L0v3ch4n

外观

Query String Parser

约 319 字大约 1 分钟

2026-02-11

题目

You're required to implement a type-level parser to parse URL query string into a object literal type.

Some detailed requirements:

  • Value of a key in query string can be ignored but still be parsed to true. For example, 'key' is without value, so the parser result is { key: true }.
  • Duplicated keys must be merged into one. If there are different values with the same key, values must be merged into a tuple type.
  • When a key has only one value, that value can't be wrapped into a tuple type.
  • If values with the same key appear more than once, it must be treated as once. For example, key=value&key=value must be treated as key=value only.

解题思路

待补充

答案

type ParseQueryString = any

验证

import type { Equal, Expect } from '@type-challenges/utils'

type cases = [
  Expect<Equal<ParseQueryString<''>, {}>>,
  Expect<Equal<ParseQueryString<'k1'>, { k1: true }>>,
  Expect<Equal<ParseQueryString<'k1&k1'>, { k1: true }>>,
  Expect<Equal<ParseQueryString<'k1&k2'>, { k1: true, k2: true }>>,
  Expect<Equal<ParseQueryString<'k1=v1'>, { k1: 'v1' }>>,
  Expect<Equal<ParseQueryString<'k1=v1&k1=v2'>, { k1: ['v1', 'v2'] }>>,
  Expect<Equal<ParseQueryString<'k1=v1&k2=v2'>, { k1: 'v1', k2: 'v2' }>>,
  Expect<Equal<ParseQueryString<'k1=v1&k2=v2&k1=v2'>, { k1: ['v1', 'v2'], k2: 'v2' }>>,
  Expect<Equal<ParseQueryString<'k1=v1&k2'>, { k1: 'v1', k2: true }>>,
  Expect<Equal<ParseQueryString<'k1=v1&k1=v1'>, { k1: 'v1' }>>,
  Expect<Equal<ParseQueryString<'k1=v1&k1=v2&k1=v1'>, { k1: ['v1', 'v2'] }>>,
  Expect<Equal<ParseQueryString<'k1=v1&k2=v1&k1=v2&k1=v1'>, { k1: ['v1', 'v2'], k2: 'v1' }>>,
  Expect<Equal<ParseQueryString<'k1=v1&k2=v2&k1=v2&k1=v3'>, { k1: ['v1', 'v2', 'v3'], k2: 'v2' }>>,
  Expect<Equal<ParseQueryString<'k1=v1&k1'>, { k1: ['v1', true] }>>,
  Expect<Equal<ParseQueryString<'k1&k1=v1'>, { k1: [true, 'v1'] }>>,
]

参考

无