Typescript通用存储库模式-带方法的返回类型

Question

When trying to create a generic repository, I ended up with an implementation that looks like this:

export class DynamoDbRepository<T extends IRepositoryItem> extends BaseRepository<T> {
    private _tableName: string = void 0;
    private _type;

    constructor(tableName: string, type: new () => T) {
    ...
    }

    ...

    findOne(appId: string, id: string): Promise<T> {
      const params = {
        Key: {
            "Id": id,
            "AppId": appId 
        },
        TableName: this._tableName
      }

      return new Promise((resolve, reject) => {
        DynamoDbClient.get(params, (error, result) => {
            // handle potential errors
            if (error) {
                Logger.error(error);
                reject(new Error(`GetItemFailed for table '${this._tableName}'`));
            }

            // no items found
            if (!result.Item) reject(new Error(`ItemNotFound in table '${this._tableName}'`));

            // create instance of correct type, map properties
            let item = new this._type();
            Object.keys(result.Item).forEach((key) => {
                item[key] = result.Item[key];
            })

            // return the item
            resolve(item);
        });
    });
}

And I use it like this, Which is less than ideal as I need to pass the class name in addition to specifying the generic type:

const userRepository = new DynamoDbRepository<User>(Tables.USERS_TABLE, User);

Is there a solution that is cleaner on one hand, and would still allow me to return the correct type?

Answer 1

There is no way to create new class instances based on generic type. Because there is no any typing information in the compiled version of your code in JavaScript, so you can't use T to create a new object.

You can do this in a non-generic way by passing the type into the constructor - this is what exactly you're doing in your example.

For more details - follow this post .