”的插件：

• 添加密钥HKLM \\ SOFTWARE \\ Wow6432 \\ Microsoft \\ Internet Explorer \\ NavigatorPluginsList \\ ABC
• 为插件支持的每种自定义MIME类型创建子项，使用MIME类型值作为子项的名称，例如“application / abc”

然后使用以下代码检查插件是否存在：

var plugin = navigator.plugins["<your plugin activex id>"];
if(plugin) {
  //plugin detected
} else {
  //plugin not found
}

更多相关信息： http ： //msdn.microsoft.com/en-us/library/ie/dn423948（v = vs。85）.aspx

Answer 4

也许这个脚本可以帮助

function detectPlugin() {
// allow for multiple checks in a single pass
var daPlugins = detectPlugin.arguments;

// consider pluginFound to be false until proven true
var pluginFound = false;

// if plugins array is there and not fake
if (navigator.plugins && navigator.plugins.length > 0) {
var pluginsArrayLength = navigator.plugins.length;

// for each plugin...
for (pluginsArrayCounter=0; pluginsArrayCounter < pluginsArrayLength; pluginsArrayCounter++ ) {

    // loop through all desired names and check each against the current plugin name
    var numFound = 0;
    for(namesCounter=0; namesCounter < daPlugins.length; namesCounter++) {

    // if desired plugin name is found in either plugin name or description
    if( (navigator.plugins[pluginsArrayCounter].name.indexOf(daPlugins[namesCounter]) >= 0) || 
        (navigator.plugins[pluginsArrayCounter].description.indexOf(daPlugins[namesCounter]) >= 0) ) {
        // this name was found
        numFound++;
    }   
    }
    // now that we have checked all the required names against this one plugin,
    // if the number we found matches the total number provided then we were successful
    if(numFound == daPlugins.length) {
    pluginFound = true;
    // if we've found the plugin, we can stop looking through at the rest of the plugins
    break;
    }
}
}
return pluginFound;} // detectPlugin

以此为例来调用它

pluginFound = detectPlugin('Shockwave','Flash');