How to get output with safeParse without checking for success

[ericmartinezr]

Hello!

My first question here.

I'm trying to get the output out of safeParse without checking directly for success but after checking for not success. Let me explain myself.

const data = ...;
const someSchema = schema(...);
const result = safeParse(someSchema, data);
if (!result.success) {
   // Throw some error
}
// This should probably work, we already went through "not success"
const MyObject = result.output

The above snipet shows that output doesn't exist if we're not checking directly for success. I have a relatively huge amount of code that should happen when it's parsed successfuly but I don't want to enclose all of it inside a if (parse.success) {}, by checking that's it's not success should be enough imho.

Am I doing something wrong?

Thanks in advance!

[fabian-hiller]

Your approach works fine for me. Did you set strict to true in your tsconfig.json?

[fabian-hiller]

[fabian-hiller]

What version are you using? Have you upgraded to v0.22.0?

[ericmartinezr]

I'm using v0.21.0 but I noticed now thanks to your example that the version is not the problem.

I'm not sure if this is normal but this is what I'm doing

but if I throw inside the if it works like just in your example

Not sure if that's intentional or not, but I will see to throw directly inside the if instead of using a function, I would prefer the function though.

Thanks for your help @fabian-hiller and sorry for the noise.

[fabian-hiller]

I think TypeScript needs to see the throw statement. So you could change your code like this:

private _rethrow() {
  return new Error('Invalid');
}

async guarder() {
  // ...
  if (!parseResult.success) {
    throw this._rethrow()
  }
  const data = parseResult.output;
}

[ericmartinezr]

Thanks a lot! That worked!