scala - 在数据帧上使用 sql 时无法解析给定的输入列

Question

平台:IntelliJ 版 2018.2.4(社区版)

SDK:1.8.0_144

操作系统:Windows 7

作为一名 future 的毕业生，我正在执行我的第一个大数据任务，我面临一个问题:

代码

//Loading my csv file here
val df = spark.read
  .format("csv")
  .option("header", "true")
  .option("delimiter",";")
  .load("/user/sfrtech/dilan/yesterdaycsv.csv")
  .toDF()


//Select required columns
val formatedDf = df.select("`TcRun.ID`", "`Td.Name`", "`TcRun.Startdate`", "`TcRun.EndDate`", "`O.Sim.MsisdnVoice`", "`T.Sim.MsisdnVoice`", "`ErrorCause`")

//Sql on DF in order to get useful data
formatedDf.createOrReplaceTempView("yesterday")
val sqlDF = spark.sql("" +
  " SELECT TcRun.Id, Td.Name, TcRun.Startdate, TcRun.EndDate, SUBSTR(O.Sim.MsisdnVoice,7,14) as MsisdnO, SUBSTR(T.Sim.MsisdnVoice,7,14) as MsisdnT", ErrorCause +
  " FROM yesterday" +
  " WHERE Td.Name like '%RING'" +
  " AND MsisdnO is not null" +
  " AND MsisdnT is not null" +
  " AND ErrorCause = 'NoError'")

出现错误

Exception in thread "main" org.apache.spark.sql.AnalysisException: cannot resolve 'Td.Name' given input columns: [TcRun.EndDate, TcRun.Startdate, O.Sim.MsisdnVoice, TcRun.ID, Td.Name, T.Sim.MsisdnVoice, ErrorCause]; line 1 pos 177;

我猜问题来自我的列名称包含“。”但我不知道如何解决这个问题，即使我使用反引号

解决方案

val newColumns = Seq("id", "name", "startDate", "endDate", "msisdnO", "msisdnT", "error")
val dfRenamed = df.toDF(newColumns: _*)

dfRenamed.printSchema
// root
// |-- id: string (nullable = false)
// |-- name: string (nullable = false)
// |-- startDate: string (nullable = false)
// |-- endDate: string(nullable = false)
// |-- msisdnO: string (nullable = false)
// |-- msisdnT: string (nullable = false)
// |-- error: string (nullable = false)

Answer 1

// Define column names of csv without "."
val schema = StructType(Array(
        StructField("id", StringType, true),
        StructField("name", StringType, true),
        // etc. etc. )

// Load csv file without headers and specify your schema
val df = spark.read
  .format("csv")
  .option("header", "false")
  .option("delimiter",";")
  .schema(schema)
  .load("/user/sfrtech/dilan/yesterdaycsv.csv")
  .toDF()

然后根据需要选择您的列

df
  .select ($"id", $"name", /*etc etc*/)