ASP。使用IConfiguration获取Json数组

在appsettings.json

{
      "MyArray": [
          "str1",
          "str2",
          "str3"
      ]
}

在Startup.cs

public void ConfigureServices(IServiceCollection services)
{
     services.AddSingleton<IConfiguration>(Configuration);
}

在HomeController

public class HomeController : Controller
{
    private readonly IConfiguration _config;
    public HomeController(IConfiguration config)
    {
        this._config = config;
    }
    
    public IActionResult Index()
    {
        return Json(_config.GetSection("MyArray"));
    }
}

上面是我的代码。结果是零。如何获取数组?

当前回答

如果你想要选择第一项的值，那么你应该这样做-

var item0 = _config.GetSection("MyArray:0");

如果你想选择整个数组的值，那么你应该这样做-

IConfigurationSection myArraySection = _config.GetSection("MyArray");
var itemArray = myArraySection.AsEnumerable();

理想情况下，您应该考虑使用官方文档建议的选项模式。这会给你带来更多的好处。

2016-12-26 11:59:45

其他回答

简式:

var myArray= configuration.GetSection("MyArray")
                        .AsEnumerable()
                        .Where(p => p.Value != null)
                        .Select(p => p.Value)
                        .ToArray();

它返回一个字符串数组:

{“str1”，“str2”，“str3”}

2019-01-09 07:05:25

.Net Core 7.x中处理对象的不同方法

在appsettings.json:

{
      "People": [
          { "FirstName": "Glen", "LastName": "Johnson", "Age": 30 },
          { "FirstName": "Matt", "LastName": "Smith", "Age": 40 },
          { "FirstName": "Fred", "LastName": "Williams", "Age": 50 }
      ]
}

Person类:

public class Person
{
    public string FirstName { get; set; }
    public string LastName { get; set; }
    public int Age { get; set; }
}

在代码中:

var appConfig = App.Current.AppConfiguration;   // Or could be passed in through DI
var children = appConfig.GetSection("People")
    .GetChildren()
    .ToList();

var people = new List<Person>();
foreach (var child in children)
{
    var rec = new Person
    {
        FirstName = appConfig[$"{child.Path}:FirstName"],
        LastName = appConfig[$"{child.Path}:LastName"],
        Age = int.Parse(appConfig[$"{child.Path}:Age"]),
    };
    people.Add(rec);
}

2023-01-27 21:48:34

对于从配置返回复杂JSON对象数组的情况，我调整了@djangojazz的答案，以使用匿名类型和动态而不是元组。

给定的设置部分:

"TestUsers": [
{
  "UserName": "TestUser",
  "Email": "Test@place.com",
  "Password": "P@ssw0rd!"
},
{
  "UserName": "TestUser2",
  "Email": "Test2@place.com",
  "Password": "P@ssw0rd!"
}],

你可以这样返回对象数组:

public dynamic GetTestUsers()
{
    var testUsers = Configuration.GetSection("TestUsers")
                    .GetChildren()
                    .ToList()
                    .Select(x => new {
                        UserName = x.GetValue<string>("UserName"),
                        Email = x.GetValue<string>("Email"),
                        Password = x.GetValue<string>("Password")
                    });

    return new { Data = testUsers };
}

2018-10-30 14:24:00

如果你有一个像这样的复杂JSON对象数组:

{
  "MySettings": {
    "MyValues": [
      { "Key": "Key1", "Value":  "Value1" },
      { "Key": "Key2", "Value":  "Value2" }
    ]
  }
}

你可以这样检索设置:

var valuesSection = configuration.GetSection("MySettings:MyValues");
foreach (IConfigurationSection section in valuesSection.GetChildren())
{
    var key = section.GetValue<string>("Key");
    var value = section.GetValue<string>("Value");
}

2017-06-06 14:21:08

在IOptions不起作用后，我发现这更简单，序列化时要做的事情也少得多。

//appsettings.json
"DatabaseModelCreationOptions": {
    "Users": [
      {
        "Id": "1",
        "UserName": "adminuser",
        "Email": "youremail@myemail.com",
        "Password": "!Ch4",
        "Player": "Admin",
        "PlayerInitials": "ADMIN",
        "ApiKey": "40753ey"
      }
    ],   
    "UserRoles": [
      {
        "Id": "af9986df",
        "Name": "Admin",
        "ConcurrencyStamp": "ddd53170"
      },
      {
        "Id": "03b82b0e",
        "Name": "Manager",
        "ConcurrencyStamp": "65da3f89"
      }
    ]
}

我需要的类，我不需要给你们看嵌套的类属性，你们可以在上面看到。

public class DatabaseModelCreationOptions
{
    public IEnumerable<User>? Users { get; set; }
    public IEnumerable<UserRole>? UserRoles { get; set; }
}

然后调用GetSection("")。得到

var dbOptions = configuration.GetSection("DatabaseModelCreationOptions")
    .Get<DatabaseModelCreationOptions>();

2022-12-18 00:33:44

ASP。使用IConfiguration获取Json数组

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