Question

仅当登录的用户具有与实体相同的ID时，我才会序列化实体User的属性“ email”。

实体用户：

@Entity()
@Exclude()
export class User {

    @Expose()
    @PrimaryGeneratedColumn("uuid")
    id: string;

    @Expose() // Only if logged user == this
    @Column({nullable: true, default: null})
    public email: string;

    @Expose()
    @Column({nullable: true, default: null})
    public username: string;

    @Column({nullable: true, default: null})
    public password: string;

    @CreateDateColumn()
    public create_at: Date;

    @UpdateDateColumn()
    public update_at: Date;
}

Answer 1

您可以执行以下操作：

@Entity()
@Exclude()
export class User {

    @Expose()
    @PrimaryGeneratedColumn("uuid")
    id: string;

    @Column({nullable: true, default: null})
    public email: string;

    public connectedUser: string?;

    @Expose({ name: 'email'}) // Only if logged user
    public get hideableEmail(): string? {
        if(this.email === this.connectedUser) {
            retun this.email;
        }
        return null;
    }

    @Expose()
    @Column({nullable: true, default: null})
    public username: string;

    @Column({nullable: true, default: null})
    public password: string;

    @CreateDateColumn()
    public create_at: Date;

    @UpdateDateColumn()
    public update_at: Date;
}

Answer 2

没有直接的方法可以基于对象本身有条件地排除属性。您可以使用分组来控制曝光：

@Expose({ groups: ['owner'] })
@Column({nullable: true, default: null})
public email: string;

，然后在控制器中根据您的条件添加组：

const groups = [];
if (isCurrentUser) {
  groups.push('owner');
}
classToPlain(user, { groups })

如何仅显示当前用户的字段？

2 个答案: