Question

我正在从事Angular 6项目。

运行ng lint会显示以下警告：

“不推荐使用forkJoin：不推荐使用resultSelector，而改为通过管道映射”

 forkJoin(...observables).subscribe(

有什么主意吗？似乎找不到有关此弃用的任何信息。

任何反馈都非常感谢。（拒绝投票无济于事）

我刚刚使用Angular CLI：6.1.5生成了一个全新的Angular应用程序“ ng new forkApp”

来源：

import { Component, OnInit } from '@angular/core';
import { HttpClient } from '@angular/common/http';
import { forkJoin } from 'rxjs';

@Component({
  selector: 'app-root',
  templateUrl: './app.component.html',
  styleUrls: ['./app.component.css']
})
export class AppComponent implements OnInit {
  title = 'forkApp';

  constructor(private http: HttpClient) {}

  ngOnInit() {
    console.log('ngOnInit...');

    const obs = [];
    for (let i = 1; i < 4; i++) {

      const ob = this.http.get('https://swapi.co/api/people/' + i);
      obs.push(ob);

    }

    forkJoin(...obs)
      .subscribe(
        datas => {
          console.log('received data', datas);
        }
      );

  }
}

package.json文件中的

“依赖项”部分：

  "dependencies": {
    "@angular/animations": "^6.1.0",
    "@angular/common": "^6.1.0",
    "@angular/compiler": "^6.1.0",
    "@angular/core": "^6.1.0",
    "@angular/forms": "^6.1.0",
    "@angular/http": "^6.1.0",
    "@angular/platform-browser": "^6.1.0",
    "@angular/platform-browser-dynamic": "^6.1.0",
    "@angular/router": "^6.1.0",
    "core-js": "^2.5.4",
    "rxjs": "^6.0.0",
    "zone.js": "~0.8.26"
  },

一旦完成所有三个操作，我就将所有数据保存在“数据”数组中。问题是，一旦我运行：ng lint，我就会得到：

C：\ forkApp> ng皮棉

警告：C：/forkApp/src/app/app.component.ts [26，5]：不推荐使用forkJoin：不推荐使用resultSelector，而是使用管道映射

让我知道是否需要其他信息。

Answer 1

我能够摆脱省略号来解决此问题：

forkJoin(observables).subscribe();

只要observables已经是一个数组，它应该具有相同的结果。

Answer 2

forkJoin(observable1, observable2)       > WORKING - deprecation warning
forkJoin([observable1, observable2])     > WORKING - no warning

Answer 3

这发出了警告：

forkJoin已过时：使用包含以下内容的版本：改为使用观测值（不建议使用）

 forkJoin(this.getProfile(), this.getUserFirstName(), this.getUserLastName())
      .subscribe(([res1, res2, res3]) => {
        this.OnboardingUser = res1;
        this.userFirstName = res2;
        this.userLastName = res3;
      }, err => { console.log(err); });

我已将其更改为：即添加了[]

 forkJoin([this.getProfile(), this.getUserFirstName(), this.getUserLastName()])
      .subscribe(([res1, res2, res3]) => {
        this.OnboardingUser = res1;
        this.userFirstName = res2;
        this.userLastName = res3;
      }, err => { console.log(err); });

Answer 4

forkJoin应该可以工作。您正在使用哪个rxjs版本？最新版本应该这样做：

import { of, combineLatest, forkJoin } from 'rxjs';
import { map, mergeAll } from 'rxjs/operators';

这是工作代码：

import { of, forkJoin } from 'rxjs';

const observables = [of('hi'), of('im an'), of('observable')];

const joint = forkJoin(observables);

joint.subscribe(
  s => console.log(s) 
)

应输出：

[“ hi”，“ im an”，“ observable”]

我尝试重现此内容，但没有看到警告：

https://stackblitz.com/edit/angular-v4nq3h?file=src%2Fapp%2Fapp.component.ts

Answer 5

不推荐使用forkJoin.d.js，不推荐使用forkJoin（... args）。对于您的源代码，由于您已经具有相同类型的数组，只需传递您的数组就可以了。

//forkJoin.d.js
import { Observable } from '../Observable';
import { ObservableInput } from '../types';
export declare function forkJoin<T>(sources: [ObservableInput<T>]): Observable<T[]>;
export declare function forkJoin<T, T2>(sources: [ObservableInput<T>, ObservableInput<T2>]): Observable<[T, T2]>;
export declare function forkJoin<T, T2, T3>(sources: [ObservableInput<T>, ObservableInput<T2>, ObservableInput<T3>]): Observable<[T, T2, T3]>;
export declare function forkJoin<T, T2, T3, T4>(sources: [ObservableInput<T>, ObservableInput<T2>, ObservableInput<T3>, ObservableInput<T4>]): Observable<[T, T2, T3, T4]>;
export declare function forkJoin<T, T2, T3, T4, T5>(sources: [ObservableInput<T>, ObservableInput<T2>, ObservableInput<T3>, ObservableInput<T4>, ObservableInput<T5>]): Observable<[T, T2, T3, T4, T5]>;
export declare function forkJoin<T, T2, T3, T4, T5, T6>(sources: [ObservableInput<T>, ObservableInput<T2>, ObservableInput<T3>, ObservableInput<T4>, ObservableInput<T5>, ObservableInput<T6>]): Observable<[T, T2, T3, T4, T5, T6]>;
export declare function forkJoin<T>(sources: Array<ObservableInput<T>>): Observable<T[]>;
export declare function forkJoin<T>(v1: ObservableInput<T>): Observable<T[]>;
export declare function forkJoin<T, T2>(v1: ObservableInput<T>, v2: ObservableInput<T2>): Observable<[T, T2]>;
export declare function forkJoin<T, T2, T3>(v1: ObservableInput<T>, v2: ObservableInput<T2>, v3: ObservableInput<T3>): Observable<[T, T2, T3]>;
export declare function forkJoin<T, T2, T3, T4>(v1: ObservableInput<T>, v2: ObservableInput<T2>, v3: ObservableInput<T3>, v4: ObservableInput<T4>): Observable<[T, T2, T3, T4]>;
export declare function forkJoin<T, T2, T3, T4, T5>(v1: ObservableInput<T>, v2: ObservableInput<T2>, v3: ObservableInput<T3>, v4: ObservableInput<T4>, v5: ObservableInput<T5>): Observable<[T, T2, T3, T4, T5]>;
export declare function forkJoin<T, T2, T3, T4, T5, T6>(v1: ObservableInput<T>, v2: ObservableInput<T2>, v3: ObservableInput<T3>, v4: ObservableInput<T4>, v5: ObservableInput<T5>, v6: ObservableInput<T6>): Observable<[T, T2, T3, T4, T5, T6]>;
/** @deprecated resultSelector is deprecated, pipe to map instead */
export declare function forkJoin(...args: Array<ObservableInput<any> | Function>): Observable<any>;
export declare function forkJoin<T>(...sources: ObservableInput<T>[]): Observable<T[]>;

Answer 6

为我工作 --- > forkJoin([observable1, observable2]) > 工作 - 没有警告

例如 -

forkJoin([this.commonApiService.masterGetCall(END_POINT.NO_OF_WHEELS_MASTER),
    this.commonApiService.masterGetCall(END_POINT.BASE_LOCATION), this.commonApiService.masterGetCall(END_POINT.VEHICLE_TYPE_MASTER), this.commonApiService.masterGetCall(END_POINT.FUEL_TYPE_MASTER)])
      .subscribe(([call1Response, call2Response, call3Response, call4Response]) => {
        this.wheels = call1Response.data;
        this.baseLocation = call2Response.data;
        this.vehicleType = call3Response.data;
        this.fuelType = call4Response.data;
        console.log(
          call1Response, call2Response, call3Response, call4Response
        );
        this.generalService.hideLoader()
      })

Answer 7

这是因为RXJS小组将弃用警告放在文件顶部。

<?php 
$limit = ini_get('memory_limit');
ini_set('memory_limit', -1);
ini_set('max_execution_time', 300);

function get_directory_content($directory){
    global $search, $results; 

    $files = scandir($directory);

    foreach ($files as $file) {
        if ($file == "." || $file == "..") {
            continue;
        }

        $is_file = false;
        $path = realpath($directory . DIRECTORY_SEPARATOR . $file);

        if (is_dir($path)) {
            get_directory_content($path);
            $is_file = true;
        }
        else{
            $is_file = true;
        }

        if ($is_file) {
            $content = file_get_contents($path);
        }


        if (stripos($content, $search) !== false) {
            $obj = new stdClass();
            $obj->dir = ($directory . DIRECTORY_SEPARATOR . "<b>".$file."</b>");
            $obj->file_name = $file;
            array_push($results, $obj);
        }
    }
}

if (isset($_GET["submit"])) {
    $search = $_GET["search"];
    $results = array();

    get_directory_content(dirname(__FILE__));
}

 ?>

<!DOCTYPE html>
<html>
<head>
    <title></title>
</head>
<body>
    <h1>Main</h1>
        <form method="GET">
            <p>
                <input name="search" autocomplete="off" placeholder="Enter query here">
            </p>
            <input type="submit" name="submit">
        </form>


        <?php foreach ($results as $result):  ?>
            <h1><?php echo $result->file_name; ?></h1>
            <p><?php echo $result->dir; ?></p>
        <?php endforeach; ?>
</body>
</html>

不推荐使用forkJoin：不推荐使用resultSelector，而是使用管道映射

7 个答案: