请看下面的代码。 当我传递分区数量的值时,我收到以下代码的错误。
def loadDataFromPostgress(sqlContext: SQLContext, tableName: String,
columnName: String, dbURL: String, userName: String, pwd: String,
partitions: String): DataFrame = {
println("the no of partitions are : "+partitions)
var dataDF = sqlContext.read.format("jdbc").options(
scala.collection.Map("url" -> dbURL,
"dbtable" -> tableName,
"driver" -> "org.postgresql.Driver",
"user" -> userName,
"password" -> pwd,
"partitionColumn" -> columnName,
"numPartitions" -> "1000")).load()
return dataDF
}
错误:
java.lang.RuntimeException: Partitioning incompletely specified
App > at scala.sys.package$.error(package.scala:27)
App > at org.apache.spark.sql.execution.datasources.jdbc.JdbcRelationProvider.createRelation(JdbcRelationProvider.scala:38)
App > at org.apache.spark.sql.execution.datasources.DataSource.resolveRelation(DataSource.scala:315)
App > at org.apache.spark.sql.DataFrameReader.load(DataFrameReader.scala:149)
App > at org.apache.spark.sql.DataFrameReader.load(DataFrameReader.scala:122)
App > at Test$.loadDataFromGreenPlum(script.scala:28)
App > at Test$.loadDataFrame(script.scala:15)
App > at Test$.main(script.scala:59)
App > at Test.main(script.scala)
App > at sun.reflect.NativeMethodAccessorImpl.invoke0(Native
Method)
App > at
答案 0 :(得分:3)
您可以在下面查看代码的具体用途。
def loadDataFromPostgress(sqlContext: SQLContext, tableName: String,
columnName: String, dbURL: String, userName: String,
pwd: String, partitions: String): DataFrame = {
println("the no of partitions are : " + partitions)
var dataDF = sqlContext.read.format("jdbc").options(
scala.collection.Map("url" -> dbURL,
"dbtable" -> "(select mod(tmp.empid,10) as hash_code,tmp.* from employee as tmp) as t",
"driver" -> "org.postgresql.Driver",
"user" -> userName,
"password" -> pwd,
"partitionColumn" -> hash_code,
"lowerBound" -> 0,
"upperBound" -> 10
"numPartitions" -> "10"
) ).load()
return dataDF
}
上面的代码将创建10个任务,包含10个查询,如下所示。 在找到工作之前
offset =(upperBound-lowerBound)/ numPartitions
此处offset = (10-0)/10 = 1
select mod(tmp.empid,10) as hash_code,tmp.* from employee as tmp where hash_code between 0 between 1
select mod(tmp.empid,10) as hash_code,tmp.* from employee as tmp where hash_code between 1 between 2
.
.
select mod(tmp.empid,10) as hash_code,tmp.* from employee as tmp where hash_code between 9 between 10
这将创建10个分区和
empid以0结尾将进入一个分区,因为mod(empid,10)总是等于0
empid以1结尾将进入一个分区,因为mod(empid,10)总是等于1
像这样,所有员工行都会被分成10个分区。你必须根据你的要求更改partitionColumn,upperBound,lowerBound,numPartitions值。
希望我的回答可以帮到你。
答案 1 :(得分:0)
分区需要:
最后两个丢失了,这就是你得到错误的原因。