类型"Observable< IProduct []>"上不存在属性"do" [英] Property 'do' does not exist on type 'Observable<IProduct[]>'

问题描述

升级到Angular 6.0和Rxjs到6.0后，我收到以下编译错误:

After upgrading to Angular 6.0 and Rxjs to 6.0 I receive the following compilation error:

Property 'do' does not exist on type 'Observable'.

这是代码:

import { Observable, of } from 'rxjs';
import 'rxjs/add/operator/do';
import 'rxjs/add/observable/throw';
import 'rxjs/add/operator/catch';
import { IProduct } from './product';

@Injectable()
export class ProductService { 
    constructor(
        private product: IProduct)
    {         
    }

    getProduct = () => { 
        return product.products
            // error on next line
            .do(data => console.log('All:' + JSON.stringify(data)))
            .catch(this.handleError);
    }

    private handleError(err: HttpErrorResponse) { 
        console.log(err.message);
        return Observable.throw(err.message);        
    }
}

有什么主意吗?

推荐答案

问题不是与angular有关，而与rxjs有关. rxjs引入了对rxjs版本6的重大更改.

The problem is not with angular but with rxjs. rxjs introduced breaking changes from rxjs version 6.

要在不更改任何代码的情况下再次使代码工作，请安装以下软件包:

To get your code working again without changing any of your code install the following package:

npm install rxjs-compat@6 --save

然后，您应该能够编译您的项目. rxjs-compat只是暂时的解决方案，因此您需要更新代码库才能使用新版本.

You should then be able to compile your project. rxjs-compat is meant to be a temporarily solution so you need to update your codebase to work with the new version.

您需要更新的内容

1. 更新来自

1. Update import statements from

import { Observable } from "rxjs/Observable";

至

import { Observable } from "rxjs";

从以下位置更新操作员导入

Update your operator imports from

import 'rxjs/add/operator/do'

至

import { do } from "rxjs/operators";

---

重命名运算符

由于与JavaScript保留字的名称冲突，一些运算符也已重命名.他们是

---

Renamed Operators

Some operators have also been renamed due to name collisions with JavaScript reserved words. They are

1. do => tap

catch => catchError

switch => switchAll

finally => finalize

---

无操作员链接

您也就无法再链接您的运算符了，例如，您需要使用pipe运算符

---

No Operator Chaining

You also then can't chain your operators anymore you need to use the pipe operator e.g.

// an operator chain
source
  .map(x => x + x)
  .mergeMap(n => of(n + 1, n + 2)
    .filter(x => x % 1 == 0)
    .scan((acc, x) => acc + x, 0)
  )
  .catch(err => of('error found'))
  .subscribe(printResult);

// must be updated to a pipe flow
source.pipe(
  map(x => x + x),
  mergeMap(n => of(n + 1, n + 2).pipe(
    filter(x => x % 1 == 0),
    scan((acc, x) => acc + x, 0),
  )),
  catchError(err => of('error found')),
).subscribe(printResult);

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