Typescript Union类型和接口

我是新的打字稿，我不明白为什么接口没有在对象中声明，当我调用数组或函数属性。如果我把它称为任何对象然后数组或函数属性获取错误。Typescript Union类型和接口

其中我使用地址属性作为字符串，然后我在makeNewEmployee对象中声明接口，然后没有错误。我对此有点困惑。

这里是我的代码

interface makeAnything{ 
    user:string; 
    address:string| string[]| (()=>string); 
} 

/// address as string 
let makeNewEmployee:makeAnything = { 
    user:"Mirajehossain", 
    address:"Dhaka,Bangladesh" 
}; 
console.log(makeNewEmployee.address);

在这里我使用makeAnything接口在我makeNewEmployee对象和申报地址财产功能，为什么我在控制台中看到错误？

///address as function 
let makeSingleEmployee:makeAnything = { 
    user:'Miraje hossain', 
    address:():any=>{ 
     return{ 
      addr:"road 4,house 3", 
      phone:18406277 
     } 
    } 
}; 
console.log(makeSingleEmployee.address()); ///getting error

来源

2017-05-16 Md Alamin

您可以使用类型警卫来帮助编译器知道address是一个函数。见type guards。

例子：

// string or string[] cannot be instances of Function. So the compiler infers that `address` must be (()=>string) 
if (makeSingleEmployee.address instanceof Function) { 
    console.log(makeSingleEmployee.address());  
} 

// typeof makeSingleEmployee.address will be "function" when address is a function 
if (typeof makeSingleEmployee.address != "string" && typeof makeSingleEmployee.address != "object") { 
    console.log(makeSingleEmployee.address()); 
}

或者，如果你肯定知道address是一个功能，您可以将其转换为any：

console.log((makeSingleEmployee as any).address());

来源

2017-05-16 15:52:38 Saravana

Typescript Union类型和接口

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