Question

冒险进入Angular 2和TypeScript的世界并遇到此错误。看起来你可以在某处设置_isScalar ......只是不确定它应该在哪里。

我收到以下错误（请参阅代码中的注释行）：

类型'Rating []'不能分配给'Observable'类型。 “评级[]”

类型中缺少属性“_isScalar”

以下代码：

import { Injectable } from '@angular/core';
import { Inject } from '@angular/core';
import { Jsonp, JsonpModule, Http } from '@angular/http';
import { Rating } from './rating.model';
import { Observable } from 'rxjs/Rx';
import { DOCUMENT } from '@angular/platform-browser'
import { IODataList } from '@microsoft/sp-odata-types';
import { SPHttpClient, SPHttpClientConfigurations, SPHttpClientConfiguration, SPHttpClientResponse, ODataVersion, ISPHttpClientConfiguration } from '@microsoft/sp-http';

@Injectable()
// See: http://www.telerik.com/kendo-angular-ui/components/dropdowns/dropdownlist/#toc-data-binding

export class RatingsService {
//constrauctor( private jsonp: Jsonp) { }
    constructor(@Inject(Http) private http: Http, @Inject(DOCUMENT) private document: any) { }

    //var url = this.context.pageContext.web.absoluteUrl + `/_api/web/lists?$filter=Hidden eq false`;

    private choices: Array<Rating> = [];

    public fetchData(): Observable<Rating[]> {

        console.log("BGW: Document location = " + document.location.href);
    // var url = this.context.pageContext.web.absoluteUrl + `/_api/web/lists?$filter=Hidden eq false`;
        //var response = this.jsonp.get(url); //.map((response => response.json()));
        console.log("BGW: Ratings choices url = " + url); 

        //var choices: Array<Rating> = new Array<Rating>();
        var choices: Array<Rating> = new Array<Rating>();
        // choices._isScalar = true;
        //let choices: Observable<Rating[]> = new Observable<Rating>();

        if (document.location.href.indexOf("localhost") > 0)
        {
            // Send dummy data to the form.
            console.log("BGW: Local workbench in use");
            choices.push(new Rating("High"));
            choices.push(new Rating("Medium"));
            choices.push(new Rating("Low"));
            return choices; // *** ISSUE HERE ***
        }
        else
        {
            var url = "https://somewhere.sharepoint.com/sites/Development/Subsite/_api/web/lists/GetByTitle('Requirements')/fields?$filter=EntityPropertyName%20eq%20%27Requirement_x0020_Rating%27";
            console.log("BGW: Ratings choices url = " + url);
            // Ref: http://chariotsolutions.com/blog/post/angular2-observables-http-separating-services-components/
            var response = this.http.get(url).map((response => response.json()));
            console.log("BGW: response = " + JSON.stringify(response));

            return response.map((choices: Array<any>) => {
                if (choices) {
                    choices.forEach((choice) => {
                        choices.push(new Rating(choice));
                    });
                }
                return choices;
            });
        }
    }
}

Answer 1

问题是因为你没有像你指定的那样返回一个observable：

public fetchData(): Observable<Rating[]>

但稍后您正在定义......

var choices: Array<Rating> = new Array<Rating>();

这就是你收到错误的原因。 Observable Array和“regular”Array之间存在不匹配。如果您没有明确需要来指定要返回的内容，我只会完全删除Observable<Rating[]>。

Answer 2

你需要返回一个类似的观察点：

return Observable.of<Rating[]>(choices);

使用静态方法of，但还有很多其他方法。

您看到编译器错误的原因是您在方法签名中指定了返回类型：

public fetchData(): Observable<Rating[]>

TypeScript SomeArray []'不能赋值为'Observable <somearray []>'

2 个答案: