Question

将通用类型投射到诸如Product，Post，Todo，Customer等实体的任何作品（E类型将是打字稿类型，类或接口）：

function test<E>(o:E):string {
  return (o as any)['property']
}

只想看看是否强制转换为任何对象？

要求提供完整的上下文。这是正在执行的全部功能：

/**
 * 
 * @param entities The entities to search
 * @param exclude Keys to exclude from each entity
 * 
 * @return E[] Array of entities with properties containing the search term.
 */
export function search<E extends WithProperty>(query:string='', entities:E[], exclude:string[]=[]) {
  const { isArray } = Array

  query = query.toLowerCase();

  let keys:string[] = []
  if (entities.length > 0) {
    keys = excludeKeys(entities[0], exclude)
  }

  return entities.filter(function (e:E) {
    return keys.some((key)=>{
      const value = (e as any)[key];
      if (isArray(value)) {
        return value.some(v => {
          return new String(v).toLowerCase().includes(query);
        });
      }
      else if (!isArray(value)) {
        return new String(value).toLowerCase().includes(query);
      }
    })
  });
}

/**
 * The method can be used to exclude keys from an instance
 * of type `E`.  
 * 
 * We can use this to exclude values when searching an object.
 * 
 * @param entity An instance of type E
 * @param eclude  The keys to exclude
 * 
 */
export function excludeKeys<E>(entity: E, exclude: string[]) {
  const keys: string[] = Object.keys(entity);
  return keys.filter((key) => {
    return exclude.indexOf(key) < 0;
  });
}

Answer 1

如果您知道类型约束具有名为属性的属性，则可以定义一个接口，该接口定义该属性，然后使用告诉E对其进行扩展的约束。然后您将可以访问该属性而无需强制转换。

interface WithProperty{
  property:string;
}

function test<E extends WithProperty>(o:E):string {
  return o.property;  // or o["property"] is valid access.
}

Playground

编辑：自从您更新了示例。还有另一种方法可以使用关键字keyof。同样，使用此不需要知识的属性。我已经修改了您的示例，如下所示：

export function search<E>(query:string='', entities:E[], exclude:string[]=[]) {
  const { isArray } = Array
  type EKey = keyof E;
  query = query.toLowerCase();

  let keys : EKey[] = []
  if (entities.length > 0) {
    keys = excludeKeys<E>(entities[0], exclude)
  }

  return entities.filter(function (e:E) {
    return keys.some((key =>{
      const value = e[key];
      if (isArray(value)) {
        return value.some(v => {
          return v.toLowerCase().includes(search);
        });
      }
      else if (!isArray(value)) {
        return new String(value).toLowerCase().includes(query);
      }
    })
  });
 }

由于这种discussion的紧迫性，对于代码的excludeKeys部分来说，这是不可避免的。

export function excludeKeys<E>(entity: E, exclude: string[]) {
 const keys: string[] = Object.keys(entity);
  return <(keyof E)[]>keys.filter((key) => {
    return exclude.indexOf(key) < 0;
   });
}

Playground

在打字稿中访问泛型类型的属性？

1 个答案: