This should work. Make sure that you use HiveContext.sql and sqlContext
correctly
This is an example in Spark, reading a CSV file, doing some manipulation,
creating a temp table, saving data as ORC file, adding another column and
inserting values to table in Hive with default values for new rows
import org.apache.spark.SparkConf
import org.apache.spark.sql.Row
import org.apache.spark.sql.hive.HiveContext
import org.apache.spark.sql.types._
import org.apache.spark.sql.SQLContext
import org.apache.spark.sql.functions._
//
val conf = new SparkConf().
setAppName("ImportCSV").
setMaster("local[12]").
set("spark.driver.allowMultipleContexts", "true").
set("spark.hadoop.validateOutputSpecs", "false")
val sc = new SparkContext(conf)
*// Create sqlContext based on HiveContext val sqlContext = new
HiveContext(sc)* import sqlContext.implicits._
*val HiveContext = new org.apache.spark.sql.hive.HiveContext(sc)* //
// Get a DF first based on Databricks CSV libraries
//
val df =
HiveContext.read.format("com.databricks.spark.csv").option("inferSchema",
"true").option("header", "true").load("/data/stg/table2")
//
// Next filter out empty rows (last colum has to be > "" and get rid of
"?" special character. Also get rid of "," in money fields
// Example csv cell £2,500.00 --> need to transform to plain 2500.00
//
val a = df.
filter(col("Total") > "").
map(x => (x.getString(0),x.getString(1),
x.getString(2).substring(1).replace(",", "").toDouble,
x.getString(3).substring(1).replace(",", "").toDouble,
x.getString(4).substring(1).replace(",", "").toDouble))
//
// convert this RDD to DF and create a Spark temporary table
//
*a.toDF.registerTempTable("tmp")* //
// Need to create and populate target ORC table t3 in database test in
Hive
//
HiveContext.sql("use test")
HiveContext.sql("DROP TABLE IF EXISTS test.t3")
var sqltext : String = ""
sqltext = """
CREATE TABLE test.t3 (
INVOICENUMBER String
,PAYMENTDATE String
,NET DOUBLE
,VAT DOUBLE
,TOTAL DOUBLE
)
COMMENT 'from csv file from excel sheet'
STORED AS ORC
TBLPROPERTIES ( "orc.compress"="ZLIB" )
"""
HiveContext.sql(sqltext)
// Note you can only see Spark temporary table in sqlContext NOT
HiveContext
val results = sqlContext.sql("SELECT * FROM tmp")
// clean up the file in HDFS directory first if exists
val hadoopConf = new org.apache.hadoop.conf.Configuration()
val hdfs = org.apache.hadoop.fs.FileSystem.get(new
java.net.URI("hdfs://rhes564:9000"), hadoopConf)
val output = "hdfs://rhes564:9000/user/hive/warehouse/test.db/t3" //
The path for Hive table just created
try { hdfs.delete(new org.apache.hadoop.fs.Path(output), true) } catch {
case _ : Throwable => { } }
* results.write.format("orc").save(output)*//
* sqlContext.sql("ALTER TABLE test.t3 ADD COLUMNS (new_col VARCHAR(30))")*
sqlContext.sql("INSERT INTO test.t3 SELECT *, 'London' FROM tmp")
HiveContext.sql("SELECT * FROM test.t3 ORDER BY
1").collect.foreach(println)
HTH
Dr Mich Talebzadeh
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http://talebzadehmich.wordpress.com
On 11 April 2016 at 01:36, Maurin Lenglart <mau...@cuberonlabs.com> wrote:
> Your solution works in hive, but not in spark, even if I use hive context.
> I tried to create a temp table and then this query:
> - sqlContext.sql("insert into table myTable select * from myTable_temp”)
> But I still get the same error.
>
> thanks
>
> From: Mich Talebzadeh <mich.talebza...@gmail.com>
> Date: Sunday, April 10, 2016 at 12:25 PM
> To: "user @spark" <user@spark.apache.org>
>
> Subject: Re: alter table add columns aternatives or hive refresh
>
> Hi,
>
> I am confining myself to Hive tables. As I stated it before I have not
> tried it in Spark. So I stand corrected.
>
> Let us try this simple test in Hive
>
>
> -- Create table
> hive>
> *create table testme(col1 int); *OK
> --insert a row
> hive> *insert into testme values(1);*
>
> Loading data to table test.testme
> OK
> -- Add a new column to testme
> hive>
> *alter table testme add columns (new_col varchar(30)); *OK
> Time taken: 0.055 seconds
>
> -- Expect one row here
> hive>
> *select * from testme; *OK
> 1 NULL
> --
> *Add a new row including values for new_col. This should work *hive>
> *insert into testme values(1,'London'); *Loading data to table test.testme
> OK
> hive>
> *select * from testme; *OK
> 1 NULL
> 1 London
> Time taken: 0.074 seconds, Fetched: 2 row(s)
> -- Now update the new column
> hive> update testme set col2 = 'NY';
> FAILED: SemanticException [Error 10297]: Attempt to do update or delete on
> table test.testme that does not use an AcidOutputFormat or is not bucketed
>
> So this is Hive. You can add new rows including values for the new
> column but cannot update the null values. Will this work for you?
>
> HTH
>
> Dr Mich Talebzadeh
>
>
>
> LinkedIn *
> https://www.linkedin.com/profile/view?id=AAEAAAAWh2gBxianrbJd6zP6AcPCCdOABUrV8Pw
> <https://www.linkedin.com/profile/view?id=AAEAAAAWh2gBxianrbJd6zP6AcPCCdOABUrV8Pw>*
>
>
>
> http://talebzadehmich.wordpress.com
>
>
>
> On 10 April 2016 at 19:34, Maurin Lenglart <mau...@cuberonlabs.com> wrote:
>
>> Hi,
>> So basically you are telling me that I need to recreate a table, and
>> re-insert everything every time I update a column?
>> I understand the constraints, but that solution doesn’t look good to me.
>> I am updating the schema everyday and the table is a couple of TB of data.
>>
>> Do you see any other options that will allow me not to move TB of data
>> everyday?
>>
>> Thanks for you answer
>>
>> From: Mich Talebzadeh <mich.talebza...@gmail.com>
>> Date: Sunday, April 10, 2016 at 3:41 AM
>> To: maurin lenglart <mau...@cuberonlabs.com>
>> Cc: "user@spark.apache.org" <user@spark.apache.org>
>> Subject: Re: alter table add columns aternatives or hive refresh
>>
>> I have not tried it on Spark but the column added in Hive to an existing
>> table cannot be updated for existing rows. In other words the new column is
>> set to null which does not require the change in the existing file length.
>>
>> So basically as I understand when a column is added to an already table.
>>
>> 1. The metadata for the underlying table will be updated
>> 2. The new column will by default have null value
>> 3. The existing rows cannot have new column updated to a non null value
>> 4. New rows can have non null values set for the new column
>> 5. No sql operation can be done on that column. For example select *
>> from <TABLE> where new_column IS NOT NULL
>> 6. The easiest option is to create a new table with the new column and
>> do insert/select from the existing table with values set for the new column
>>
>> HTH
>>
>> Dr Mich Talebzadeh
>>
>>
>>
>> LinkedIn *
>> https://www.linkedin.com/profile/view?id=AAEAAAAWh2gBxianrbJd6zP6AcPCCdOABUrV8Pw
>> <https://www.linkedin.com/profile/view?id=AAEAAAAWh2gBxianrbJd6zP6AcPCCdOABUrV8Pw>*
>>
>>
>>
>> http://talebzadehmich.wordpress.com
>>
>>
>>
>> On 10 April 2016 at 05:06, Maurin Lenglart <mau...@cuberonlabs.com>
>> wrote:
>>
>>> Hi,
>>> I am trying to add columns to table that I created with the
>>> “saveAsTable” api.
>>> I update the columns using sqlContext.sql(‘alter table myTable add
>>> columns (mycol string)’).
>>> The next time I create a df and save it in the same table, with the new
>>> columns I get a :
>>> “ParquetRelation
>>> requires that the query in the SELECT clause of the INSERT
>>> INTO/OVERWRITE statement generates the same number of columns as its
>>> schema.”
>>>
>>> Also thise two commands don t return the same columns :
>>> 1. sqlContext.table(‘myTable’).schema.fields <— wrong result
>>> 2. sqlContext.sql(’show columns in mytable’) <—— good results
>>>
>>> It seems to be a known bug :
>>> https://issues.apache.org/jira/browse/SPARK-9764 (see related bugs)
>>>
>>> But I am wondering, how else can I update the columns or make sure that
>>> spark take the new columns?
>>>
>>> I already tried to refreshTable and to restart spark.
>>>
>>> thanks
>>>
>>>
>>
>
