I would like to insert the table, and the value of the partition column
to be inserted must be from temporary registered table/dataframe.
Patcharee
On 16. mars 2015 15:26, Cheng Lian wrote:
Not quite sure whether I understand your question properly. But if you
just want to read the partition columns, it’s pretty easy. Take the
“year” column as an example, you may do this in HiveQL:
|hiveContext.sql("SELECT year FROM speed")
|
or in DataFrame DSL:
|hiveContext.table("speed").select("year")
|
Cheng
On 3/16/15 9:59 PM, patcharee wrote:
Hi,
I tried to insert into a hive partitioned table
val ZONE: Int = Integer.valueOf(args(2))
val MONTH: Int = Integer.valueOf(args(3))
val YEAR: Int = Integer.valueOf(args(4))
val weightedUVToDF = weightedUVToRecord.toDF()
weightedUVToDF.registerTempTable("speeddata")
hiveContext.sql("INSERT OVERWRITE table speed partition (year=" +
YEAR + ",month=" + MONTH + ",zone=" + ZONE + ")
select key, speed, direction from speeddata")
First I registered a temporary table "speeddata". The value of the
partitioned column (year, month, zone) is from user input. If I would
like to get the value of the partitioned column from the temporary
table, how can I do that?
BR,
Patcharee