Typescript中的 GUID UUID 类型

发布于05月11日

我有这个功能:

function getProduct(id: string){    
    //return some product 
}

其中id实际上是GUID.Typescript没有guid类型.是否可以手动创建类型GUID？

function getProduct(id: GUID){    
    //return some product 
}

所以如果'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'是'notGuidbutJustString'，那么我会看到typescript编译错误.

Update:正如David Sherret所说:没有办法确保编译时基于regex或其他函数的字符串值，但有可能在运行时在一个地方执行所有判断.

推荐答案

您可以围绕字符串创建一个包装器，并将其传递:

class GUID {
    private str: string;

    constructor(str?: string) {
        this.str = str || GUID.getNewGUIDString();
    }

    toString() {
        return this.str;
    }

    private static getNewGUIDString() {
        // your favourite guid generation function could go here
        // ex: http://stackoverflow.com/a/8809472/188246
        let d = new Date().getTime();
        if (window.performance && typeof window.performance.now === "function") {
            d += performance.now(); //use high-precision timer if available
        }
        return 'xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx'.replace(/[xy]/g, (c) => {
            let r = (d + Math.random() * 16) % 16 | 0;
            d = Math.floor(d/16);
            return (c=='x' ? r : (r & 0x3 | 0x8)).toString(16);
        });
    }
}

function getProduct(id: GUID) {    
    alert(id); // alerts "xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx"
}

const guid = new GUID("xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx");
getProduct(guid); // ok
getProduct("notGuidbutJustString"); // errors, good

const guid2 = new GUID();
console.log(guid2.toString()); // some guid string

Update

另一种方法是使用品牌:

type Guid = string & { _guidBrand: undefined };

function makeGuid(text: string): Guid {
  // todo: add some validation and normalization here
  return text as Guid;
}

const someValue = "someString";
const myGuid = makeGuid("ef3c1860-5ce6-47af-a13d-1ed72f65b641");

expectsGuid(someValue); // error, good
expectsGuid(myGuid); // ok, good

function expectsGuid(guid: Guid) {
}