【问题标题】:Angular returns error when building for Production but not DevelopmentAngular 在为生产而不是开发构建时返回错误
【发布时间】:2018-04-22 15:55:29
【问题描述】:

github 链接到我的 repo here

相关代码:

import { Http } from '@angular/http';
import { Observable } from 'rxjs/Observable';
import 'rxjs/add/operator/catch';
import 'rxjs/add/operator/map';
import 'rxjs/add/observable/throw';

@Injectable()
export class DataService {
  constructor (private url: string, private http: Http) { }
  getAll() {
    return this.http.get(this.url)
        .map(response => response.json())
        .catch(this.errorHandler);
  }


  private errorHandler(error: Response) {
    // tslint:disable-next-line:curly
    if (error.status === 400)
      return Observable.throw(new BadInputError(error));
    // tslint:disable-next-line:curly
    else if (error.status === 404)
      return Observable.throw(new NotFoundError(error));
    return Observable.throw(new AppError(error));
  }

}

我找到了similar question,它表明问题与覆盖错误处理有关,但它并没有真正提出解决方案。

当我运行ng serve --prodng build --prod 时,我得到以下异常:

ERROR in Error: Can't resolve all parameters for DataService in C:/Users/Patrick/Desktop
/Angular/trial-app/src/app/services/data.service.ts: (?, [object Object]).
    at Error (native)
    at syntaxError (C:\Users\Patrick\Desktop\Angular\trial-app\node_modules\@angular\com
piler\bundles\compiler.umd.js:1729:34)
    at CompileMetadataResolver._getDependenciesMetadata (C:\Users\Patrick\Desktop\Angula
r\trial-app\node_modules\@angular\compiler\bundles\compiler.umd.js:15816:35)
    at CompileMetadataResolver._getTypeMetadata (C:\Users\Patrick\Desktop\Angular\trial-
app\node_modules\@angular\compiler\bundles\compiler.umd.js:15684:26)
    at CompileMetadataResolver._getInjectableMetadata (C:\Users\Patrick\Desktop\Angular\
trial-app\node_modules\@angular\compiler\bundles\compiler.umd.js:15670:21)
    at CompileMetadataResolver.getProviderMetadata (C:\Users\Patrick\Desktop\Angular\tri
al-app\node_modules\@angular\compiler\bundles\compiler.umd.js:15961:40)
    at C:\Users\Patrick\Desktop\Angular\trial-app\node_modules\@angular\compiler\bundles
\compiler.umd.js:15890:49
    at Array.forEach (native)
    at CompileMetadataResolver._getProvidersMetadata (C:\Users\Patrick\Desktop\Angular\t
rial-app\node_modules\@angular\compiler\bundles\compiler.umd.js:15850:19)
    at CompileMetadataResolver.getNgModuleMetadata (C:\Users\Patrick\Desktop\Angular\tri
al-app\node_modules\@angular\compiler\bundles\compiler.umd.js:15505:50)
    at addNgModule (C:\Users\Patrick\Desktop\Angular\trial-app\node_modules\@angular\com
piler\bundles\compiler.umd.js:24268:58)
    at C:\Users\Patrick\Desktop\Angular\trial-app\node_modules\@angular\compiler\bundles
\compiler.umd.js:24279:14
    at Array.forEach (native)
    at _createNgModules (C:\Users\Patrick\Desktop\Angular\trial-app\node_modules\@angula
r\compiler\bundles\compiler.umd.js:24278:26)
    at analyzeNgModules (C:\Users\Patrick\Desktop\Angular\trial-app\node_modules\@angula
r\compiler\bundles\compiler.umd.js:24153:14)
    at analyzeAndValidateNgModules (C:\Users\Patrick\Desktop\Angular\trial-app\node_modu
les\@angular\compiler\bundles\compiler.umd.js:24163:35)

【问题讨论】:

  • 我不确定,但是:Angular 在构造函数中使用依赖注入。也许发送字符串不算作 DI ?如果你改用 setter 会发生什么?可以试试吗?
  • 不仅仅是字符串。我已经定义了一个类型(链接)并将 url 作为类型的属性传入。还是不行。

标签: angular build production dev-to-production


【解决方案1】:

会将此添加为评论,但没有足够的代表。但是,我同意 trichetriche 的观点,我怀疑一个字符串不算作依赖注入,因此会绊倒你。检查这里给出的答案:Angular 2 passing parameters to constructor throws DI exception

就我个人而言,我使用服务作为一种全局变量来实现这一点。例如:

//in another class called URL-Service.ts you will need to add this 
//to providers at the root level in app.module

export class URLService {
  public savedURL: string = '';
}

//in another class where you have access to the url

export class SomeClass {
  constructor(private urlService = URLService){}
  someMethod(){
    this.urlService.savedURL = whateverHasTheURL;
  }    
}


export class DataService {
  constructor (private urlService = URLService, private http: Http) { 
    this.url = this.urlService.savedURL;
  }
  ...
}

【讨论】:

  • 当有权访问 URL 的类实际上是来自 DataService 的派生类时,我将如何实现这一点:export class PostsService extends DataService { constructor(http: Http) { super('http://jsonplaceholder.typicode.com/posts', http); } }
  • 明白了!在基类中创建一个受保护的道具“url”,然后是:export class PostsService extends DataService { constructor(http: Http) { super(http); this.url = 'http://jsonplaceholder.typicode.com/posts'; } } 其中“url”是在基类上声明的属性。 @fleadram
  • 太好了,是的,如果您正在扩展一个运行良好的基类。
猜你喜欢
  • 2019-08-01
  • 2018-01-03
  • 2020-07-30
  • 1970-01-01
  • 1970-01-01
  • 1970-01-01
  • 2021-12-10
  • 1970-01-01
  • 2019-07-07
相关资源
最近更新 更多