根据TypeOrm文档:https://github.com/typeorm/typeorm/blob/master/docs/select-query-builder.md#joining-relations
我们可以查询连接的实体的字段,该字段将填充所有的字段到响应中。我不确定如何限制只有几个选定的字段(单个/多个),我尝试添加'select([])‘,但它在生成的SQL查询中不起作用,我可以看到它正在查询所有字段。
代码:
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;
@Column()
alt: string;
@ManyToOne(type => User, user => user.photos)
user: User;
}在代码中:
const user = await createQueryBuilder("user")
.leftJoinAndSelect("user.photos", "photo")
.where("user.name = :name", { name: "Timber" })
.getOne();上面的代码给出的输出是-
{
id: 1,
name: "Timber",
photos: [{
id: 1,
url: "me-with-chakram.jpg",
alt: "Me With Chakram"
}, {
id: 2,
url: "me-with-trees.jpg",
alt: "Me With Trees"
}]
}有没有办法我只能查询'url‘和'alt’,这样输出就会像这样--
{
id: 1,
name: "Timber",
photos: [{
url: "me-with-chakram.jpg",
alt: "Me With Chakram"
}, {
url: "me-with-trees.jpg",
alt: "Me With Trees"
}]
}发布于 2020-10-09 00:44:00
const user = await createQueryBuilder("user")
.leftJoinAndSelect("user.photos", "photo")
.select(['user', 'photo.url', 'photo.alt'])
.where("user.name = :name", { name: "Timber" })
.getOne();或
const user = await createQueryBuilder("user")
.leftJoinAndSelect("user.photos", "photo")
.addSelect(['photo.url', 'photo.alt'])
.where("user.name = :name", { name: "Timber" })
.getOne();(对第二个不太确定)
发布于 2021-11-04 18:31:31
Art Olshansky给出的第一个答案是正确的(第二个答案不是),所以,你必须站在“基本”实体上,但是,如果你不需要所有/任何用户字段,你可以直接应用
const user = await createQueryBuilder("user")
.leftJoinAndSelect("user.photos", "photo")
.select(['user.id', 'photo.url', 'photo.alt'])
.where("user.name = :name", { name: "Timber" })
.getOne();
https://stackoverflow.com/questions/64260584
复制相似问题