如何使用TypeORM

Question

Note与Subject有多对多关系

最好的查询方式是什么？我会喜欢写以下内容，以便将所有主题都记录下来：

  const subjectRepo = connection.getRepository(Subject);
  const response = await subjectRepo.find({
    relations: ['notes'],
    where: { note }
  });

但这会返回所有主题，而不仅仅是笔记上的主题。

Reln定义为：

  @ManyToMany(() => Subject, (subject: Subject) => subject.notes)
  subjects: Subject[];

-和-

  @ManyToMany(() => Note, note => note.subjects)
  @JoinTable()
  notes: Note[];

执行的查询是：

SELECT "Subject"."id" AS "Subject_id", "Subject"."name" AS "Subject_name", "Subject"."description" AS "Subject_description", "Subject"."createdDate" AS "Subject_createdDate", "Subject"."updatedDate" AS "Subject_updatedDate", "Subject"."notebookId" AS "Subject_notebookId", "Subject"."measurementsId" AS "Subject_measurementsId", "Subject_notes"."id" AS "Subject_notes_id", "Subject_notes"."content" AS "Subject_notes_content", "Subject_notes"."notedAt" AS "Subject_notes_notedAt", "Subject_notes"."createdDate" AS "Subject_notes_createdDate", "Subject_notes"."updatedDate" AS "Subject_notes_updatedDate", "Subject_notes"."notebookId" AS "Subject_notes_notebookId" FROM "subject" "Subject" LEFT JOIN "subject_notes_note" "Subject_Subject_notes" ON "Subject_Subject_notes"."subjectId"="Subject"."id" LEFT JOIN "note" "Subject_notes" ON "Subject_notes"."id"="Subject_Subject_notes"."noteId"

---

注意：您可以执行以下操作：

  return subjectRepo
    .createQueryBuilder('subject')
    .leftJoin('subject.notes', 'note')
    .where('note.id = :id', { id: note.id })
    .getMany();

但是我希望使用更少的字符串并显式加入的方法

Answer 1

您试图让TypeORM生成的SQL大致如下

SELECT *
FROM subject
JOIN subject_note AS jt on jt.subject_id = subject.id
WHERE jt.note_id = :id

1。 repo.find

无法实现

在撰写本文时，无法使用where在联接表上创建repo.find(...)子句。您可以join（doc），但是where子句仅影响存储库的实体。

TypeORM还默默地忽略无效的where子句，因此请小心那些子句。

2。重新选择笔记实体

如果您要给定subject中的所有note，则需要使用查询生成器（如您提到的那样），或者需要重新选择具有其关系的注释对象。

note = await noteRepo.find({
    relations: ['subjects'],
    where: { id: note.id }
});
const subjects = note.subjects

3。使用TypeORM惰性关系

如果要避免重新选择，则需要使用TypeORM Lazy relations，但这会迫使您将两个实体的类型都更改为Promise

// note entity
@ManyToMany(() => Subject, (subject: Subject) => subject.notes)
subjects: Promise<Subject[]>;

// subject entity
@ManyToMany(() => Note, note => note.subjects)
@JoinTable()
notes: Promise<Note[]>;

使用这种惰性关系，您将需要await以便在每次使用之前加载链接的注释，但是您将不需要提供与find方法的关系数组。

const note = await noteRepo.find({
    where: { id: someId }
});
const subjects = await note.subjects