Scala 使用另一个 Json 的键排序
Scala Sorting with Another Json's Key
我是 Scala 的新手,请帮助我。
我有 2 Json 个文件。我想用第二个 json.
中的键对第一个 json 进行排序
例如:
首先Json
{
"id": 1,
"response" : [{
"user_id" : 1,
"products" : [
{
"product_id": 10,
"price": 200
},
{
"product_id": 13,
"price": 210
},
{
"product_id": 9,
"price": 320
}
]
},{
"user_id" : 2,
"products" : [
{
"product_id": 15,
"price": 200
},
{
"product_id": 13,
"price": 210
},
{
"product_id": 8,
"price": 320
}
]
}
]
}
还有我的第二个Json
{
"sort": [
{
"user_id": 1,
"products": [
{
"id": 8,
"rank": 5
},
{
"id": 9,
"rank": 1
},
{
"id": 10,
"rank": 3
},
{
"id": 13,
"rank": 2
},{
"id": 15,
"rank": 6
},{
"id": 17,
"rank": 4
},{
"id": 20,
"rank": 7
},{
"id": 21,
"rank": 8
},{
"id": 23,
"rank": 9
}
]
},{
"user_id": 2,
"products": [
{
"id": 8,
"rank": 5
},
{
"id": 9,
"rank": 1
},
{
"id": 10,
"rank": 3
},
{
"id": 13,
"rank": 2
},{
"id": 15,
"rank": 6
},{
"id": 17,
"rank": 4
},{
"id": 20,
"rank": 7
},{
"id": 21,
"rank": 8
},{
"id": 23,
"rank": 9
}
]
}
]
}
我想根据第二 Json 的排名对第一 json 进行排序。
输出应该像每个用户应该根据在第二个 JSON.
上为每个用户指定的排名对其产品进行排序
这是目前为止我尝试过的方法
def sortedRes() = Action.async {
val url = //1st json url
val sortUrl = //2nd json url
ws.url(url).get().map { response =>
val value: JsValue = Json.parse(response.body)
val result: Either[Exception, SampleResponses] = value.validate[SampleResponses] match {
case JsSuccess(searchResponse, _) =>
Right(searchResponse)
case JsError(errors) =>
Left(new Exception("Couldn't parse Search API response"))
}
val values: List[SampleResponse] = result.right.get.responses
ws.url(sortUrl).get().map { response =>
val sorted: JsValue = Json.parse(response.body)
val sortRespResult: Either[Exception, Sort] = sorted.validate[Sort] match {
case JsSuccess(sortResponse, _) =>
Right(sortResponse)
case JsError(errors) =>
Left(new Exception("Couldn't parse because of these errors : " + errors))
}
val prodRankDetails: List[SampleRank] = sortRespResult.right.get.sort
println("prod = " + prodRankDetails.head.products.sortWith(_.rank > _.rank))
}
Ok(Json.toJson(result.right.get))
}
}
在最后一个打印语句中,我对第二个 json 的第一个用户产品进行了排序。我想要得到的是我的第一个 json 根据第二个用户排序。
这是我的模型class
package models
import play.api.libs.functional.syntax._
import play.api.libs.json.Reads._
import play.api.libs.json._ // Combinator syntax
object sample {
case class SampleProduct(productId:Int, price: Int)
case class SampleResponse(userId: Int, products: List[SampleProduct])
case class SampleResponses(id: Int, responses: List[SampleResponse])
case class SampleRankedProduct(id: Int, rank: Int)
case class SampleRank(userId: Int, products: List[SampleRankedProduct])
case class Sort(sort: List[SampleRank])
implicit val productReads: Reads[SampleProduct] = (
(JsPath \ "product_id").read[Int] and
(JsPath \ "price").read[Int]
)(SampleProduct.apply _)
implicit val productWrites: Writes[SampleProduct] = (
(JsPath \ "product_id").write[Int] and
(JsPath \ "price ").write[Int]
)(unlift(SampleProduct.unapply))
implicit val responseReads: Reads[SampleResponse] = (
(JsPath \ "user_id").read[Int] and
(JsPath \ "products").read[List[SampleProduct]]
)(SampleResponse.apply _)
implicit val responseWrites: Writes[SampleResponse] = (
(JsPath \ "user_id").write[Int] and
(JsPath \ "products").write[List[SampleProduct]]
)(unlift(SampleResponse.unapply))
implicit val responsesReads: Reads[SampleResponses] = (
(JsPath \ "id").read[Int] and
(JsPath \ "response").read[List[SampleResponse]]
)(SampleResponses.apply _)
implicit val responsesWrites: Writes[SampleResponses] = (
(JsPath \ "id").write[Int] and
(JsPath \ "response").write[List[SampleResponse]]
)(unlift(SampleResponses.unapply))
implicit val rankedProductReads: Reads[SampleRankedProduct] = (
(JsPath \ "id").read[Int] and
(JsPath \ "rank").read[Int]
)(SampleRankedProduct.apply _)
implicit val rankedProductWrites: Writes[SampleRankedProduct] = (
(JsPath \ "id").write[Int] and
(JsPath \ "rank ").write[Int]
)(unlift(SampleRankedProduct.unapply))
implicit val rankReads: Reads[SampleRank] = (
(JsPath \ "user_id").read[Int] and
(JsPath \ "products").read[List[SampleRankedProduct]]
)(SampleRank.apply _)
implicit val rankWrites: Writes[SampleRank] = (
(JsPath \ "user_id").write[Int] and
(JsPath \ "products").write[List[SampleRankedProduct]]
)(unlift(SampleRank.unapply))
implicit val sortReads: Reads[Sort] = (JsPath \ "sort").read[List[SampleRank]].map(x ⇒ Sort(x))
implicit val sortWrites: Writes[Sort] = (__ \ "sort").write[List[SampleRank]].contramap { (person: Sort) => person.sort }
}
使用第二个的排名对第一个 JSON 进行排序的方法是这样的:
....
val prodRankDetails: List[SampleRank] = sortRespResult.right.get.sort
val sortedResults = values.copy(
responses = values.responses.map { resultForUser =>
val rankingForUser = prodRankDetails
.find(_.userId == resultForUser.userId)
.getOrElse(throw new RuntimeException(s"Rank not found for user ${resultForUser.userId}"))
.products
.map(product => (product.id, product.rank))
.toMap
resultForUser.copy(
products = resultForUser.products.sortWith((a, b) =>
rankingForUser.getOrElse(a.productId, Int.MaxValue) < rankingForUser.getOrElse(b.productId, Int.MaxValue))
)
}
)
Ok(Json.toJson(sortedResults))
应该在你得到 prodRankDetails
之后立即执行。而且当您管理 Future
时,响应将在 .map
内可用。
因此,您还必须将该地图内部的 Ok(...)
移动到外部 Future
.
的 flatMap
希望对您有所帮助。
我是 Scala 的新手,请帮助我。 我有 2 Json 个文件。我想用第二个 json.
中的键对第一个 json 进行排序例如: 首先Json
{
"id": 1,
"response" : [{
"user_id" : 1,
"products" : [
{
"product_id": 10,
"price": 200
},
{
"product_id": 13,
"price": 210
},
{
"product_id": 9,
"price": 320
}
]
},{
"user_id" : 2,
"products" : [
{
"product_id": 15,
"price": 200
},
{
"product_id": 13,
"price": 210
},
{
"product_id": 8,
"price": 320
}
]
}
]
}
还有我的第二个Json
{
"sort": [
{
"user_id": 1,
"products": [
{
"id": 8,
"rank": 5
},
{
"id": 9,
"rank": 1
},
{
"id": 10,
"rank": 3
},
{
"id": 13,
"rank": 2
},{
"id": 15,
"rank": 6
},{
"id": 17,
"rank": 4
},{
"id": 20,
"rank": 7
},{
"id": 21,
"rank": 8
},{
"id": 23,
"rank": 9
}
]
},{
"user_id": 2,
"products": [
{
"id": 8,
"rank": 5
},
{
"id": 9,
"rank": 1
},
{
"id": 10,
"rank": 3
},
{
"id": 13,
"rank": 2
},{
"id": 15,
"rank": 6
},{
"id": 17,
"rank": 4
},{
"id": 20,
"rank": 7
},{
"id": 21,
"rank": 8
},{
"id": 23,
"rank": 9
}
]
}
]
}
我想根据第二 Json 的排名对第一 json 进行排序。
输出应该像每个用户应该根据在第二个 JSON.
上为每个用户指定的排名对其产品进行排序这是目前为止我尝试过的方法
def sortedRes() = Action.async {
val url = //1st json url
val sortUrl = //2nd json url
ws.url(url).get().map { response =>
val value: JsValue = Json.parse(response.body)
val result: Either[Exception, SampleResponses] = value.validate[SampleResponses] match {
case JsSuccess(searchResponse, _) =>
Right(searchResponse)
case JsError(errors) =>
Left(new Exception("Couldn't parse Search API response"))
}
val values: List[SampleResponse] = result.right.get.responses
ws.url(sortUrl).get().map { response =>
val sorted: JsValue = Json.parse(response.body)
val sortRespResult: Either[Exception, Sort] = sorted.validate[Sort] match {
case JsSuccess(sortResponse, _) =>
Right(sortResponse)
case JsError(errors) =>
Left(new Exception("Couldn't parse because of these errors : " + errors))
}
val prodRankDetails: List[SampleRank] = sortRespResult.right.get.sort
println("prod = " + prodRankDetails.head.products.sortWith(_.rank > _.rank))
}
Ok(Json.toJson(result.right.get))
}
}
在最后一个打印语句中,我对第二个 json 的第一个用户产品进行了排序。我想要得到的是我的第一个 json 根据第二个用户排序。
这是我的模型class
package models
import play.api.libs.functional.syntax._
import play.api.libs.json.Reads._
import play.api.libs.json._ // Combinator syntax
object sample {
case class SampleProduct(productId:Int, price: Int)
case class SampleResponse(userId: Int, products: List[SampleProduct])
case class SampleResponses(id: Int, responses: List[SampleResponse])
case class SampleRankedProduct(id: Int, rank: Int)
case class SampleRank(userId: Int, products: List[SampleRankedProduct])
case class Sort(sort: List[SampleRank])
implicit val productReads: Reads[SampleProduct] = (
(JsPath \ "product_id").read[Int] and
(JsPath \ "price").read[Int]
)(SampleProduct.apply _)
implicit val productWrites: Writes[SampleProduct] = (
(JsPath \ "product_id").write[Int] and
(JsPath \ "price ").write[Int]
)(unlift(SampleProduct.unapply))
implicit val responseReads: Reads[SampleResponse] = (
(JsPath \ "user_id").read[Int] and
(JsPath \ "products").read[List[SampleProduct]]
)(SampleResponse.apply _)
implicit val responseWrites: Writes[SampleResponse] = (
(JsPath \ "user_id").write[Int] and
(JsPath \ "products").write[List[SampleProduct]]
)(unlift(SampleResponse.unapply))
implicit val responsesReads: Reads[SampleResponses] = (
(JsPath \ "id").read[Int] and
(JsPath \ "response").read[List[SampleResponse]]
)(SampleResponses.apply _)
implicit val responsesWrites: Writes[SampleResponses] = (
(JsPath \ "id").write[Int] and
(JsPath \ "response").write[List[SampleResponse]]
)(unlift(SampleResponses.unapply))
implicit val rankedProductReads: Reads[SampleRankedProduct] = (
(JsPath \ "id").read[Int] and
(JsPath \ "rank").read[Int]
)(SampleRankedProduct.apply _)
implicit val rankedProductWrites: Writes[SampleRankedProduct] = (
(JsPath \ "id").write[Int] and
(JsPath \ "rank ").write[Int]
)(unlift(SampleRankedProduct.unapply))
implicit val rankReads: Reads[SampleRank] = (
(JsPath \ "user_id").read[Int] and
(JsPath \ "products").read[List[SampleRankedProduct]]
)(SampleRank.apply _)
implicit val rankWrites: Writes[SampleRank] = (
(JsPath \ "user_id").write[Int] and
(JsPath \ "products").write[List[SampleRankedProduct]]
)(unlift(SampleRank.unapply))
implicit val sortReads: Reads[Sort] = (JsPath \ "sort").read[List[SampleRank]].map(x ⇒ Sort(x))
implicit val sortWrites: Writes[Sort] = (__ \ "sort").write[List[SampleRank]].contramap { (person: Sort) => person.sort }
}
使用第二个的排名对第一个 JSON 进行排序的方法是这样的:
....
val prodRankDetails: List[SampleRank] = sortRespResult.right.get.sort
val sortedResults = values.copy(
responses = values.responses.map { resultForUser =>
val rankingForUser = prodRankDetails
.find(_.userId == resultForUser.userId)
.getOrElse(throw new RuntimeException(s"Rank not found for user ${resultForUser.userId}"))
.products
.map(product => (product.id, product.rank))
.toMap
resultForUser.copy(
products = resultForUser.products.sortWith((a, b) =>
rankingForUser.getOrElse(a.productId, Int.MaxValue) < rankingForUser.getOrElse(b.productId, Int.MaxValue))
)
}
)
Ok(Json.toJson(sortedResults))
应该在你得到 prodRankDetails
之后立即执行。而且当您管理 Future
时,响应将在 .map
内可用。
因此,您还必须将该地图内部的 Ok(...)
移动到外部 Future
.
flatMap
希望对您有所帮助。