spark-cassandra-connector -从Dataframe创建表格- StructType？

Question

我正试着从Spark数据帧中写给Cassandra。当我有一个简单的数据帧模式时，如示例中所示，它可以工作：

root
 |-- id: string (nullable = true)
 |-- url: string (nullable = true)

但是，当我尝试编写包含StructTypes的数据帧时，其模式如下：

root
 |-- crawl: struct (nullable = true)
 |    |-- id: string (nullable = true)

然后我得到以下异常：

Exception in thread "main" java.lang.IllegalArgumentException: Unsupported type: StructType(StructField(id,StringType,true))
    at com.datastax.spark.connector.types.ColumnType$.unsupportedType$1(ColumnType.scala:132)
    at com.datastax.spark.connector.types.ColumnType$.fromSparkSqlType(ColumnType.scala:155)
    at com.datastax.spark.connector.mapper.DataFrameColumnMapper$$anonfun$1.apply(DataFrameColumnMapper.scala:18)
    at com.datastax.spark.connector.mapper.DataFrameColumnMapper$$anonfun$1.apply(DataFrameColumnMapper.scala:16)
    at scala.collection.TraversableLike$$anonfun$map$1.apply(TraversableLike.scala:244)
    at scala.collection.TraversableLike$$anonfun$map$1.apply(TraversableLike.scala:244)
    at scala.collection.immutable.List.foreach(List.scala:318)
    at scala.collection.TraversableLike$class.map(TraversableLike.scala:244)
    at scala.collection.AbstractTraversable.map(Traversable.scala:105)
    at com.datastax.spark.connector.mapper.DataFrameColumnMapper.newTable(DataFrameColumnMapper.scala:16)
    at com.datastax.spark.connector.cql.TableDef$.fromDataFrame(Schema.scala:215)
    at com.datastax.spark.connector.DataFrameFunctions.createCassandraTable(DataFrameFunctions.scala:26)

我的代码如下所示：

val df = sqlContext.read.parquet(input)
df.createCassandraTable(keyspace, table)

df.write
  .format("org.apache.spark.sql.cassandra")
  .options(Map("table" -> table, "keyspace" -> keyspace))
  .save()

帮助?

Answer 1

看起来连接器目前还不支持从DataFrame结构中动态创建UDT类型。您应该向Spark Cassandra Connector Jira添加一个工单，并将其作为功能请求。在此之前，您始终可以手动创建一个新类型来匹配您的结构类型。