【发布时间】:2018-12-25 03:17:35
【问题描述】:
我有一组类似以下示例的记录
|ACCOUNTNO|VEHICLENUMBER|CUSTOMERID|
+---------+-------------+----------+
| 10003014| MH43AJ411| 20000000|
| 10003014| MH43AJ411| 20000001|
| 10003015| MH12GZ3392| 20000002|
我想解析成 JSON,它应该是这样的:
{
"ACCOUNTNO":10003014,
"VEHICLE": [
{ "VEHICLENUMBER":"MH43AJ411", "CUSTOMERID":20000000},
{ "VEHICLENUMBER":"MH43AJ411", "CUSTOMERID":20000001}
],
"ACCOUNTNO":10003015,
"VEHICLE": [
{ "VEHICLENUMBER":"MH12GZ3392", "CUSTOMERID":20000002}
]
}
我已经编写了程序,但未能实现输出。
package com.report.pack1.spark
import org.apache.spark.SparkConf
import org.apache.spark.SparkContext
import org.apache.spark.sql.SQLContext
import org.apache.spark.sql._
object sqltojson {
def main(args:Array[String]) {
System.setProperty("hadoop.home.dir", "C:/winutil/")
val conf = new SparkConf().setAppName("SQLtoJSON").setMaster("local[*]")
val sc = new SparkContext(conf)
val sqlContext = new SQLContext(sc)
import sqlContext.implicits._
val jdbcSqlConnStr = "jdbc:sqlserver://192.168.70.88;databaseName=ISSUER;user=bhaskar;password=welcome123;"
val jdbcDbTable = "[HISTORY].[TP_CUSTOMER_PREPAIDACCOUNTS]"
val jdbcDF = sqlContext.read.format("jdbc").options(Map("url" -> jdbcSqlConnStr,"dbtable" -> jdbcDbTable)).load()
jdbcDF.registerTempTable("tp_customer_account")
val res01 = sqlContext.sql("SELECT ACCOUNTNO, VEHICLENUMBER, CUSTOMERID FROM tp_customer_account GROUP BY ACCOUNTNO, VEHICLENUMBER, CUSTOMERID ORDER BY ACCOUNTNO ")
res01.coalesce(1).write.json("D:/res01.json")
}
}
如何以给定的格式序列化?提前致谢!
【问题讨论】:
标签: json scala apache-spark