使用 switchmap 和 forkjoin 链接 observables 不能按预期工作

Question

我目前正在创建尝试创建一个系统，如果我成功更新了一个请求，我会在 forkjoin 中触发 3 个可观察对象，然后一旦这些对象完成 console.log('complete/or finish') 否则如果它不是什么都不做。 目前虽然可观察对象在 forkjoin 完成之前就完成了，我完全不知道发生了什么，

 copyAttachments(): Observable<any> { if (this.model.attachments.length > 0) { return this.attachmentService.copyAttachments(this.model.attachments, this.model.id); } return empty(); } uploadAttachments(): Observable<any> { this.showAttachmentsUploadingModal = true; const formData = new FormData(); if (this.formAttachments.length > 0) { this.showAttachmentsUploadingModal = true; this.formAttachments.forEach(file => { formData.append('files', file, file.name); }); } // uploading empty formData will still trigger creation of default folder structure return this.attachmentService.uploadAttachments(formData, this.model.id); } uploadCustomerData(): Observable<any> { this.showAttachmentsUploadingModal = true; const formData = new FormData(); if (this.formCustomerData.length > 0) { this.showAttachmentsUploadingModal = true; this.formCustomerData.forEach(file => { formData.append('files', file, file.name); }); return this.attachmentService.uploadCustomerData(formData, this.model.id); } return empty(); }

 handleAttachments(): Observable<any> { return forkJoin ( this.copyAttachments(), this.uploadCustomerData(), this.uploadAttachments() ) } updateRequest() { this.myservice.updateRequest(this.model).pipe( switchMap((saveResult: boolean) => { this.showSubmittingModal = false; if (saveResult === true) { return this.handleAttachments(); } else { return empty(); } }), ).pipe(finalize(() => { console.log('finish') })).subscribe( response => { // todo: do anything with this response from uploadAttachments()? this.showAttachmentsUploadingModal = false; }, error => { location.reload(); }, () => { console.log('complete') } ); }

预期的结果是我的服务完成，然后我的 forkjoin observables 完成，然后它将 console.log('finish')

目前的结果是服务完成。 它 console.logs('finish') 然后我的 forkjoin 触发器

Answer 1

我发现这个问题的解决方案不是返回 empty() 或 of() 你需要返回一个空的 object 作为可观察的，这样 Forkjoin 仍然保持值所以 return of({}) 一切都很好