Add Solid JS support #24
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🦋 Changeset detectedLatest commit: 1a6ee1e The changes in this PR will be included in the next version bump. This PR includes changesets to release 2 packages
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| return ( | ||
| <Show when={fieldState()}> |
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I'm not sure we need this Show check if we're throwing above? We could also remove the function wrapper on fieldState without this check I think
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I agree but if I do this, it says it can be undefined 😭
Should I add the !?
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And if I do this, it loses reactivity
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Do you have other ideas? I could change the tsconfig for this template to strict instead of strictest and it will work in this approach
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Ahh thanks, I see the issue now. I'd like to keep strictest since this will be generated code in the user's project, and I want to pass the strictest config possible to ensure we don't raise type errors in your project.
I figured out the problem though! We need to hoist the state inside the state function. That way it can guarantee the function never returns undefined, since it's always mapped to an error. The solution you showed above does not guarantee that. You can check fieldState() !== undefined, but what if calling that function a second time returns a different value? TypeScript unfortunately isn't smart enough to see that.
| sanitizeFormResult | ||
| ].forEach((formResult) => { | ||
| if (formResult?.data) { | ||
| console.log(formResult.data); |
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