Question

我试图通过滥用打字稿来强制执行自定义类型规则。我已经开始使用它了，但是我在接口上遇到了麻烦。这是有效的类版本：

export class HeaderElement extends WebElement { 
  // 'dummy' type: won't create any instances of it at runtime.
  // type-filler member that doesn't exist in other WebElement.
  _HEADER_ELEMENT_TYPE_FILLER_6897689FSAFDSF: string; 
  //  -> dummy member: not even read/written/set in code.
}
const export function getHeaderElement(): HeaderElement {
  // findElement: returns WebElement type
  const header = browser.findElement( ... );
  return header as HeaderElement; 
  // **tricking typescript (actual runtime type would be 'WebElement')**
}
export function clickWowPageMenu(
  headerElem: HeaderElement
  // (headerElem's runtime-type is WebElement. 
  // It's 'HeaderElem' only in compilation phase.)
) { ... }
const headerElem = getHeaderElement();
const saveButtonElem = getSaveButtonElement();
clickWowPageMenu( headerElem ); // ok
clickWowPageMenu( saveButtonElem ); // type-error.

但是对于interfaces版本，以下代码会导致typescript类型错误：

interface IContext {
   __EXTENDS_IContextElem123124: string,
}
interface IChatContext extends IContext {
   name: string,
}
const test: IChatContext = { name: 'namehere' } as IChatContext;
// want: no type error here. (but 'missing __EXTENDS_IContextElem123124')
const test2: IChatContext = {} as IChatContext;
// want: type error here. (missing property 'name')

我如何滥用＆＃39;打字机类型检查接口我想要的方式？还是有其他方式（例如官方的东西）？

谢谢！

---------------- EDITS --------- 抱歉打字错误

const test: IContextElem ... --> IContext

添加缺失的详细信息（想要/想要的）

Answer 1

TypeError是正确的，因为您有// URL scheme: http://gist.github.com/* // API endpoint: http://gist.github.com/services/oembed/ // Example call: http://gist.github.com/services/oembed/?url=http%3A//gist.github.com/54099而不是IContextElem。

更改后，您的示例会在playground

中再次编译

欺骗打字稿编译器为变量分配错误接口

1 个答案: