What is the solution for resolving the no-unsafe-any rule?

Question

What is the solution for resolving the no-unsafe-any rule?

Currently incorporating TSLint for maintaining the quality of my Angular TypeScript code. I've opted to activate the 'no-unsafe-any' rule from TSLint, as it appears beneficial to avoid making assumptions about properties with type 'any'.

An issue has arisen where this rule is flagging errors in some sections of my code that I am struggling to rectify without disabling the rule altogether. An example snippet of code that violates this rule is provided below.

public intercept(request: HttpRequest<{}>, next: HttpHandler): Observable<HttpEvent<{}>> {
  return next
    .handle(request)
    .pipe(
      catchError(error => {
        if (error && error.status === httpCodeUnauthorized) {
          // Auto logout if unathorized
          this.authenticationService.logout();
        }

        const errorMessage = (error.error && error.error.message) || error.statusText;

        return throwError(errorMessage);
      }),
    );
}

The linter is showing 4 errors across 2 lines:

ERROR: /home/robert/programming/npc/gui/src/app/core/authentication/unauthorized.interceptor.ts[24, 24]: Unsafe use of expression of type 'any'.
ERROR: /home/robert/programming/npc/gui/src/app/core/authentication/unauthorized.interceptor.ts[29, 33]: Unsafe use of expression of type 'any'.
ERROR: /home/robert/programming/npc/gui/src/app/core/authentication/unauthorized.interceptor.ts[29, 48]: Unsafe use of expression of type 'any'.
ERROR: /home/robert/programming/npc/gui/src/app/core/authentication/unauthorized.interceptor.ts[29, 72]: Unsafe use of expression of type 'any'

The problematic lines are:

if (error && error.status === httpCodeUnauthorized) {

const errorMessage = (error.error && error.error.message) || error.statusText;

The main challenge originates from the fact that the error parameter in the handler passed to the catchError function from the RxJS library has a type of 'any'. While acknowledging that error could be of any type, I am taking precautions by verifying the existence of its properties before utilizing them, which I believe makes the code safe.

Is there a recommended approach to persuading the linter and TypeScript compiler that the code adheres to safety standards and satisfies the rule?

typescript tslint typescript-types

Answer 1

Answer №1

When dealing with Angular errors, it is crucial to ensure that the error type is always an instance of HttpErrorResponse.

catchError((error: HttpErrorResponse) => {
//...
}

Furthermore, when accessing error.error in your code, keep in mind that it is defined as type any within HttpErrorResponse. Therefore, it is advisable to use a type guard to validate and cast it to an Error object. It is unnecessary to explicitly define the Error type as it is already part of TypeScript's base types.

function isError(value: any | undefined): value is Error {
  return error && ((error as Error).message !== undefined);
}

Subsequently, you can utilize this function in your code:

const errorMessage = isError(error.error) ? error.error.message : error.statusText;

Answer 2

When dealing with Angular errors, it is crucial to ensure that the error type is always an instance of HttpErrorResponse.

catchError((error: HttpErrorResponse) => {
//...
}

Furthermore, when accessing error.error in your code, keep in mind that it is defined as type any within HttpErrorResponse. Therefore, it is advisable to use a type guard to validate and cast it to an Error object. It is unnecessary to explicitly define the Error type as it is already part of TypeScript's base types.

function isError(value: any | undefined): value is Error {
  return error && ((error as Error).message !== undefined);
}

Subsequently, you can utilize this function in your code:

const errorMessage = isError(error.error) ? error.error.message : error.statusText;

Answer 3

Answer №2

When faced with the dilemma of determining the specific type of an error, there are two paths you can take. The first option is to simply annotate the type if you are confident in its specificity. However, if uncertainty looms, utilizing a type guard might be the better approach.

Utilizing Type Annotation

Type annotation allows you to clearly indicate to the compiler the expected type of the error. This method entirely eliminates the need for utilizing the "any" type:

interface Error {
    status: string,
    statusText: string,
    error: { message: string | undefined } | undefined;
}

catchError((error: Error | undefined) => {
    //...
}

Implementing a Type Guard

A type guard comes into play when a value could potentially be of a certain type but does not exclusively have to be of that type. By using a type guard, the type will be checked and within the subsequent block, the variable will assume the identified type:

function isError(value: any | undefined): value is Error {
    return error && ((error as Error).status !== undefined);
}

catchError(error => {
    if (isError(error)) {
        //Within this block, the variable 'error' will be recognized as type Error
    }
}

Answer 4

When faced with the dilemma of determining the specific type of an error, there are two paths you can take. The first option is to simply annotate the type if you are confident in its specificity. However, if uncertainty looms, utilizing a type guard might be the better approach.

Utilizing Type Annotation

Type annotation allows you to clearly indicate to the compiler the expected type of the error. This method entirely eliminates the need for utilizing the "any" type:

interface Error {
    status: string,
    statusText: string,
    error: { message: string | undefined } | undefined;
}

catchError((error: Error | undefined) => {
    //...
}

Implementing a Type Guard

A type guard comes into play when a value could potentially be of a certain type but does not exclusively have to be of that type. By using a type guard, the type will be checked and within the subsequent block, the variable will assume the identified type:

function isError(value: any | undefined): value is Error {
    return error && ((error as Error).status !== undefined);
}

catchError(error => {
    if (isError(error)) {
        //Within this block, the variable 'error' will be recognized as type Error
    }
}

What is the solution for resolving the no-unsafe-any rule?

Answer №1

Answer №2

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