是否可以为angular 2 http创建略有不同版本的subscribe()方法?

时间:2017-05-20 20:38:29

标签: angular typescript rxjs angular2-services

我正在尝试angular2。我发现我们应该使用subscribe()方法来检索getpost方法的结果:

this.http.post(path, item).subscribe(
  (response: Response)=> {console.log(response)},
  (error: any)=>{console.log(error)}
);

但是我想创建一个自定义版本的subscribe()方法,它有一个错误回调函数,它的错误参数不是any并且是强类型的。因此,我们将能够以这种方式订阅Observable:

this.http.post(path, item).subscribe(
  (response: Response)=> {console.log(response)},
  (error: HttpError)=>{console.log(error.body)}
);

我声明HttpError如下:

import { ModelState } from "app/Base/model-state";
import { ModelStateDictionary } from "app/Base/model-state-dictionary";
import { ErrorBody } from "app/Base/error-body";

export class HttpError {
    public ok: boolean;
    public status: number;
    public statusText: string;
    public type: number;
    public url: string;
    public body: ErrorBody;
    public static create(error: any): HttpError {
        let errorBody: ErrorBody = new ErrorBody();
        let body = JSON.parse(error._body)
        errorBody.message = body.message == null ? "" : body.message;
        errorBody.modelStateDictionary = new ModelStateDictionary();
        if (body.modelState != null) {
            for (let key in body.modelState) {
                let modelState: ModelState = new ModelState();
                modelState.Value = key;
                for (let value in body.modelState[key]) {
                    modelState.Error.push(value);
                }
                errorBody.modelStateDictionary.push(modelState);
            }
        }
        let httpError: HttpError = new HttpError();
        httpError.body = errorBody;
        httpError.ok = error.ok;
        httpError.status = error.status;
        httpError.statusText = error.statusText;
        httpError.type = error.type;
        httpError.url = error.url;
        return httpError;
    }
}

但在某些地方我需要在订阅之前调用create()方法,并将java对象错误转换为HttpError 。我想我需要创建一个自定义的Observable,或者可以使用map()。我是TypeScriptReactive programming的新手。能否请您解释一下如何进行此转换,因此用户可以使用更好的subscribe()版本?

1 个答案:

答案 0 :(得分:1)

首先,我相信Angular中的http.post只是封装了HTTP响应。以下是Angular源代码。

function httpRequest(backend: ConnectionBackend, request: Request): Observable<Response> {
  return backend.createConnection(request).response;
}

此处,响应或错误实际上是HTTP响应对象。

this.http.post(path, item).subscribe(
  (response: Response)=> {console.log(response)},
  (error: any)=>{console.log(error)}
);

我认为您可以在Observable中扩展您想要的方法来处理响应并获取定义的Object。类似于this.http.post(path,item).castHttpError()。subscribe()

https://github.com/Reactive-Extensions/RxJS/blob/master/doc/api/core/operators/subscribe.md

但是,我不确定这是否安全,因为您正在扩展可靠的库。至少,Angular不建议开发人员创建他/她自己的Angular分发版。你可以尝试一下。

最后,我建议你编写一个实用程序方法来实现它,这可能是多余的但不会有害。

this.http.post(path, item).subscribe(
  response => { console.log(response); },
  error => {
    console.log(toHttpError(error).body);
  }
);