从关系中获取字段而不是对象(typeorm)

Question

我对typeorm很陌生，当我想检索具有关系的表时，该关系的列将在一个具有该表名称的对象中检索。下面是一个例子

import { Entity, PrimaryGeneratedColumn, Column, OneToMany } from "typeorm"
import { Photo } from "./Photo"

@Entity()
export class User {
    @PrimaryGeneratedColumn()
    id: number

    @Column()
    name: string

    @OneToMany((type) => Photo, (photo) => photo.user)
    photos: Photo[]
}

import { Entity, PrimaryGeneratedColumn, Column, ManyToOne } from "typeorm"
import { User } from "./User"

@Entity()
export class Photo {
    @PrimaryGeneratedColumn()
    id: number

    @Column()
    url: string

    @ManyToOne((type) => User, (user) => user.photos)
    user: User
}

const user = await createQueryBuilder("user")
    .leftJoinAndSelect("user.photos", "photo")
    .where("user.name = :name", { name: "Timber" })
    .getOne()

结果

{
    id: 1,
    name: "Timber",
    photos: [{
        id: 1,
        url: "me-with-chakram.jpg"
    }, {
        id: 2,
        url: "me-with-trees.jpg"
    }]
}

我希望将关系的字段与表中的字段放在相同的级别上。

预期结果

{
   {
    id: 1,
    name: "Timber",
    id: 1,
    url: "me-with-chakram.jpg"
    }
   {
    id: 1,
    name: "Timber",
    id: 2,
    url: "me-with-trees.jpg"
    }
}

有办法这样做吗？我在他们的文档里什么都没找到

Answer 1

我试着用“*”，它起了作用

  const user = await createQueryBuilder("user")
        .leftJoinAndSelect("user.photos", "photo")
        .select('*')
        .where("user.name = :name", { name: "Timber" })
        .getOne();

Answer 2

你可以试试这个，

const user = await createQueryBuilder("user")
    .leftJoinAndSelect("user.photos", "photo")
    .select([
            'user.id',
            'user.name',
            'photos.id',
            'photos.url'
        ])
    .where("user.name = :name", { name: "Timber" })
    .getOne();