Can't access error property on safeParse() result?

[nemanjam]

I have type check that success is false but still can't access error property. When I log the result I see that error property exists.

const validationResult = postsGetSchema.safeParse(query);
if (!validationResult.success) {
  throw ApiError.ZodErrorToApiError(validationResult.error);

Example from Readme:

const result = stringSchema.safeParse("billie");
if (!result.success) {
  // handle error then return
  result.error;
} else {
  // do something
  result.data;
}

[sylvestertarnowski]

There was an update in Typescript 4.4 that improved type guard narrowing, maybe bumping typescript version to this or above could help?
https://www.typescriptlang.org/docs/handbook/release-notes/typescript-4-4.html#control-flow-analysis-of-aliased-conditions-and-discriminants

[nemanjam]

It works when I set "strict": true in tsconfig.json.

[sylvestertarnowski]

I assume it enabled Strict Null Checks, which would allow type narrowing to know when error is undefined, and when it is defined:
https://www.typescriptlang.org/tsconfig#strictNullChecks

[nemanjam]

Zod Readme says strict mode must be enabled.

https://github.com/colinhacks/zod#installation