DynamoDB / Scanamo:提供的关键元素与架构不匹配

问题描述 投票:0回答:1

我一直在尝试通过 Scanamo 库使用 DynamoDB。我的 scala 代码如下所示:

package my.package

import com.amazonaws.ClientConfiguration
import com.amazonaws.regions.{Region, Regions}
import com.amazonaws.services.dynamodbv2.AmazonDynamoDBClient
import com.gu.scanamo._
import com.gu.scanamo.syntax._
import com.amazonaws.auth.{AWSStaticCredentialsProvider, BasicAWSCredentials}
import com.amazonaws.services.dynamodbv2.datamodeling._

object MusicService {

  def main(args: Array[String]): Unit = {
    val musicService = new MusicService
    musicService.getAlbums()
  }

}

class MusicService {

  def getAlbums() {

    val awsCreds = new BasicAWSCredentials("my","creds")
    val client = AmazonDynamoDBClient
      .builder()
      .withRegion(Regions.EU_WEST_2)
      .withCredentials(new AWSStaticCredentialsProvider(awsCreds))
      .build();


    case class Music(@DynamoDBIndexRangeKey(attributeName = "Artist") 
        artist: String, @DynamoDBIndexHashKey(attributeName = "SongTitle") songTitle: String);

    val table = Table[Music]("Music")

    val putOp = table.putAll(Set(
      Music("The Killers", "Sam's Town"),
      Music("The Killers", "Spaceman")
    ))
    Scanamo.exec(client)(putOp)

}

我在执行 putOp 时遇到此错误:

Exception in thread "main" com.amazonaws.services.dynamodbv2.model.AmazonDynamoDBException: The provided key element does not match the schema (Service: AmazonDynamoDBv2; Status Code: 400; Error Code: ValidationException; Request ID: 0KAFH90JO39COO143LC5H6RPPNVV4KQNSO5AEMVJF66Q9ASUAAJG)
at com.amazonaws.http.AmazonHttpClient$RequestExecutor.handleErrorResponse(AmazonHttpClient.java:1638)
at com.amazonaws.http.AmazonHttpClient$RequestExecutor.executeOneRequest(AmazonHttpClient.java:1303)
at com.amazonaws.http.AmazonHttpClient$RequestExecutor.executeHelper(AmazonHttpClient.java:1055)
at com.amazonaws.http.AmazonHttpClient$RequestExecutor.doExecute(AmazonHttpClient.java:743)
at com.amazonaws.http.AmazonHttpClient$RequestExecutor.executeWithTimer(AmazonHttpClient.java:717)
at com.amazonaws.http.AmazonHttpClient$RequestExecutor.execute(AmazonHttpClient.java:699)
at com.amazonaws.http.AmazonHttpClient$RequestExecutor.access$500(AmazonHttpClient.java:667)
at com.amazonaws.http.AmazonHttpClient$RequestExecutionBuilderImpl.execute(AmazonHttpClient.java:649)
at com.amazonaws.http.AmazonHttpClient.execute(AmazonHttpClient.java:513)
at com.amazonaws.services.dynamodbv2.AmazonDynamoDBClient.doInvoke(AmazonDynamoDBClient.java:2186)
at com.amazonaws.services.dynamodbv2.AmazonDynamoDBClient.invoke(AmazonDynamoDBClient.java:2162)
at com.amazonaws.services.dynamodbv2.AmazonDynamoDBClient.executeBatchWriteItem(AmazonDynamoDBClient.java:575)
at com.amazonaws.services.dynamodbv2.AmazonDynamoDBClient.batchWriteItem(AmazonDynamoDBClient.java:551)
at com.gu.scanamo.ops.ScanamoInterpreters$$anon$1.apply(ScanamoInterpreters.scala:51)
at com.gu.scanamo.ops.ScanamoInterpreters$$anon$1.apply(ScanamoInterpreters.scala:30)
at cats.free.Free.$anonfun$foldMap$1(Free.scala:126)
at cats.package$$anon$1.tailRecM(package.scala:41)
at cats.free.Free.foldMap(Free.scala:124)
at cats.free.Free.$anonfun$foldMap$1(Free.scala:127)
at cats.package$$anon$1.tailRecM(package.scala:41)
at cats.free.Free.foldMap(Free.scala:124)
at com.gu.scanamo.Scanamo$.exec(Scanamo.scala:17)
at my.package.MusicService.getAlbums(MusicService.scala:39)
at my.package.MusicService$.main(MusicService.scala:14)
at my.package.MusicService.main(MusicService.scala)

我在 DynamoDB 上的表结构非常简单,如下所示:

Table name: Music
Partition key: Artist
Sort key: SongTitle

仅此而已。

请给我一些指导,为什么会失败以及我能做些什么来解决它?

amazon-web-services scala amazon-dynamodb scanamo
1个回答
0
投票

首先,您需要交换

@DynamoDBIndexHashKey
@DynamoDBIndexRangeKey
(因为
@DynamoDBIndexHashKey
应该用于散列键 - 艺术家,而
@DynamoDBIndexRangeKey
则用于排序键 - 歌曲标题)。

您还提到 Artist 是分区键,SongTitle 是排序键。那么为什么要使用

@DynamoDBIndexHashKey
@DynamoDBIndexRangeKey
?我想您需要
@DynamoDBHashKey
@DynamoDBRangeKey
(以防艺术家和歌曲标题不是索引)。

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