df.write.saveAsTable("db_name.tbl_name") // works, spark-shell, latest spark
version 1.6.0
df.write.saveAsTable("db_name.tbl_name") // NOT work, spark-shell, old spark
version 1.4
--
Jacky Wang
At 2016-02-21 17:35:53, "Mich Talebzadeh" <[email protected]> wrote:
I looked at doc on this. It is not clear what goes behind the scene. Very
little documentation on it
First in Hive a database has to exist before it can be used so sql(“use
mytable”) will not create a database for you.
Also you cannot call your table mytable in database mytable!
Remember in Hive the hierarchy is the database followed by tables.
If you want to create a Hive database (I would not for every table) for this
purpose you can do
scala> sql("create database if not exists mytable_db")
res10: org.apache.spark.sql.DataFrame = [result: string]
scala> sql("use mytable_db")
res12: org.apache.spark.sql.DataFrame = [result: string]
This puts you in the context of mytable_db database
If you do
hdfs dfs -ls /user/hive/warehouse
You will see a directory called mytable_db.db is created
/user/hive/warehouse/mytable.db
Then you can create a table in Hive in mytable_db if you wish. The way I do it
personally is to register your DF as a temporary table and do insert/select
into Hive table
scala> sql("use mytable_db")
res21: org.apache.spark.sql.DataFrame = [result: string]
scala> """
| CREATE TABLE mytable (
| INVOICENUMBER INT
| ,PAYMENTDATE timestamp
| ,NET DECIMAL(20,2)
| ,VAT DECIMAL(20,2)
| ,TOTAL DECIMAL(20,2)
| )
| COMMENT 'a test table'
| STORED AS ORC
| TBLPROPERTIES ( "orc.compress"="ZLIB" )
| """
res22: String =
"
CREATE TABLE mytable (
INVOICENUMBER INT
,PAYMENTDATE timestamp
,NET DECIMAL(20,2)
,VAT DECIMAL(20,2)
,TOTAL DECIMAL(20,2)
)
COMMENT 'a test table'
STORED AS ORC
TBLPROPERTIES ( "orc.compress"="ZLIB" )
"
scala> sql(sqltext)
res6: org.apache.spark.sql.DataFrame = [result: string]
My DF is called “a” below so I register it as a temp table called tmp
a.toDF.registerTempTable("tmp")
Then just insert/select into Hicve table mytable from tmp
scala> sqltext = "INSERT INTO mytable SELECT * FROM tmp"
sqltext: String = INSERT INTO mytable SELECT * FROM tmp
scala> sql(sqltext)
res10: org.apache.spark.sql.DataFrame = []
scala> sql("select count(1) from mytable").show
+---+
|_c0|
+---+
| 65|
|
|
HTH
Dr Mich Talebzadeh
LinkedIn
https://www.linkedin.com/profile/view?id=AAEAAAAWh2gBxianrbJd6zP6AcPCCdOABUrV8Pw
http://talebzadehmich.wordpress.com
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From: Glen [mailto:[email protected]]
Sent: 21 February 2016 03:26
To: gen tang <[email protected]>
Cc:[email protected]
Subject: Re: how to set database in DataFrame.saveAsTable?
Any example code?
In pyspark:
sqlContex.sql("use mytable")
my_df.saveAsTable("tmp_spark_debug", mode="overwrite")
1. The code above seems not register the table in hive. I have to create table
from hdfs in hive, it reports some format error: rcformat and parquet.
2. Rerun the saveAsTable using mode="overwrite" in saveAsTable, it reports the
table already exists.
3. Sometimes it creates a directory in hive/warehouse/tmp_spark_debug, not in
hive/warehouse/mytable/tmp_spark_debug.
My goal is simple:
df.saveAsTable('blablabla') // create a hive table in some database, then it
can be visited by hive.
I tried lots of time, it seems there are lots of bug in pyspark. Or my mehtod
is wrong?
2016-02-21 10:04 GMT+08:00 gen tang <[email protected]>:
Hi,
You can use
sqlContext.sql("use <your database>")
before use dataframe.saveAsTable
Hope it could be helpful
Cheers
Gen
On Sun, Feb 21, 2016 at 9:55 AM, Glen <[email protected]> wrote:
For dataframe in spark, so the table can be visited by hive.
--
Jacky Wang
--
Jacky Wang
