Question

我正在尝试保存具有许多类别的产品。同时，一个类别可以有许多产品。因此，我认为最好的解决方案是对数据进行建模的ManyToMany关系。

当我尝试创建新产品时，发生错误，提示无法插入具有空值的新类别。疯狂的是，我不需要创建新类别，而是想要采用现有类别并将其分配给新产品。

这是控制台返回的SQL消息：

SQLITE_CONSTRAINT: NOT NULL constraint failed: category.name',
errno: 19,
code: 'SQLITE_CONSTRAINT',
name: 'QueryFailedError',
query: 'INSERT INTO "category"("id", "parent_id", "name", "description", "createdAt", "updatedAt") VALUES (NULL, NULL, NULL, NULL, datetime(\'now\'), datetime(\'now\'))

按照 typeorm documentation example ，我推断我的代码必须类似于下一个：

Product.ts

@Entity()
@Unique(["code"])
export class Product {
  @PrimaryGeneratedColumn()
  id: number;

  @Column({
    length: 20
  })
  code: string;

  @Column({
    length: 100
  })
  name: string;

  @Column()
  description: string;

  @Column()
  stock: number;

  @Column({
    type: "real"
  })
  price: number;

  @Column()
  public_price: boolean;

  @Column()
  public: boolean;

  @ManyToMany(type => Category, category => category.products, {
    cascade: true
  })
  @JoinTable()
  categories: Category[];
}

Category.ts

@Entity()
export class Category {
  @PrimaryGeneratedColumn()
  id: number;

  @Column({ nullable: true })
  parent_id: number;

  @Column()
  @IsNotEmpty()
  name: string;

  @Column()
  description?: string;

  @Column()
  @CreateDateColumn()
  createdAt: Date;

  @Column()
  @UpdateDateColumn()
  updatedAt: Date;

  @ManyToMany(type => Product, product => product.categories)
  products: Product[];
}

ProductController.ts（创建方法）

static create = async (req: Request, res: Response) => {
  const { categories, ...data } = req.body

  const categoryRepository = getRepository(Category);
  const productRepository = getRepository(Product);

  const cats = []
  for (const id of categories) {
    const cat = await categoryRepository.find(id)
    cats.push(cat)
  }
    
  const product = new Product();
  for (const index in data) {
    product[index] = data[index]
  }
  product.categories = cats

  try {
    await productRepository.save(product);
    res.send('Product successfuly created');
  } catch (err) {
    console.log(err)
    res.status(409).send('Error on product creation');
  }
}

这是我用来通过Postman测试我的应用程序的JSON对象：

{
    "code": "1234",
    "name": "TESTING PRODUCT CREATION",
    "description": "Product sended through postman",
    "stock": "99",
    "price": "255.88",
    "public_price": true,
    "public": true,
    "categories": [1]
}

Answer 1

替换

await categoryRepository.find(id)

使用

await categoryRepository.findOne(id)

find将返回Array中的项目。

如何正确地创建一个与ManyToMany关系的注册表？

1 个答案: