如何定义的一个定稿的定义的一系列其中的所有元素是数量,除了第一次

0

的问题

这些阵列在的问题是SVG路段,例如 ['L', 0, 0] 我基本上使用这一定义这些阵列:

// doSomethingToSegment.js

/** @type {Object.<string, number>} */
const paramsCount = {
  a: 7, c: 6, h: 1, l: 2, m: 2, r: 4, q: 4, s: 4, t: 2, v: 1, z: 0,
};
  
/**
 * This definition is WRONG, FIX ME!!
 *
 * @typedef {(string|number)[]} segment
 */

/**
 * Check segment validity.
 *
 * @param {segment} seg input segment
 * @return {boolean} segment is/not valid
 */
function checkSegment(seg) {
  const [pathCommand] = seg;
  const LK = pathCommand.toLowerCase();
  const UK = pathCommand.toUpperCase();
  const segmentValues = seg.slice(1);
  const expectedAmount = paramsCount[LK];

  return checkPathCommand(UK) && checkPathValues(segmentValues, expectedAmount);
}

/**
 * @param {string} ch input character
 * @returns {boolean} true when `ch` is path command
 */
function checkPathCommand(ch) {
  return ('ACHLMRQSTVZ').includes(ch);
}

/**
 * @param {Number[]} values input values
 * @param {Number} expected amount
 * @return {boolean} segment has/not the right amount of valid numbers
 */
function checkPathValues(values, expected) {
  return values.length === expected && values.every(x => !Number.isNaN(x));
}

现在 pathCommand.toLowerCase() 呼叫抛出这个错误:

Property 'toLowerCase' does not exist on type 'string | number'.
  Property 'toLowerCase' does not exist on type 'number'.

segmentValues 引发这一:

Argument of type '(string | number)[]' is not assignable to parameter of type 'number[]'.
  Type 'string | number' is not assignable to type 'number'.
    Type 'string' is not assignable to type 'number'.

因此,如何定义的一个定义类型的定义 @type {WHAT} segment) 满足这种具体的需要?

ecmascript-6 javascript typescript
2021-11-22 15:33:56
1

最好的答案

2
type A = [string, ...number[]];

更多信息有关的其他元素元组种类型: https://www.typescriptlang.org/docs/handbook/2/objects.html#tuple-types

这里是例子文档:

组可能还有其他元素,这必须是一系列/tuple 类型。

type StringNumberBooleans = [string, number, ...boolean[]];
type StringBooleansNumber = [string, ...boolean[], number];
type BooleansStringNumber = [...boolean[], string, number];
2021-11-22 15:49:59

快速的问题:这个得到采用Tuple的对象? 喜欢的东西 {type: string, ...{string, number}[]}我是测试的一切没有得到它的工作。
thednp

对不起,我不太理解它。 也许你可以发布一个新的堆溢问题与更详细一点?
Anastasia

同样的事情但是对于对象,而不是的 [string, ...number[]] 是否有可能喜欢的东西 {myString: string, ...{string, number}[]].
thednp

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