小编Mos*_*ito的帖子

我有3个可观察链,我需要链接,第一个用于另一个2.它们必须按顺序运行,每个必须等待前一个完成.是否有可能在没有嵌套的情况下进行链接？

public observable1(): Observable<Response> {
    let headers = new Headers();
    headers.append("Content-Range", "bytes */*");
    let requestOptions = new RequestOptions({headers: headers});
    return this.http.put(url, "", requestOptions);
}

public observable2(data1url): Observable<Response> {
    let headers = new Headers();
    headers.append("Content-Range", "bytes */*");
    let requestOptions = new RequestOptions({headers: headers});
    return this.http.put(data1url, "", requestOptions);
}

public observable3(data1url): Observable<Response> {
    let headers = new Headers();
    headers.append("Content-Range", "bytes */*");
    let requestOptions = new RequestOptions({headers: headers});
    return this.http.put(data1url, "", requestOptions);
}

这是我发现链接它们的唯一方法,但它们仍然是嵌套的,是否可以这样做而不是嵌套它们？

public doHttpProcess() {
    return this.observable1().concatMap(result1 => {
        return this.observable2(result1.json().data1) …