scala—如何参数化将Dataframe写入配置单元表

xnifntxz  于 2021-05-27  发布在  Spark
关注(0)|答案(1)|浏览(447)

我在rbdms中有一个表列表(跨不同的类别),我想提取并保存在hive中,并且我想以这样一种方式进行参数化,以便能够将类别名称附加到hive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”的格式保存rdbms中提取的表
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)

我有以下第一个表格的结果。同样的结果也适用于其他表格。。

STARTING THE EXTRACTION PROCESSING FOR TABLE: table_1
Query to extract table table_1 is: (select * from table_1) tble
+---------+--------------------+
|EMPLOY_ID|EMPLOYE_NM          |
+---------+--------------------+
|1        |WELLINGTON          |
|2        |SMITH               |
|3        |CURLEY              |
|4        |PENDRAGON           |
|5        |KEESLER             |
+---------+--------------------+
only showing top 5 rows

hive_db.employee_table_extracted_for_table_1

[EMPLOY_ID: int, EMPLOYE_NM: string].write.bucketBy(80, "EMPLOY_ID", "EMPLOYE_NO").sortBy("EMPLOY_ID").format("parquet").saveAsTable("hive_db.employee_table_extracted_for_table_1")

COMPLETED EXTRACTION PROCESS FOR TABLE: table_1

它没有将提取的Dataframe写入配置单元,而是打印列名

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

如何将df写入hive表?

tpgth1q7

tpgth1q71#

你能试试这个方法吗,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),

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

并更换 df.bucketBy() 有足够的论据 (numBuckets: Int, colName: String, colNames: String*) ```
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)
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")

相关问题