Question

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|

我想解析为JSON，它应该看起来像这样：

{
    "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")      
  }
}

如何以给定格式进行序列化？预先感谢！

Answer 1

您可以使用struct和groupBy获得所需的结果。下面是相同的代码。我已在需要时注释了代码。

val df = Seq((10003014,"MH43AJ411",20000000),
  (10003014,"MH43AJ411",20000001),
  (10003015,"MH12GZ3392",20000002)
).toDF("ACCOUNTNO","VEHICLENUMBER","CUSTOMERID")

df.show
//output
//+---------+-------------+----------+
//|ACCOUNTNO|VEHICLENUMBER|CUSTOMERID|
//+---------+-------------+----------+
//| 10003014|    MH43AJ411|  20000000|
//| 10003014|    MH43AJ411|  20000001|
//| 10003015|   MH12GZ3392|  20000002|
//+---------+-------------+----------+

//create a struct column then group by ACCOUNTNO column and finally convert DF to JSON
df.withColumn("VEHICLE",struct("VEHICLENUMBER","CUSTOMERID")).
  select("VEHICLE","ACCOUNTNO"). //only select reqired columns
  groupBy("ACCOUNTNO"). 
  agg(collect_list("VEHICLE").as("VEHICLE")). //for the same group create a list of vehicles
  toJSON. //convert to json
  show(false)

//output
//+------------------------------------------------------------------------------------------------------------------------------------------+
//|value                                                                                                                                     |
//+------------------------------------------------------------------------------------------------------------------------------------------+
//|{"ACCOUNTNO":10003014,"VEHICLE":[{"VEHICLENUMBER":"MH43AJ411","CUSTOMERID":20000000},{"VEHICLENUMBER":"MH43AJ411","CUSTOMERID":20000001}]}|
//|{"ACCOUNTNO":10003015,"VEHICLE":[{"VEHICLENUMBER":"MH12GZ3392","CUSTOMERID":20000002}]}                                                   |
//+------------------------------------------------------------------------------------------------------------------------------------------+

您也可以使用与您提到的语句相同的语句将此dataframe写入文件。

使用Apache Spark将表序列化为嵌套JSON

1 个答案: