php - 原则插入看不到外键字段

Question

An exception occurred while executing 'INSERT INTO service (title, code, price, deleted) VALUES (?, ?, ?, ?)' with params ["asdf", "234", "234", 0]:

SQLSTATE[23000]: Integrity constraint violation: 1452 Cannot add or update a child row: a foreign key constraint fails 

实体$serviceCategory中还有一个字段，它引用service_category_id字段。

为什么它不在插入运算符中？

我认为它应该看起来像插入服务(标题，代码，价格，已删除，service_category_id)....

这是 yml orm:

Service:
    type: entity
    table: service
    id:
        id:
          type: integer
          generator: {strategy: AUTO}
    fields:
        title:
          type: string
          length: 255
        code:
          type: string
          length: 50
        price:
          type: decimal
          precision: 12
          scale: 2
        deleted:
          type: boolean
    ManyToOne:
        serviceCategoryId:
            targetEntity: ServiceCategory
            joinColumn:
                name: service_category_id
                referencedColumnName: id

这是实体(仅 Prop ):

class Service {
    /**
     * @var integer
     */

    protected $id;

    /**
     * @var string
     */
    protected $title;

    /**
     * @var string
     */
    protected $code;

    /**
     * @var integer
     */
    protected $price;

    /**
     * @var boolean
     */
    protected $deleted = false;

    protected $serviceCategoryId;

}

Answer 1

正如您所说，“服务”表中还有一个引用另一个表的字段“serviceCategory”，因此您没有传递该字段的值。您必须传递该值的正确值，并确保传递的值必须是引用表主键值的值。