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'
将是一些 'notGuidbutJustString'
那么我将看到 typescript 编译错误。
更新:正如 David Sherret 所说:无法在编译时确保基于正则表达式或其他一些函数的字符串值,但可以在运行时在一个地方进行所有检查.
最佳答案
您可以围绕一个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
更新
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) {
}
关于types - typescript 中的 GUID/UUID 类型,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37144672/