得到像需要结构类型的错误，但在简单的结构类型的spark scala中得到了字符串

Question

这是我的架构

root
 |-- DataPartition: string (nullable = true)
 |-- TimeStamp: string (nullable = true)
 |-- PeriodId: long (nullable = true)
 |-- FinancialAsReportedLineItemName: struct (nullable = true)
 |    |-- _VALUE: string (nullable = true)
 |    |-- _languageId: long (nullable = true)
 |-- FinancialLineItemSource: long (nullable = true)
 |-- FinancialStatementLineItemSequence: long (nullable = true)
 |-- FinancialStatementLineItemValue: double (nullable = true)
 |-- FiscalYear: long (nullable = true)
 |-- IsAnnual: boolean (nullable = true)
 |-- IsAsReportedCurrencySetManually: boolean (nullable = true)
 |-- IsCombinedItem: boolean (nullable = true)
 |-- IsDerived: boolean (nullable = true)
 |-- IsExcludedFromStandardization: boolean (nullable = true)
 |-- IsFinal: boolean (nullable = true)
 |-- IsTotal: boolean (nullable = true)
 |-- ParentLineItemId: long (nullable = true)
 |-- PeriodPermId: struct (nullable = true)
 |    |-- _VALUE: long (nullable = true)
 |    |-- _objectTypeId: long (nullable = true)
 |-- ReportedCurrencyId: long (nullable = true)

从上面的模式我试图这样做

val temp = tempNew1
      .withColumn("FinancialAsReportedLineItemName", $"FinancialAsReportedLineItemName._VALUE")
      .withColumn("FinancialAsReportedLineItemName_languageId", $"FinancialAsReportedLineItemName._languageId")
      .withColumn("PeriodPermId", $"PeriodPermId._VALUE")
      .withColumn("PeriodPermId_objectTypeId", $"PeriodPermId._objectTypeId").drop($"AsReportedItem").drop($"AsReportedItem")

我不知道我在这里失踪了什么。我得到以下错误

线程“main”中的异常org.apache.spark.sql.AnalysisException：无法从FinancialAsReportedLineItemName＃2262中提取值：需要struct类型但是得到了字符串;

Answer 1

问题是，当FinancialAsReportedLineItemName._languageId列已被FinancialAsReportedLineItemName取代时，您正试图访问FinancialAsReportedLineItemName._VALUE

你应该改变以下两行

.withColumn("FinancialAsReportedLineItemName", $"FinancialAsReportedLineItemName._VALUE")
.withColumn("FinancialAsReportedLineItemName_languageId", $"FinancialAsReportedLineItemName._languageId")

至

.withColumn("FinancialAsReportedLineItemName_value", $"FinancialAsReportedLineItemName._VALUE")
.withColumn("FinancialAsReportedLineItemName_languageId", $"FinancialAsReportedLineItemName._languageId")

如果FinancialAsReportedLineItemName_value列名称应该是FinancialAsReportedLineItemName那么你应该交换withColumns作为

.withColumn("FinancialAsReportedLineItemName_languageId", $"FinancialAsReportedLineItemName._languageId")    
.withColumn("FinancialAsReportedLineItemName", $"FinancialAsReportedLineItemName._VALUE")