How to Query virtual column in Typeorm

I have this entity declaration with the virtual column slugWithUuid. How can I query this kind of column within a query.

@Entity()
export class Course extends DefaultCredential {
  @Column({ type: 'varchar', length: 75 })
  title: string;

  @Column({ type: 'varchar', length: 50 })
  slug: string;

  @Column({ type: 'json', nullable: true })
  additionalInfoJSON: any;

  slugWithUuid: string;
  @AfterLoad()
  setCustomProperty() {
    this.slugWithUuid =
      this.slug + '-' + JSON.parse(this.additionalInfoJSON)?.uuid;
  }

  @BeforeInsert()
  insertShortUUID() {
    const shortUUID = generateUUID(6);
    const additionalInfo = {
      ...this.additionalInfoJSON,
      uuid: shortUUID,
    };
    this.additionalInfoJSON = JSON.stringify(additionalInfo);
  }
}

This is an example of my query:

async findBySlug(slug: string) {
    const entity = await this.repos.findOne({
      where: { slugWithUuid: slug }
    });

    if (!entity) {
      throw new NotFoundException('Entity not found with this slug');
    }
    return entity;
  }

Is there any other way to declare this virtual column to be able to filter it within a query, because this query return me this error for the moment:

[Nest] 637270 - 10/03/2024, 9:37:42 AM ERROR [ExceptionsHandler] Property "slugWithUuid" was not found in "Course". Make sure your query is correct. EntityPropertyNotFoundError: Property "slugWithUuid" was not found in "Course". Make sure your query is correct.

After a long research, I found out this solution:

@VirtualColumn({
    query: (alias) => {
      const uuidSelect = `JSON_UNQUOTE(JSON_EXTRACT(JSON_UNQUOTE(${alias}.additionalInfoJSON), '$.uuid'))`;
      return `CONCAT(
        ${alias}.slug,
        '-',
        ${uuidSelect}
      )`;
    },
  })
  slugWithUuid: string;

It re-declare slugWithUuid with @VirtualColumn and now I can query it.