[英]Observable.zip is not a function
VM95422:27 ORIGINAL EXCEPTION: WEBPACK_IMPORTED_MODULE_3_rxjs_Observable .Observable.zip is not a function
VM95422:27 ORIGINAL EXCEPTION: WEBPACK_IMPORTED_MODULE_3_rxjs_Observable .Observable.zip不是函数
Tried various imports 试过各种进口
// import 'rxjs/add/operator/zip';
// import 'rxjs/add/observable/zip-static';
// import 'rxjs/add/operator/zip';
import 'rxjs/operator/zip';
Trying to use it like that: 试着像这样使用它:
const zippedUsers: Observable<User[]> = Observable.zip<User>(this.usersObservable);
Angular 4, TypeScript 2.1.6 Angular 4,TypeScript 2.1.6
package.json: 的package.json:
"rxjs": "^5.1.0",
maybe something like 也许是这样的
import {Observable} from "rxjs/Observable";
import "rxjs/add/observable/zip";
then something like: 然后像:
Observable.zip(this.someProvider.getA(), this.someProvider.getB())
.subscribe(([a, b]) => {
console.log(a);
console.log(b);
});
Starting from RxJS 6... 从RxJS 6开始......
Observable
creation functions Observable
创建功能 such as from()
, fromPromise()
, of()
, zip()
should be imported like this: 例如
from()
, fromPromise()
, of()
, zip()
应该像这样导入:
import { from, fromPromise, of, zip } from 'rxjs';
and used as a plain function call: 并用作普通函数调用:
const data: Observable<any> = fromPromise(fetch('/api/endpoint'));
should be imported like this: 应该像这样导入:
import { map, filter, scan } from 'rxjs/operators';
and used as pipe()
method arguments: 并用作
pipe()
方法参数:
const someObservable: Observable<number> = ...;
const squareOddVals = someObservable.pipe(
filter((n: number) => n % 2 !== 0),
map(n => n * n))
.subscribe((n: number): void => ...);
5.5 rxjs: 5.5 rxjs:
import {zip} from "rxjs/observable/zip";
const zippedUsers: Observable<User[]> = zip(this.usersObservable);
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