Vue 3 script setup prop validation typescript

Question

I'm trying to replace my Vue 2 options API props object code with Vue 3 script setup syntax in typescript.

Existing:

type: {
  type: String,
  default: 'button',
  validator: (prop) => ['button', 'submit', 'reset'].includes(prop)
}

I have this so far:

<script lang="ts" setup>
interface Props {
  type?: string;
}

const props = withDefaults(defineProps<Props>(), { type: 'button' });
</script>

but I can't find any info on how to handle prop validators in the script setup syntax

Answer 1

You can use in defineProps the same structure as in Options API. (DOCS)

<script lang="ts" setup>
  type Type = 'button' | 'submit' | 'reset';

  interface Props {
    type: Type;
  }

  const props = defineProps<Props>({ 
    type: {
      type: String,
      default: 'button',
      validator: (prop: Type) => ['button', 'submit', 'reset'].includes(prop)
    }
  });
</script>

Answer 2

For those also arriving from Google, and since the Vue docs aren't very clear on this, you can do a straight conversion using the Vue 2 style options format as well:

const props = defineProps({
  type: {
    type: String,
    default: 'button',
    validator: (prop) => ['button', 'submit', 'reset'].includes(prop)
  }
})

Answer 3

I believe I've worked it out, still new to typescript but this should be the equivalent (will be validated in typescript rather than vue runtime)

interface Props {
  type?: 'button' | 'submit' | 'reset';
}

Answer 4

Following up @Titan 's answer I like to write it in a way in which I get direct access to the prop in the script.

<script setup lang="ts">
type Type = 'button' | 'submit' | 'reset';

const { type } = withDefaults(defineProps<{
  type?: Type
}>(),{
  type: 'button'
})
//As it is already deconstructed, you can use it as `type` instead of `props.type`
</script>

Although I would recommend changing the prop name from type to something different since typescript might misunderstand the prop name for an actual type declaration.