Pass command-line arguments to Startup class in AS

问题:

I have arguments passed in via the command-line

private static int Main(string[] args)
{

    const string PORT = "12345"    ;

    var listeningUrl = $"http://localhost:{PORT}";

    var builder = new WebHostBuilder()
        .UseStartup<Startup>()
        .UseKestrel()
        .UseUrls(listeningUrl);

    var host = builder.Build();
    WriteLine($"Running on {PORT}");
    host.Run();

    return 0;
}

One of these arguments is a logging output directory. How do I get this value into my Startup class so I can write out to this directory when I receive a request?

I'd like to avoid using a static class. Would a service that supplies the value be the right way? If so, how do I get services injected into my middleware?

回答1:

You should be able to use the AddCommandLine() extension. First install the Nuget package Microsoft.Extensions.Configuration.CommandLine and ensure you have the correct import:

using Microsoft.Extensions.Configuration;

Now update your Main method to include the new config:

var config = new ConfigurationBuilder()
    .AddJsonFile("hosting.json", optional: true) //this is not needed, but could be useful
    .AddCommandLine(args)
    .Build();

var builder = new WebHostBuilder()
    .UseConfiguration(config)  //<-- Add this
    .UseStartup<Startup>()
    .UseKestrel()
    .UseUrls(listeningUrl);

Now you treat the command line options as configuration:

dotnet run /MySetting:SomeValue=123

And read in code:

var someValue = Configuration.GetValue<int>("MySetting:SomeValue");

回答2:

DavidG's answer is correct, but there were still some pieces of the puzzle missing for me.

There are two Nuget packages you need:

1. Microsoft.Extensions.Configuration.Binder
2. Microsoft.Extensions.Configuration.CommandLine

Because we want the command line arguments, we need to create the configuration in the Main(string[]).

using Microsoft.Extensions.Configuration;

class Program
{
    private static int Main(string[] args)
    {
        const string PORT = "12345";

        var listeningUrl = $"http://localhost:{PORT}";
        var configuration = new ConfigurationBuilder()
                            .AddCommandLine(args)
                            .Build();
        // Set the `static` `Configuration` property on the `Startup` class.
        Startup.Configuration = configuration;

        var builder = new WebHostBuilder()
            .UseStartup<Startup>()
            .UseKestrel()
            .UseSetting("Message", "Hello World")
            .UseUrls(listeningUrl);

        var host = builder.Build();
        WriteLine($"Running on {listeningUrl}");
        host.Run();

        return SUCCESS_EXIT_CODE;
    }
}

The Startup class is:

using Microsoft.Extensions.Configuration;

public class Startup
{
    public static IConfiguration Configuration { get; set; }


    public void Configure(IApplicationBuilder app)
    {
        foreach (var c in Configuration.AsEnumerable())
            Console.WriteLine($"{c.Key,-15}:{c.Value}");
    }
}

Is the command argument are --port 6000 outputDirectory C:\Temp then this will output:

port            :6000
outputDirectory :C:\Temp

回答3:

ASP.NET Core 2 answer:

Change the default Program.cs to be:

using System;
using System.IO;
using Microsoft.AspNetCore;
using Microsoft.AspNetCore.Hosting;
using Microsoft.Extensions.Configuration;

public class Program
{
    public static void Main(string[] args)
    {
        BuildWebHost(args).Run();
    }

    public static IWebHost BuildWebHost(string[] args) =>
        WebHost.CreateDefaultBuilder(args)
            .ConfigureAppConfiguration((hostingContext, config) =>
            {
                config.SetBasePath(Directory.GetCurrentDirectory());
                config.AddJsonFile("appsettings.json", optional: true, reloadOnChange: true)
                      .AddJsonFile($"appsettings.{hostingContext.HostingEnvironment.EnvironmentName}.json", optional: true, reloadOnChange: true);
                config.AddEnvironmentVariables();
                config.AddCommandLine(args);
            })
            .UseStartup<Startup>()
            .Build();
}

I removed other bits just to make the configuration explanation easier.

Note the .AddCommandLine(args) line in the configuration builder.

Unlike @BanksySan's answer you DON'T need to create a static property, instead let DI inject the IConfiguration into the startup class.

public class Startup
{
    public Startup(IConfiguration configuration)
    {
        Configuration = configuration;
    }

    private IConfiguration Configuration { get; }

You can now use the configuration entries from any of the config providers, file, env variables and commandline.

Example:

dotnet run --seed true

    public void Configure(IApplicationBuilder app)
    {
        app.UseExceptionHandler("/Home/Error");

        var seed = Configuration.GetValue<bool>("seed");
        if (seed)
            SeedData.Initialize(app);

        app.UseStaticFiles();
        app.UseMvcWithDefaultRoute();
    }

Hope this helps someone further.

回答4:

Dotnet Core 2

You don't have need most of the code as in dotnet core 1.0 to achieve this. AddCommandLinearguments are automatically injected when you BuildWebHost using the following syntax

Step 1.

 public static IWebHost BuildWebHost(string[] args)
        {
            return WebHost.CreateDefaultBuilder(args)
                .UseStartup<Startup>()
                .Build();
        }

Step 2. Inject configurations to Startip.cs using DI using the following code

public class Startup
    {
        private readonly IConfiguration Configuration;

        public Startup(IConfiguration configuration)
        {
            Configuration = configuration;
        }

Step 3. Access your arguments using configurations object.

 var seed = Configuration.GetValue<bool>("seed");
        Console.WriteLine("seed :" +seed);

Step 4. Start application with arguments

 dotnet run seed=True