来自Linux的.Net核心Microsoft.Web.Administration ServerManager

Question

.net核心版本: 2.1.500

操作系统: Debian 9(4.9.0-8-AMD 64 #1 SMP Debian 4.9.130-2 (2018-10-27) x86_64 GNU/Linux )

我试图通过ServerManager从Linux停止/启动应用程序池，但没有成功。根据api目录，它应该工作：https://apisof.net/catalog/Microsoft.Web.Administration.ServerManager

ServerManager.OpenRemote()试图在尝试NullReference时抛出ApplicationPools异常：

// also trying ip adress
using (ServerManager manager = ServerManager.OpenRemote("someserver"))
{
    try
    {
        if (manager.ApplicationPools == null) /* <= Here is exception!!! */
        {
            Console.WriteLine("ApplicationPool is null!");
        }
        else
        {
            foreach (var item in manager.ApplicationPools)
            {
                Console.WriteLine("  " + item.Name);
            }
        }
    }
    catch (Exception ex)
    {
        Console.WriteLine("Exception: " + ex.Message);
        Console.WriteLine(" Trace=" + ex.StackTrace);
    }
}

结果：

异常:对象引用未设置为对象的实例。 Microsoft.Web.Administration.ConfigurationManager.CreateWritableAdminManager(WebConfigurationMap webConfigMap的Trace=，Microsoft.Web.Administration.ConfigurationManager.CreateConfiguration(WebConfigurationMap configMap的字符串configPathToEdit，布尔isAdminConfig，布尔isRedirectionConfig)，Microsoft.Web.Administration.ConfigurationManager.GetConfiguration(String rawConfigurationPath的字符串configPathToEdit，布尔isAdminConfig，布尔isRedirectionConfig)，字符串cacheKey，布尔isAdminConfig，( Microsoft.Web.Administration.ServerManager.ApplicationPoolsSectionCreator() at Microsoft.Web.Administration.Lazy.InitializeT at Microsoft.Web.Administration.ServerManager.ApplicationPoolCollectionCreator() at Microsoft.Web.Administration.Lazy.InitializeT at SomeMethod() in some_file.cs:line 123

当ServerManager通过接受字符串的构造函数创建时，ApplicationPools抛出System.PlatformNotSupportedException异常：

using (ServerManager manager = new ServerManager(@"\\someserver\IISSharedConfig\applicationHost.config"))
{
    try
    {
        if (manager.ApplicationPools == null) /* <= Here is exception!!! */
        {
            Console.WriteLine("ApplicationPool is null!");
        }
        else
        {
            foreach (var item in manager.ApplicationPools)
            {
                Console.WriteLine("  " + item.Name);
            }
        }
    }
    catch (Exception ex)
    {
        Console.WriteLine("Exception: " + ex.Message);
        Exception e = ex.InnerException;
        while (e != null)
        {
            Console.WriteLine("InnerException: " + e.Message);
            e = e.InnerException;
        }
        Console.WriteLine(" Trace=" + ex.StackTrace);
    }
}

结果：

异常:调用的目标引发了异常。COM不支持System.RuntimeTypeHandle.CreateInstance(RuntimeType类型的Trace=，布尔型publicOnly、布尔型wrapExceptions、布尔& canBeCached、RuntimeMethodHandleInternal& ctor)在Microsoft.Web.Administration.ConfigurationManager.CreateConfiguration(WebConfigurationMap、String、String、Microsoft.Web.Administration.ConfigurationManager.CreateConfiguration(WebConfigurationMap、Microsoft.Web.Administration.ConfigurationManager.CreateConfiguration(WebConfigurationMap、Microsoft.Web.Administration.ConfigurationManager.CreateConfiguration(WebConfigurationMap、Microsoft.Web.Administration.ConfigurationManager.CreateConfiguration(WebConfigurationMap，System.RuntimeType.CreateInstanceSlow(布尔publicOnly、布尔wrapExceptions、布尔skipCheckThis、布尔fillCache)的System.Activator.CreateInstanceT ( Microsoft.Web.Administration.ConfigurationManager.CreateWritableAdminManager(WebConfigurationMap webConfigMap )不支持InnerException: COM。Microsoft.Web.Administration.ConfigurationManager.GetConfiguration(String rawConfigurationPath中的布尔isAdminConfig，布尔isRedirectionConfig)，some_file.cs:line 123中Microsoft.Web.Administration.Lazy.InitializeT at Microsoft.Web.Administration.ServerManager.ApplicationPoolCollectionCreator() at Microsoft.Web.Administration.Lazy.InitializeT at SomeMethod()的字符串cacheKey，布尔isAdminConfig，布尔isRedirectionConfig)

如何在Linux中使用ServerManager？

Answer 1

我相信在Linux上不支持您想要做的事情。请参阅您在问题中发布的链接的底部：

平台 .NET标准+平台扩展

据我所知，平台扩展不是跨平台的添加，因此不能在运行Linux的应用程序中使用。

您还可以在发布的第二个错误中看到一个提示：

不支持COM。

这在Linux上是正确的。

希望能帮上忙！