Angular 角度5中的并行请求

我试图提出并行请求，获取并合并结果。 为此，我使用以下功能：

    getStudent(query): Observable<any> {
    const code = this.http.get(
        `http://localhost:8090/etudiantFiltre?codEtudiant=${query}`
    );
    const prenom = this.http.get(
        `http://localhost:8090/etudiantFiltre?prenom1=${query}`
    );
    const nom = this.http.get(
        `http://localhost:8090/etudiantFiltre?patronyme=${query}`
    );
    return Observable.forkJoin([code, nom, prenom]).map(responses => {
        console.log(`code : ${code}
                     nom : ${nom}
                     prenom : ${prenom}`);
        return [].concat(...responses);
    });
}

但是，当我试图打印该方法的结果时，以下是我得到的结果，我真的不明白为什么：

{
    "_isScalar": false,
    "source": {
    "_isScalar": false,
    "sources": [
        {
        "_isScalar": false,
        "source": {
            "_isScalar": false,
            "source": {
            "_isScalar": false,
            "source": {
                "_isScalar": true,
                "value": {
                "url": "http://localhost:8090/etudiantFiltre?codEtudiant=firstname",
                "body": null,
                "reportProgress": false,
                "withCredentials": false,
                "responseType": "json",
                "method": "GET",
                "headers": {
                    "normalizedNames": {},
                    "lazyUpdate": null,
                    "headers": {}
                    ...

---

如果需要并行请求，也可以尝试使用

forkJoin

studentObservables。此外，您还可以使用

map

而不是

forEach

来获得更清晰的代码

getAllStudent(query) {
    let studentObservables = query.split(' ')
        .map(element => this.getStudent(element)));
    return Observable.forkJoin(studentObservable);
}

ngOnInit() {
    this.getAllStudent('name firstname')
        .subscribe(data => console.log(data));
}

---

您可以继续使用forkJoin并使用

combinelateest

将所有请求统一在一起。如果你想为你的最终观察得到一个简单的答案。您可以使用

mergeMap

谁将接收

Array

并可以返回简单的

Response[]

const keywords = ['yanis','git'];
const request$ = [];

keywords.forEach((keyword) => {
        //Create array of ForkJoin Observable.
        request$.push(forkJoin([
            this.http.get('https://randomuser.me/api/?query='+keyword),
            this.http.get('https://randomuser.me/api/?query='+keyword),
            this.http.get('https://randomuser.me/api/?query='+keyword)
        ]));
});
// We combine all together and provide mergeMap strategy.
// Possible Observable here : combineLatest, forkJoin, combineAll ...
Observable.combineLatest(request$).mergeMap(e => {
    //All is merged to returned array. Here you can also deduplicate result.
    const returned = [];
    e.forEach(t => {
        t.forEach(i => returned.push(i));
    });
    return returned;

}).subscribe(e => {
    //You will receive each result returned by mergeMap.
    console.log(e);
});

---

在线示例：

您好，您能为每种请求提供json输出示例吗。实际上，我认为你的可观察对象试图递归地将每个响应对象合并在一起。为了防止这种情况，请尝试使用“Observable.merge（[code，nom，prenom]）”，然后对于“统一”部分，我将在询问json输出时为您提供实现。如果您询问三个请求的结果，它们都具有相同的json体。至于结果，这就是我使用它时得到的结果，

getAllStudent(query) {
    let studentObservables = query.split(' ')
        .map(element => this.getStudent(element)));
    return Observable.forkJoin(studentObservable);
}

ngOnInit() {
    this.getAllStudent('name firstname')
        .subscribe(data => console.log(data));
}

const keywords = ['yanis','git'];
const request$ = [];

keywords.forEach((keyword) => {
        //Create array of ForkJoin Observable.
        request$.push(forkJoin([
            this.http.get('https://randomuser.me/api/?query='+keyword),
            this.http.get('https://randomuser.me/api/?query='+keyword),
            this.http.get('https://randomuser.me/api/?query='+keyword)
        ]));
});
// We combine all together and provide mergeMap strategy.
// Possible Observable here : combineLatest, forkJoin, combineAll ...
Observable.combineLatest(request$).mergeMap(e => {
    //All is merged to returned array. Here you can also deduplicate result.
    const returned = [];
    e.forEach(t => {
        t.forEach(i => returned.push(i));
    });
    return returned;

}).subscribe(e => {
    //You will receive each result returned by mergeMap.
    console.log(e);
});