safeParse discriminated union doesn't have 'error' attribute

const parsedUser = userZ.safeParse(user.toObject())

if (!parsedUser.success) {
      console.error(parsedUser.error)
      return null
}

Gives me this error:

Property 'error' does not exist on type 'SafeParseReturnType<{ telegramId?: number; telegramHandle?: string; twitterId?: number; twitterHandle?: string; address?: string; ens?: string; telegramFirstName?: string; twitterUsername?: string; isMetabotEnabled?: boolean; }, { ...; }>'.
  Property 'error' does not exist on type 'SafeParseSuccess<{ telegramId?: number; telegramHandle?: string; twitterId?: number; twitterHandle?: string; address?: string; ens?: string; telegramFirstName?: string; twitterUsername?: string; isMetabotEnabled?: boolean; }>'.ts(2339)

when success===true it doesn’t give an error when i try to access data. Example:

return parsedUser.success ? parsedUser.data : null

[Daniel-Griffiths]

I found a solution:

This fails

 const result = validator.safeParse(example);

if (!result.success) {
    result.error // this shows a TS error
}

This works

 const result = validator.safeParse(example);

if (result.success === false) {
    result.error // it works!
}

Perhaps the docs should update the example so users can infer the correct types.

[aaronmcadam]

Found the issue.

In order to do this, if (!result.success), strictNullChecks must be set to true.

Alternatively, you can do this: if (result.success === false)

or

if (!result.success) {
  const { error } = result as SafeParseError;

I have strict: true, but I’m still seeing errors when doing !result.success. I’m still having to do a manual comparison with ===.