Question

我正在尝试从数据库中获取用户。当传递用户名作为参数时，控制器服务应返回控制器。控制器只是一种特定的用户。

当我尝试将已解析的用户解析为Controller时遇到的错误是

Argument of type 'Observable<Controller>' is not assignable to parameter of type 'string'

这是发生错误的部分

this.currentController = JSON.parse（controllerService.getController（token.getUsername（）））;

下面是我的代码。

控制器服务

import {Injectable} from '@angular/core';
import { HttpClient, HttpHeaders } from '@angular/common/http';
import { Controller } from '../models/controller.model';
import {catchError} from 'rxjs/operators';
import { throwError} from 'rxjs';
import { FilterSearchControllerDTO } from 
'../models.dto/filterSearchController.dto';

const httpOptions = {
headers: new HttpHeaders({ 'Content-Type': 'application/json' })
};


@Injectable()
export class ControllerService{

constructor(private http:HttpClient) {}

private controllerUrl = 'http://localhost:8080/api/controller/';

public getControllers() {
    return this.http.get<Controller[]>(this.controllerUrl);
  }

public getController(username: string) {
  return this.http.get<Controller>(`${this.controllerUrl + 'getController'}/${username}`);
}

public deleteController(id: BigInteger){
  const url = `${this.controllerUrl + 'deleteController'}/${id}`;
  return this.http.get<boolean>(url);

}

public addController(controller: Controller){
  return this.http.post(this.controllerUrl + 'addController',controller)
  .pipe(
    catchError(e => throwError(e))
  );                 

}

public filterSearch(values: FilterSearchControllerDTO){
  return this.http.post<Controller[]>(this.controllerUrl + 'searchFilter',values);                 

}

}

ControllerView组件

import { Component, OnInit, ViewChild } from '@angular/core';
import { HeaderComponent } from '../header/header.component';
import { ControllerService } from '../services/controller.service';
import { Controller } from '../models/controller.model';
import { Router } from '@angular/router';
import { TokenStorageService } from '../auth/token-storage.service';

@Component({
  selector: 'app-controller-view',
  templateUrl: './controller-view.component.html',
  styleUrls: ['./controller-view.component.css']
})
export class ControllerViewComponent implements OnInit {

  currentController: Controller;

  @ViewChild("header") header: HeaderComponent;
  showView: string = 'home';

  constructor(private controllerService: ControllerService, private router: Router, private token: TokenStorageService) { 
    this.currentController = JSON.parse(controllerService.getController(token.getUsername()));
  }

  ngOnInit() {
    this.header.controllerView();
  }

  onNavigate(feature: string){
    this.showView = feature;
    if(feature == 'logout') {
      window.sessionStorage.clear();
      this.router.navigate(['mainPage']);
      window.alert("Successfully Logged out!");
    }
  }

}

Answer 1

在您的服务getController()方法中返回一个Observable<Controller>。因此，您基本上不能从中返回Controller。相反，您需要订阅它并获得如下值：

ControllerView组件

constructor(private controllerService: ControllerService, private router: Router, private token: TokenStorageService) { 

    this.controllerService.getController(token.getUsername()).subscribe((response) => {
        this.currentController = response;
    })

}

无法将类型'Observable <controller>'的参数分配给类型'string'的参数

1 个答案: