Question

我想使用一个typescript变量，允许根据其值假定不同的类型，并进行类型检查（因此我需要避免使用“ any”）。

例如：

class Car
{
    public string LicensePlate;
}
class Driver
{
    public string Name;
}

function GetCar(licensePlate: string): Car { ... }
function GetDriver(name: string): Driver { ... }

let licensePlate: string;
let name: string;

let result = GetCar("123");
licensePlate = result.LicensePlate;

result = GetDriver("Bob");
name = result.Name;

基本上，我需要“重用”具有不同类型值的变量，以进行类型检查。

打字稿可以允许吗？

谢谢

Answer 1

只要您不进行任何类型的初始化，就可以将其转换
只需声明没有类型的变量即可，例如let test1

class A {
  val1: string
}
class B {
  val2: number
}

// OK
let test1
test1 = new A()
console.log('test1', (test1 as A).val1) // requires type cast
test1 = new B()
console.log('test1', (test1 as B).val2) // requires type cast

// Error
let test2 = new A()
console.log('test2', test2.val1)        // dont need type cast
test2 = new B()                         // Property 'val1' is missing in type 'B' but required in type 'A'
console.log('test2', test2.val2)        // Property 'val2' does not exist on type 'A'
console.log('test2', (test2 as B).val2) // Conversion of type 'A' to type 'B' may be a mistake because neither type sufficiently overlaps with the other

Answer 2

我找到了解决方法：

将Typescript版本更新为3.5.3
设置noImplicitAny标志

现在我可以做：

let result; //Here result type is any   
result = new A(); //Here result type is A   
result = new B(); //Here result type is B

智能感知和类型检查就像基于变量赋值的魅力一样工作

是否可以根据其值类型来动态更改变量类型？

2 个答案: