sangria graphql 查询返回 1 个元素列表

Posted 2023-03-08

【中文标题】sangria graphql 查询返回 1 个元素列表

【英文标题】：sangria graphql query returning 1 element list

【发布时间】：2016-12-18 19:49:37

【问题描述】：

我使用 sangria 作为 GraphQL 服务器。架构的相关部分是：

  val Account =
    ObjectType(
      "Account",
      "An account with a municipal unit",
      fields[Unit, Account](
        Field("id", StringType, Some("The account id"), resolve = _.value.id),
        Field("mu", OptionType(MunicipalUnit), Some("The municipal unit this account is with"), resolve = ctx => ctx.ctx.asInstanceOf[ObjectResolver].resolve[MunicipalUnit](ctx.value.mu)),
        Field("eu", OptionType(EconomicUnit), Some("The economic unit this account belongs to"), resolve = ctx => ctx.ctx.asInstanceOf[ObjectResolver].resolve[EconomicUnit](ctx.value.eu)),
        Field("location", OptionType(Location), Some("The physical location associated with this account"), resolve = ctx => ctx.ctx.asInstanceOf[ObjectResolver].resolve[Location](ctx.value.location)),
        Field("amountDue", BigDecimalType, Some("The amount currently due"), resolve = _.value.amountDue)
      ))

  val Citizen =
    ObjectType(
      "Citizen",
      "A Citizen",
      interfaces[Unit, Citizen](EconomicUnit),
      fields[Unit, Citizen](
        Field("id", StringType, Some("The ID of the citizen"), resolve = _.value.id),
        Field("name", StringType, Some("The name of the citizen"), resolve = _.value.id),
        Field("delegates", OptionType(ListType(OptionType(EconomicUnit))), Some("The delegates of the citizen"), resolve = ctx => DeferDelegates(ctx.value.delegates)),
        Field("locations", OptionType(ListType(OptionType(Location))), Some("The locations of the citizen"), resolve = ctx => DeferLocations(ctx.value.locations)),
        Field("accounts", OptionType(ListType(OptionType(Account))), Some("The accounts of the citizen"), resolve = ctx => DeferAccounts(ctx.value.accounts))
      )
    )

延期代码是

  def resolveByType[T](ids: List[Any])(implicit m: Manifest[T]) = ids map (id => resolver.resolve[T](id))

  override def resolve(deferred: Vector[Deferred[Any]], ctx: Any) = deferred flatMap 
    case DeferAccounts(ids) => resolveByType[Account](ids)
    case DeferLocations(ids) => resolveByType[Location](ids)
    case DeferDelegates(ids) => resolveByType[EconomicUnit](ids)
    case DeferMUs(ids) => resolveByType[MunicipalUnit](ids)

    case _ =>
      List(Future.fromTry(Try(List[Any]())))
  

一切都适用于单个对象，但是当我尝试请求一个对象及其子对象时，我只会得到一个孩子回来

查询：

    citizen(id: "12345") 
    name
    accounts 
      id
      amountDue
    
  

回复：

  "data": 
    "citizen": 
      "name": "12345",
      "accounts": [
        
          "id": "12345",
          "amountDue": 12.34
        
      ]
    
  

所以 - 这是正确的，我可以在后端看到列表的所有元素都在加载，但它们似乎没有被返回。

【参考方案1】：

问题是您正在使用flatMap 并将不相关列表的所有元素合并到一个结果列表中。

我认为这些小改动会达到理想的效果：

def resolveByType[T](ids: List[Any])(implicit m: Manifest[T]): Future[Seq[T]] = 
  Future.sequence(ids map (id => resolver.resolve[T](id)))

override def resolve(deferred: Vector[Deferred[Any]], ctx: Any) = deferred map 
  case DeferAccounts(ids) => resolveByType[Account](ids)
  case DeferLocations(ids) => resolveByType[Location](ids)
  case DeferDelegates(ids) => resolveByType[EconomicUnit](ids)
  case DeferMUs(ids) => resolveByType[MunicipalUnit](ids)

  case _ =>
    List(Future.fromTry(Try(List[Any]())))

重要的是要确保对于deferred 向量中的每个Deferred 值，结果列表中只有一个Future 元素（并且它应该在列表中的相同位置）。

它是针对性能进行了优化的低级 API，因此resolve 方法的签名中没有太多类型安全性。在这种情况下，我只是 created an issue 来改进错误报告。