3 replies

Dynamic page omit assertion of router.query

Hey , I am using an eslint rule that prevents me from invalid template literals. Is there anything I can do to omit

pid as string

pid as string

import { useRouter } from 'next/router'

const Post = () => {
  const router = useRouter()
  const { pid } = router.query

  return <Link href={`/somepath/${pid}`}>Click</Link> // @typescript-eslint/restrict-template-expressions pid - string | string[] | undefined
}

export default Post

import { useRouter } from 'next/router'

const Post = () => {
  const router = useRouter()
  const { pid } = router.query

  return <Link href={`/somepath/${pid}`}>Click</Link> // @typescript-eslint/restrict-template-expressions pid - string | string[] | undefined
}

export default Post

TTC

Theo's Typesafe Cult•4y ago•

3 replies

WOLG

Dynamic page omit assertion of router.query

Hey , I am using an eslint rule that prevents me from invalid template literals. Is there anything I can do to omit

pid as string

pid as string

import { useRouter } from 'next/router'

const Post = () => {
  const router = useRouter()
  const { pid } = router.query

  return <Link href={`/somepath/${pid}`}>Click</Link> // @typescript-eslint/restrict-template-expressions pid - string | string[] | undefined
}

export default Post

import { useRouter } from 'next/router'

const Post = () => {
  const router = useRouter()
  const { pid } = router.query

  return <Link href={`/somepath/${pid}`}>Click</Link> // @typescript-eslint/restrict-template-expressions pid - string | string[] | undefined
}

export default Post

Dynamic page omit assertion of router.query

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