Prisma 可以從關系表中創建一對多關系嗎？

Question

我正在使用 Prisma 和 MySQL 創建一個鍛煉應用程序。

我在用戶和指定程序之間存在多對多關系，如 model 'ProgramEnrollment' 所示。 我遇到的問題是我希望用戶記錄他們的個人鍛煉集。

我的想法是創建一個新的 model“LoggedWorkoutSet”，然后通過一對多關系將其連接到 model“ProgramEnrollment”，這意味着一個“ProgramEnrollment”可以有多個“LoggedWorkoutSet”。 這似乎對我不起作用，問題在於嘗試在這里定義關系時：

program   ProgramEnrollment @relation(fields: [programId], references: [id])
  programId Int // relation scalar field  (used in the `@relation` attribute above)

我得到的錯誤是“驗證錯誤：參數references必須僅引用相關 model ProgramEnrollment中的現有字段。 相關model中不存在以下字段：id'

我不知道該怎么做，任何幫助將不勝感激。

model ProgramEnrollment {
  program                 Program     @relation(fields: [programId], references: [id])
  programId               Int // relation scalar field (used in the `@relation` attribute above)
  user                    User @relation(fields: [userId], references: [id])
  userId                  Int // relation scalar field (used in the `@relation` attribute above)
  assignedAt              DateTime @default(now())

  loggedWorkoutSet        LoggedWorkoutSet[]

  @@id([programId, userId])
}

model LoggedWorkoutSet {
  id              Int  @id @default(autoincrement())
  reps            Int
  weight          Int
  completed       Boolean     @default(false)

  author   ProgramEnrollment @relation(fields: [programId], references: [id])
  programId Int // relation scalar field  (used in the `@relation` attribute above)
}

Answer 1

CREATE TABLE Users (
    user_id ...,
    ...
    PRIMARY KEY (user_id)
) ENGINE=InnoDB;

CREATE TABLE Programs (
    pgm_id ...,
    part_of_body ... -- legs/shoulders/...
    ...
    PRIMARY KEY (pgm_id)
) ENGINE=InnoDB;

-- Many-to-many, with stats:
CREATE TABLE UserPgm (
    user_id ...,
    pgm_id ...,
    latest_date DATE ...,
    reps SMALLINT UNSIGNED NOT NULL,
    weight SMALLINT UNSIGNED NOT NULL,  -- in the normal units for this device
    PRIMARY KEY(user_id, pgm_id),
    INDEX(pgm_id, user_id)
) ENGINE=InnoDB;

使用所有 3 個表的查詢：

-- The latest workout for a user:
SELECT u.name,
       up.latest_date,
       up.reps,
       p.device_name
    FROM Users    AS u
    JOIN UserPgm  AS up  USING(user_id)
    JOIN Programs AS p   USING(pgm_id)
    WHERE user_id = 123