Scala sangria graphql查询返回1个元素列表
我使用sangria作为GraphQL服务器。模式的相关部分是:Scala sangria graphql查询返回1个元素列表,scala,graphql,sangria,Scala,Graphql,Sangria,我使用sangria作为GraphQL服务器。模式的相关部分是: 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"
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))
)
)
延迟代码为
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]())))
}
每件事都适用于单个对象,但是当我试图请求一个对象及其子对象时,我只得到一个子对象返回
查询:
{
citizen(id: "12345") {
name
accounts {
id
amountDue
}
}
}
答复:
{
"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,但它们似乎没有返回。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
我认为这些小小的改变会达到理想的效果:
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]())))
}
它的引入是为了确保对于递延
向量中的每个递延
值,在结果列表中只有一个未来
元素(它应该在列表中的相同位置)
它是针对性能优化的低级API,因此
resolve
方法的签名中没有太多类型安全性。我只是想改进这种情况下的错误报告 问题在于您正在使用flatMap
并将不相关列表的所有元素合并到一个结果列表中
我认为这些小小的改变会达到理想的效果:
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]())))
}
它的引入是为了确保对于递延
向量中的每个递延
值,在结果列表中只有一个未来
元素(它应该在列表中的相同位置)
它是针对性能优化的低级API,因此resolve
方法的签名中没有太多类型安全性。我只是想改进这种情况下的错误报告