Scala 如何参数化将数据帧写入配置单元表

我在RBDM中有一个表列表（跨不同类别），我希望提取并保存在配置单元中，并且我希望以这样一种方式进行参数化，以便能够将类别名称附加到配置单元中的输出位置。 例如，我有一个类别“employee”，我希望能够以“hive\u db.employee\u some\u other\u random\u name”格式保存从RDBMS提取的表

我有如下代码

    val category = "employee"
    val tableList = List("schema.table_1", "schema.table_2", "schema.table_3")

    val tableMap = Map("schema.table_1" -> "table_1",
    "schema.table_2" -> "table_2",
    "schema.table_3" -> "table_3")

    val queryMap = Map("table_1" -> (select * from table_1) tble,
    "table_2" -> (select * from table_2) tble,
    "table_3" -> (select * from table_3) tble)

    val tableBucketMap = Map("table_1" -> "bucketBy(80,\"EMPLOY_ID\",\"EMPLOYE_ST\").sortBy(\"EMPLOY_ST\").format(\"parquet\")",
    "table_2" -> "bucketBy(80, \"EMPLOY_ID\").sortBy(\"EMPLOY_ID\").format(\"parquet\")",
    "table_3" -> "bucketBy(80, \"EMPLOY_ID\", \"SAL_ID\", \"DEPTS_ID\").sortBy(\"EMPLOY_ID\").format(\"parquet\")")

     for (table <- tableList){
       val tableName = tableMap(table)
       val print_start = "STARTING THE EXTRACTION PROCESSING FOR TABLE: %s"
       val print_statement = print_start.format(tableName)
       println(print_statement)

       val extract_query = queryMap(table)
       val query_statement_non = "Query to extract table %s is: "
       val query_statement = query_statement_non.format(tableName)
       println(query_statement + extract_query)


       val extracted_table = spark.read.format("jdbc")
         .option("url", jdbcURL)
         .option("driver", driver_type)
         .option("dbtable", extract_query)
         .option("user", username)
         .option("password", password)
         .option("fetchsize", "20000")
         .option("queryTimeout", "0")
         .load()

       extracted_table.show(5, false)
       //saving extracted table in hive
       val tableBucket = tableBucketMap(table)
       val output_loc = "hive_db.%s_table_extracted_for_%s"
       val hive_location = output_loc.format(category, tableName)
       println(hive_location)

       val saving_table = "%s.write.%s.saveAsTable(\"%s\")"
       saving_table.format(extracted_table, tableBucket, hive_location)
       println(saving_table.format(extracted_table, tableBucket, hive_location))
  
       val print_end = "COMPLETED EXTRACTION PROCESS FOR TABLE: %s"
       val print_end_statement = print_end.format(tableName)
       println(print_end_statement)

它只打印列名，而不是将提取的数据帧写入配置单元

[EMPLOY_ID: int, EMPLOYE_NM: String].write............saveAsTable("hive_db.employee_table_extracted_for_table_1")

如何将DF写入hive表？

您能试试这种方法吗， 像这样改变你的桶图（我已经为t1做了，请为t2和t3做同样的事情）

并用足够的参数替换

df.bucketBy（）

，如

（numBuckets:Int，colName:String，colNames:String*）

这种方法将解决上述问题

[EMPLOY_ID: int, EMPLOYE_NM: String].write............saveAsTable("hive_db.employee_table_extracted_for_table_1")

---

bucketBy（80，\'EMPLOY\u ID\，\'ACP\u ID\，\'DEPTS\u ID\”

这能满足这个

bucketBy（numBuckets:Int，colName:String，colNames:String*）

方法谢谢聪明的程序员。我会尝试你建议的方法，并给你反馈。聪明的程序员，我尝试了你建议的方法。我仍然面临同样的问题。Smart_coder，谢谢你，我最终能够使用你建议的方法解决这个问题，并对我的要求进行了一些修改。非常感谢。。

    val tableBucketMap = Map("table_1" -> "80,\"employe_st\"")

   
           val stringArr=tableBucket.split(",")
           val numBuckets=stringArr(0).toInt
           val colName=stringArr(1)
           
           extracted_table.write.mode("append").bucketBy(numBuckets,colName).format("parquet").saveAsTable(hive_location)
           

[EMPLOY_ID: int, EMPLOYE_NM: String].write............saveAsTable("hive_db.employee_table_extracted_for_table_1")