typescript 我如何使用 catchError() 并仍然使用 rxJs 6.0 返回类型化的 Observable？
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How do I use catchError() and still return a typed Observable with rxJs 6.0?
提问by Rick Strahl
cdSo I'm trying to migrate some of my Angular 5 code to 6, and I understand most of the changes required for rxjs to work using the .pipe() operator. It works as you would expect for 'pipable' operations.
cdSo 我正在尝试将我的一些 Angular 5 代码迁移到 6，并且我了解使用 .pipe() 运算符使 rxjs 工作所需的大部分更改。它的工作原理与您对“可管道”操作的期望相同。
However, the behavior of catchError()is different than the .catch()operator. Prior to rxjs 6 I used the .catch()operator to transform the error input into a a consistent error object that can then be caught in the `.subscribe().
但是，的行为catchError()与.catch()运算符不同。在 rxjs 6 之前，我使用.catch()运算符将错误输入转换为一致的错误对象，然后可以在`.subscribe() 中捕获该对象。
getAlbums(): Observable<Album[]> {
return this.httpClient.get<Album[]>(this.config.urls.url("albums"))
.map(albumList => this.albumList = albumList)
.catch(new ErrorInfo().parseObservableResponseError);
}
new ErrorInfo().parseObservableResponseErroris a function that takes a error object as input and parses the input error into a simpler error object with a normalized object. Catch returns Observable<any>:
new ErrorInfo().parseObservableResponseError是一个函数，它将一个错误对象作为输入，并将输入的错误解析为一个带有规范化对象的更简单的错误对象。捕获返回Observable<any>：
parseObservableResponseError(response): Observable<any> {
let err = new ErrorInfo();
...
return Observable.throw(err);
}
This worked great for easily handling errors up rxjs5 - errors basically are captured as part of the pipeline and then throw a well known error structure that can be captured .subscribe()error function.
这对于轻松处理 rxjs5 上的错误非常有用 - 错误基本上作为管道的一部分被捕获，然后抛出一个众所周知的错误结构，可以捕获.subscribe()错误函数。
Moving that same code to rxjs 6 however and using .pipe()I am trying to do the following:
然而，将相同的代码移动到 rxjs 6 并使用.pipe()我正在尝试执行以下操作：
getAlbums(): Observable<Album[]> {
return this.httpClient.get<Album[]>(this.config.urls.url("albums"))
.pipe(
map(albumList => this.albumList = albumList),
catchError(new ErrorInfo().parseObservableResponseError)
);
}
However this doesn't work because catchError now returns a result of Observable<any>into the pipeline, which is not what I want here. In essence I just want to re-throw any errors as I did before.
但是这不起作用，因为 catchError 现在将结果返回Observable<any>到管道中，这不是我在这里想要的。本质上，我只想像以前一样重新抛出任何错误。
error TS2322: Type 'Observable<{} | Album[]>' is not assignable to type 'Observable'
错误 TS2322：输入“Observable<{} | 专辑 []>' 不可分配到类型 'Observable'
How do simulate the old operator .catch()behavior?
如何模拟旧的操作员.catch()行为？
Update:
更新：
After mucking around with this a bit more, the code as shown just started working without giving me a build error. Honestly I have no idea why it failed with the error message before, but is working now. The suggestion by @cartant in the comments is an explicit way to specify the result and that works as well.
在稍微解决这个问题之后，所示的代码开始工作而没有给我一个构建错误。老实说，我不知道为什么它之前因错误消息而失败，但现在正在工作。@cartant 在评论中的建议是指定结果的明确方式，并且也有效。
回答by cartant
You should be able to explicitly specify the type parameters to catchErrorto indicate that it won't return an observable that emits a value - that is, it will return Observable<never>- and will always throw:
您应该能够明确指定类型参数以catchError指示它不会返回发出值的 observable - 也就是说，它将返回Observable<never>- 并且将始终抛出：
getAlbums(): Observable<Album[]> {
return this.httpClient.get<Album[]>(this.config.urls.url("albums"))
.pipe(
map(albumList => this.albumList = albumList),
catchError<Album[], never>(new ErrorInfo().parseObservableResponseError)
);
}
Doing so will see the type inferred from the pipecall to be Observable<Album[]>.
这样做将看到从pipe调用推断的类型为Observable<Album[]>.
