根据参数化SQL查询在Spark Dataframe中添加列，这取决于数据帧某些字段的值。

Question

我有几个Spark Dataframes（我们可以称它们为Table a，Table b等）。 我想根据对其中一个表的查询结果向表a添加一列，但此表每次都会根据表a的某个字段的值进行更改。所以这个查询应该是参数化的。 下面我列出一个例子来解决问题：

每个表都有OID列和TableName列以及当前表的名称加上其他列。

This is the fixed query to be performed on Tab A to add new column: 

    Select $ColumnName from $TableName where OID=$oids 

    Tab A 
    | oids|TableName |ColumnName | other fields|New Column: ValueOidDb 
    ================================================================ 
    | 2 | Book  | Title  |  x  |result query:harry potter 
    | 8 | Book  | Isbn  |  y  |result query: 556 
    | 1 | Author | Name  |  z  |result query:Tolkien 
    | 4 | Category |Description|  b  |result query: Commedy 


    Tab Book 
    | OID |TableName |Title  |Isbn |other fields| 
    ================================================================ 
    | 2 | Book  |harry potter| 123 | x   | 
    | 8 | Book  | hobbit  | 556 | y   | 
    | 21 | Book  | etc  | 8942 | z   | 
    | 5 | Book  | etc2  | 984 | b   | 

    Tab Author 
    | OID |TableName  |Name  |nationality |other fields| 
    ================================================================ 
    | 5 | Author  |J.Rowling | eng  | x   | 
    | 2 | Author  |Geor. Martin| us   | y   | 
    | 1 | Author  | Tolkien | eng  | z   | 
    | 13 | Author  | Dan Brown | us   | b   | 


    | OID | TableName |Description | 
    ===================================== 
    | 12 | Category | Fantasy | 
    | 4 | Category | Commedy | 
    | 9 | Category | Thriller | 
    | 7 | Category | Action  | 

我试着用这个UDF

def setValueOid = (oid: Int,TableName: String, TableColumn: String) => { 

    try{ 
     sqlContext.sql(s"Select $currTableColumn from $currTableName where OID = $curroid ").first().toString() 
     } 
    catch{ 
     case x: java.lang.NullPointerException => "error" 
     } 

     } 
    sqlContext.udf.register("setValueOid", setValueOid) 

    val FinalRtxf = sqlContext.sql("SELECT all the column of TAB A ," 
       + " setValueOid(oid, Table,AttributeDatabaseColumn) as  ValueOidDb" 
       + " FROM TAB A") 

我把代码中的一个尝试捕捉，否则它给了我一个NullPointerException异常，但它不工作，因为它总是会返回一个“问题” 。 如果我尝试这个功能没有SQL查询的只是路过一些手动参数它完美的作品：

  val try=setValueOid(8,"BOOK","ISBN") 
      try: String = [0977326403 ]     FINISHED 
      Took 4 sec. Last updated by anonymous at November 20 2016, 3:29:28 AM. 

我读到这里，是不是可以做一个查询UDF内 Trying to execute a spark sql query from a UDF

因此，如何能我解决了我的问题？我不知道如何进行参数化连接。我尝试这样做：

 %sql 
     Select all attributes TAB A,  
     FROM TAB A as a 
     join (Select $AttributeDatabaseColumn ,TableName from $Table where OID=$oid) as b 
     on a.Table=b.TableName 

，但它给了我这个例外：

org.apache.spark.sql.AnalysisException: cannot recognize input near '$' 'AttributeDatabaseColumn' ',' in select clause; line 3 pos 1  at org.apache.spark.sql.hive.HiveQl$.createPlan(HiveQl.scala:318) 
at org.apache.spark.sql.hive.ExtendedHiveQlParser$$anonfun$hiveQl$1.apply(ExtendedHiveQlParser.scala:41) 
at org.apache.spark.sql.hive.ExtendedHiveQlParser$$anonfun$hiveQl$1.apply(ExtendedHiveQlParser.scala:40) 

Answer 1

一个选项：

• 变换每个Book，Author，Category到窗体：

  root 
  |-- oid: integer (nullable = false) 
  |-- tableName: string (nullable = true) 
  |-- properties: map (nullable = true) 
  | |-- key: string 
  | |-- value: string (valueContainsNull = true) 
  

  例如在Book第一记录：

  val book = Seq((2L, "Book", 
      Map("title" -> "harry potter", "Isbn" -> "123", "other field" -> "x") 
  )).toDF("oid", "title", "properties") 
  
  +---+---------+---------------------------------------------------------+ 
  |oid|tableName|properties            | 
  +---+---------+---------------------------------------------------------+ 
  |2 |Book  |Map(title -> harry potter, Isbn -> 123, other field -> x)| 
  +---+---------+---------------------------------------------------------+ 
  

• 工会Book，Author，Category作为属性。

  val properties = book.union(author).union(category) 
  

• 加入与基表：

  val comb = properties.join(table, Seq($"oid", $"tableName")) 
  

• 使用case when ...基于tableName从properties字段添加新列。