小编ril*_*009的帖子

我在使用angular 4和observable时遇到错误.

/ Users//backend/src/app/app.component.ts(15,55):输入'()=> any'不能赋值为'State []'./ Users / /backend/src/app/app.component.ts(15,55):输入'()=> any'不能赋值为'State []'.类型'()=>任何'中缺少属性'includes'.

我究竟做错了什么

我的模特

export class State {
  id: number;
  state: string;
  code: string;
}

我的服务

import { Injectable } from '@angular/core';
import { Http, Response } from '@angular/http';
import 'rxjs/add/operator/map';
import 'rxjs/add/operator/do';
import 'rxjs/add/operator/toPromise';

import {environment} from '../../../environments/environment';

@Injectable()

export class StateService {
  private baseUrl: string = environment.baseUrl;

  constructor( private http: Http) {}

  /**
   * Get all States
   */
  GetStates() {
   return this.http.get(this.baseUrl + 'v1/states')
      .map((res: Response) => res.json); …

如何为MVC CRUD操作创建前缀路由。我正在开发需要管理员和前端的应用程序。对于管理员，我希望所有路线都指向localhost:5000/admin/....

我有不同的控制器

public class RoomsController : Controller
{
    // GET: Rooms        
    public async Task<IActionResult> Index()
    {

        return View(await _context.Rooms.ToListAsync());
    }

    //...
}

和

public class SlidersController : Controller
{
    private readonly ApplicationDbContext _context;

    public SlidersController(ApplicationDbContext context)
    {
        _context = context;
    }

    // GET: Sliders
    public async Task<IActionResult> Index()
    {
        return View(await _context.Sliders.ToListAsync());
    }

    //...
}

现在我希望管理路线为

public class RoomsController : Controller
{
    // GET: Rooms        
    public async Task<IActionResult> Index()
    {

        return View(await _context.Rooms.ToListAsync());
    }

    //...
}

而其他路线仍然存在

public …