成员的多种类型签名，TypeScript 中的联合类型
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Multiple type signatures for members, Union Types in TypeScript
提问by basarat
If I have a property that might be a string or a boolean how do I define it:
如果我有一个可能是字符串或布尔值的属性，我该如何定义它：
interface Foo{
bar:string;
bar:boolean;
}
I don't want to resort to:
我不想诉诸：
interface Foo{
bar:any;
}
I don't think its possible without any. You can answer any of these:
我认为没有any. 您可以回答以下任一问题：
Have I overlooked a spec and its possible right now? Is something like this planned? Has a feature request been logged: http://typescript.codeplex.com/workitem/list/basic? (UPDATEthis is the issue report you can vote on https://typescript.codeplex.com/workitem/1364)
我现在是否忽略了规范及其可能？有这样的计划吗？是否记录了功能请求：http: //typescript.codeplex.com/workitem/list/basic？（更新这是您可以在https://typescript.codeplex.com/workitem/1364 上投票的问题报告）
I would imagine something like this:
我会想象这样的事情：
interface Foo{
bar:string;
bar:boolean;
bar:any;
}
var x:Foo = <any>{};
x.bar="asdf";
x.bar.toUpperCase(); // intellisence only for string
采纳答案by Ryan Cavanaugh
This is usually referred to as "union types". The TypeScript type system from 1.4 does allow for this.
这通常称为“联合类型”。1.4 中的 TypeScript 类型系统确实允许这样做。
See: Advanced Types
请参阅：高级类型
回答by MarcG
As of 2015, union-types work:
截至 2015 年，联合类型工作：
interface Foo {
bar:string|boolean;
}
回答by Damian
Not saying this answers your question, but could you resort to something like this?
不是说这回答了你的问题，但你能诉诸于这样的事情吗？
interface Foo<T>{
bar:T;
}
function createFoo<T>(bar:T) : Foo<T>{
return {bar:bar};
}
var sFoo = createFoo("s");
var len = sFoo.bar.length;
var bFoo = createFoo(true);
var result = bFoo.bar === true;
