可能返回一个嵌套的可观察对象：_scalar: false;明确地返回未定义的对象

Question

我遇到一个问题，当我应该 return 整个对象时，我正在 returning 对象中的值的 _scalar。我对象中的这个属性正在停止我的代码。参数都是return正确值

我正在尝试记录尽可能多的代码以帮助理解问题。我是最年轻的我可以这么赤裸。

相关视图模型（class应该是我稍后会更改的界面）：

    export class StandardWaiver {
    id?: number;
    waiverType?: WaiverType;
    duplicateFound?: boolean;
    notFoundInCAP?: boolean;
    policyNumber?: string;
    policyType?: number;
    dateOfLoss?: Date;

豁免服务：

   export class WaiverSevice {
   privat baseURL = '/xxx/XXX/xxxxxx/';
   constructor(private http: HttpClient) {}

   searchForWaiver(waiverTypeId: number, policyNumber: string, dateOfLoss: Date): any {
    let options = this.setOptionHeaders();

    if (isNotNullOrUndefined(policyNumber) && isNotNullOrUndefined(dateOfLoss)) {
      const duplicateCheck: any = {
        policyNumber,
        dateOfLoss,
      };
      if (waiverTypeId === 1) { return this.http.post(this.baseURL + 
    'Standard/CheckForDuplicateOrCAP', duplicateCheck, options); }
      else if (waiverTypeId === 2) { return this.http.post(this.baseURL + 
    'ICC/CheckForDuplicateOrCAP', duplicateCheck, options); }
      else {
        return null;
       }
     }

   }

    export const standardWaiverFromJSON = (data: any) => {
    let standardWaiver: StandardWaiver;
   standardWaiver.id = isNotNullOrUndefined(data?.id) ? data.id : null;

this is where the code breaks:

    export class CreateNewWaiverModalComponent {

     public opened = false;
 
    standardWaiver: StandardWaiver = new StandardWaiver(); 
    dateOfLoss: Date = null;
    policyNumber: string;
    waiverTypeId: number;
    waiverData: any;

    constructor(
    private activatedRoute: ActivatedRoute,
    private router: Router,
    private waiverService: WaiverService) { }

    async searchForDuplicateAndCAP() {
     if (this.waiverTypeId === 1)// standard
    {
      this.waiverData = this.waiverService.searchForWaiver(this.waiverTypeId, this.policyNumber, 
    this.dateOfLoss);

    console.log('DATA: ' + this.waiverData, '; WAIVERTYPEID: ' + this.waiverTypeId, '; 
    POLICYNUMBER: ' + this.policyNumber, '; DATEOFLOSS: ' + this.dateOfLoss, '; STANDARDWAIVER 
    OBJECT: ' + this.standardWaiver);

    this.standardWaiver = standardWaiverFromJSON(this.waiverData);
    }

这就是我 return 从 console.log:

    DATA: [object Object] ; WAIVERTYPEID: 1 ; POLICYNUMBER: 1234556654 ; DATEOFLOSS: Thu Jan 31 
    2019 00:00:00 GMT-0500 (Eastern Standard Time) ; STANDARDWAIVER OBJECT: [object Object] 

    ERROR Error: Uncaught (in promise): TypeError: standardWaiver is undefined - firefox browser
    ERROR Error: Uncaught (in promise): TypeError: Cannot set property 'id' of undefined - chrome browser

我必须 return this.standardWaiver 作为属性，参数中的值具有 (this.waiverTypeId, this.policyNumber, this.dateOfLoss)。到目前为止，它并没有给我 [object Object] 并且我不明白为什么。我欢迎任何建议，并提前谢谢你。

Answer 1

基本上 searchForWaiver 方法（包含可观察的）进行 ajax 调用。这意味着你必须 .subscribe 到 searchForWaiver 方法，这样 ajax 调用才会发生。然后你可以从 .subscribe 成功回调中的 searchForWaiver 方法获取数据 returned。

async searchForDuplicateAndCAP() {
    if (this.waiverTypeId === 1)// standard
    {
      this.waiverService.searchForWaiver(this.waiverTypeId, this.policyNumber, 
    this.dateOfLoss).subscribe(
        data => {
            this.waiverData = data;
                console.log('DATA: ' + this.waiverData, '; WAIVERTYPEID: ' + this.waiverTypeId, '; 
    POLICYNUMBER: ' + this.policyNumber, '; DATEOFLOSS: ' + this.dateOfLoss, '; STANDARDWAIVER 
    OBJECT: ' + this.standardWaiver);

            this.standardWaiver = standardWaiverFromJSON(this.waiverData);
       });
    }

再更正一次，searchForWaiver 方法应该 return Observable 来自所有可能的地方。否则打字稿将不允许您订阅该方法。

因此改变

return null

到

return of(null)

可能返回一个嵌套的可观察对象：_scalar: false;明确地返回未定义的对象

Possibly returning a nested observable: _scalar: false; Definetly returning undefined objects

typescript

angular

rxjs-observables