Spark SQL doesn't support Chinese column name in sql query - apache-spark

I am running the following spark sql:
test("SparkSQLTest") {
val spark = SparkSession.builder().master(s"local[1]").config("spark.sql.cbo.enabled", true).appName("SparkSQLTest").getOrCreate()
spark.createDataFrame(Seq(Person("Tom", 31), Person("Jack", 22))).createOrReplaceTempView("t")
val df = spark.sql("select name, '' XYZ, '' 水果, age from t ")
df.show(false)
}
I add a constant, name it to 水果, but Spark SQL doesn't recognize it

While creating the spark session object set the encoding option as SJIS. It’s a character encoding

Related

Spark to SparkSQL equivqlent syntax

I have this two line in spark, I want to get the equivalent in SparkSQL (im working with python env)
df = spark_df.filter(col["col_name".lower()].rlike("[0-9]{9}$")).count()
spark_df = spark_df.withColumn(columnname, F.to_date(F.col((columnname, ),"yyyyMMdd"))
For spark sql first convert dataframe to temp view then run sql.
Example:
spark_df.createOrReplaceTempView("tmp")
df=spark.sql("""select count(*) from tmp where lower(col_name) rlike("[0-9]{9}$") """).collect()[0][0]
spark_df = spark.sql("""select *, to_date(columnname,"yyyyMMdd") columnname from tmp """)

Spark SQL 'Show table extended from db like table gives different result' from Hive

spark.sql("SHOW TABLE EXTENDED IN DB LIKE 'TABLE'")
Beeline >>SHOW TABLE EXTENDED IN DB LIKE 'TABLE';
Both queries have different results.
If I run the same query in Spark it is giving different result than Hive. Format and lastUpdatedTime is missing in Spark SQL.
If anyone have idea then please let me know how to see lastUpdatedTime of Hive table from Spark SQL
Try this -
scala> val df = spark.sql(s"describe extended ${db}.${table_name}").select("data_type").where("col_name == 'Table Properties'")
df: org.apache.spark.sql.Dataset[org.apache.spark.sql.Row] = [data_type: string]
scala> df.map(r => r.getString(0).split(",")(1).trim).collect
res39: Array[String] = Array(last_modified_time=1539848078)
scala> df.map(r => r.getString(0).split(",")(1).trim.split("=")(1)).collect.mkString
res41: String = 1539848078

Does Spark SQL 2.1 support write temp table or hive table to Mysql/Oracle?

I am using Spark SQL 2.1,
I am able to read data from mysql/oracle as a temporary view in Spark SQL:
val url = "jdbc:mysql://localhost:3306/db1"
val user = "root"
val pass = "root"
val sql =
s"""
CREATE OR REPLACE TEMPORARY VIEW foobar
USING org.apache.spark.sql.jdbc
OPTIONS (url '$url', dbtable '(select id, name, now() from db1.company) as x', user '$user', password '$pass', numPartitions '3')
""".stripMargin(' ')
val session = SparkSession.builder().appName("SparkSQL003").master("local[4]").getOrCreate()
session.sql(sql)
My question is whether Spark SQL 2.1 supports to write temporary view or Hive table back to oracle/mysql with pure sql
I post my findings here, Spark SQL supports this functionality in 2.1.0
test("SparkSQL006") {
val url = "jdbc:mysql://localhost:3306/db1"
val user = "root"
val pass = "root"
val sql =
s"""
CREATE OR REPLACE TEMPORARY VIEW foobar
USING org.apache.spark.sql.jdbc
OPTIONS (url '$url', dbtable 'db1.company', user '$user', password '$pass', numPartitions '3')
""".stripMargin(' ')
val session = SparkSession.builder().appName("SparkSQL003").master("local[4]").getOrCreate()
import session.implicits._
session.createDataset(Seq((100,"IBM"), (200, "Apply"))).createOrReplaceTempView("t")
session.sql(sql)
session.sql(
"""
insert into foobar select * from t
""".stripMargin(' '))
}

spark Dataframe string to Hive varchar

I read data from Oracle via spark JDBC connection to a DataFrame. I have a column which is obviously StringType in dataframe.
Now I want to persist this in Hive, but as datatype Varchar(5). I know the string would be truncated but it is ok.
I tried using UDFs which didn't work since dataframe does not have varchar or char types. I also created a temporary view in Hive using:
val tv = df.createOrReplaceTempView("t_name")
val df = spark.sql("select cast(col_name as varchar(5)) from tv")
But then when i printSchema, i still see a string type.
How can I make I save it as a varchar column in Hive table ?
Try creating Hive table("dbName.tableName") with required schema (varchar(5) in this case) and insert into the table directly from Dataframe like below.
df.write.insertInto("dbName.tableName" ,overwrite = False)

Overwrite specific partitions in spark dataframe write method

I want to overwrite specific partitions instead of all in spark. I am trying the following command:
df.write.orc('maprfs:///hdfs-base-path','overwrite',partitionBy='col4')
where df is dataframe having the incremental data to be overwritten.
hdfs-base-path contains the master data.
When I try the above command, it deletes all the partitions, and inserts those present in df at the hdfs path.
What my requirement is to overwrite only those partitions present in df at the specified hdfs path. Can someone please help me in this?
Finally! This is now a feature in Spark 2.3.0:
SPARK-20236
To use it, you need to set the spark.sql.sources.partitionOverwriteMode setting to dynamic, the dataset needs to be partitioned, and the write mode overwrite. Example:
spark.conf.set("spark.sql.sources.partitionOverwriteMode","dynamic")
data.write.mode("overwrite").insertInto("partitioned_table")
I recommend doing a repartition based on your partition column before writing, so you won't end up with 400 files per folder.
Before Spark 2.3.0, the best solution would be to launch SQL statements to delete those partitions and then write them with mode append.
This is a common problem. The only solution with Spark up to 2.0 is to write directly into the partition directory, e.g.,
df.write.mode(SaveMode.Overwrite).save("/root/path/to/data/partition_col=value")
If you are using Spark prior to 2.0, you'll need to stop Spark from emitting metadata files (because they will break automatic partition discovery) using:
sc.hadoopConfiguration.set("parquet.enable.summary-metadata", "false")
If you are using Spark prior to 1.6.2, you will also need to delete the _SUCCESS file in /root/path/to/data/partition_col=value or its presence will break automatic partition discovery. (I strongly recommend using 1.6.2 or later.)
You can get a few more details about how to manage large partitioned tables from my Spark Summit talk on Bulletproof Jobs.
spark.conf.set("spark.sql.sources.partitionOverwriteMode","dynamic")
data.toDF().write.mode("overwrite").format("parquet").partitionBy("date", "name").save("s3://path/to/somewhere")
This works for me on AWS Glue ETL jobs (Glue 1.0 - Spark 2.4 - Python 2)
Adding 'overwrite=True' parameter in the insertInto statement solves this:
hiveContext.setConf("hive.exec.dynamic.partition", "true")
hiveContext.setConf("hive.exec.dynamic.partition.mode", "nonstrict")
df.write.mode("overwrite").insertInto("database_name.partioned_table", overwrite=True)
By default overwrite=False. Changing it to True allows us to overwrite specific partitions contained in df and in the partioned_table. This helps us avoid overwriting the entire contents of the partioned_table with df.
Using Spark 1.6...
The HiveContext can simplify this process greatly. The key is that you must create the table in Hive first using a CREATE EXTERNAL TABLE statement with partitioning defined. For example:
# Hive SQL
CREATE EXTERNAL TABLE test
(name STRING)
PARTITIONED BY
(age INT)
STORED AS PARQUET
LOCATION 'hdfs:///tmp/tables/test'
From here, let's say you have a Dataframe with new records in it for a specific partition (or multiple partitions). You can use a HiveContext SQL statement to perform an INSERT OVERWRITE using this Dataframe, which will overwrite the table for only the partitions contained in the Dataframe:
# PySpark
hiveContext = HiveContext(sc)
update_dataframe.registerTempTable('update_dataframe')
hiveContext.sql("""INSERT OVERWRITE TABLE test PARTITION (age)
SELECT name, age
FROM update_dataframe""")
Note: update_dataframe in this example has a schema that matches that of the target test table.
One easy mistake to make with this approach is to skip the CREATE EXTERNAL TABLE step in Hive and just make the table using the Dataframe API's write methods. For Parquet-based tables in particular, the table will not be defined appropriately to support Hive's INSERT OVERWRITE... PARTITION function.
Hope this helps.
Tested this on Spark 2.3.1 with Scala.
Most of the answers above are writing to a Hive table. However, I wanted to write directly to disk, which has an external hive table on top of this folder.
First the required configuration
val sparkSession: SparkSession = SparkSession
.builder
.enableHiveSupport()
.config("spark.sql.sources.partitionOverwriteMode", "dynamic") // Required for overwriting ONLY the required partitioned folders, and not the entire root folder
.appName("spark_write_to_dynamic_partition_folders")
Usage here:
DataFrame
.write
.format("<required file format>")
.partitionBy("<partitioned column name>")
.mode(SaveMode.Overwrite) // This is required.
.save(s"<path_to_root_folder>")
I tried below approach to overwrite particular partition in HIVE table.
### load Data and check records
raw_df = spark.table("test.original")
raw_df.count()
lets say this table is partitioned based on column : **c_birth_year** and we would like to update the partition for year less than 1925
### Check data in few partitions.
sample = raw_df.filter(col("c_birth_year") <= 1925).select("c_customer_sk", "c_preferred_cust_flag")
print "Number of records: ", sample.count()
sample.show()
### Back-up the partitions before deletion
raw_df.filter(col("c_birth_year") <= 1925).write.saveAsTable("test.original_bkp", mode = "overwrite")
### UDF : To delete particular partition.
def delete_part(table, part):
qry = "ALTER TABLE " + table + " DROP IF EXISTS PARTITION (c_birth_year = " + str(part) + ")"
spark.sql(qry)
### Delete partitions
part_df = raw_df.filter(col("c_birth_year") <= 1925).select("c_birth_year").distinct()
part_list = part_df.rdd.map(lambda x : x[0]).collect()
table = "test.original"
for p in part_list:
delete_part(table, p)
### Do the required Changes to the columns in partitions
df = spark.table("test.original_bkp")
newdf = df.withColumn("c_preferred_cust_flag", lit("Y"))
newdf.select("c_customer_sk", "c_preferred_cust_flag").show()
### Write the Partitions back to Original table
newdf.write.insertInto("test.original")
### Verify data in Original table
orginial.filter(col("c_birth_year") <= 1925).select("c_customer_sk", "c_preferred_cust_flag").show()
Hope it helps.
Regards,
Neeraj
As jatin Wrote you can delete paritions from hive and from path and then append data
Since I was wasting too much time with it I added the following example for other spark users.
I used Scala with spark 2.2.1
import org.apache.hadoop.conf.Configuration
import org.apache.hadoop.fs.Path
import org.apache.spark.SparkConf
import org.apache.spark.sql.{Column, DataFrame, SaveMode, SparkSession}
case class DataExample(partition1: Int, partition2: String, someTest: String, id: Int)
object StackOverflowExample extends App {
//Prepare spark & Data
val sparkConf = new SparkConf()
sparkConf.setMaster(s"local[2]")
val spark = SparkSession.builder().config(sparkConf).getOrCreate()
val tableName = "my_table"
val partitions1 = List(1, 2)
val partitions2 = List("e1", "e2")
val partitionColumns = List("partition1", "partition2")
val myTablePath = "/tmp/some_example"
val someText = List("text1", "text2")
val ids = (0 until 5).toList
val listData = partitions1.flatMap(p1 => {
partitions2.flatMap(p2 => {
someText.flatMap(
text => {
ids.map(
id => DataExample(p1, p2, text, id)
)
}
)
}
)
})
val asDataFrame = spark.createDataFrame(listData)
//Delete path function
def deletePath(path: String, recursive: Boolean): Unit = {
val p = new Path(path)
val fs = p.getFileSystem(new Configuration())
fs.delete(p, recursive)
}
def tableOverwrite(df: DataFrame, partitions: List[String], path: String): Unit = {
if (spark.catalog.tableExists(tableName)) {
//clean partitions
val asColumns = partitions.map(c => new Column(c))
val relevantPartitions = df.select(asColumns: _*).distinct().collect()
val partitionToRemove = relevantPartitions.map(row => {
val fields = row.schema.fields
s"ALTER TABLE ${tableName} DROP IF EXISTS PARTITION " +
s"${fields.map(field => s"${field.name}='${row.getAs(field.name)}'").mkString("(", ",", ")")} PURGE"
})
val cleanFolders = relevantPartitions.map(partition => {
val fields = partition.schema.fields
path + fields.map(f => s"${f.name}=${partition.getAs(f.name)}").mkString("/")
})
println(s"Going to clean ${partitionToRemove.size} partitions")
partitionToRemove.foreach(partition => spark.sqlContext.sql(partition))
cleanFolders.foreach(partition => deletePath(partition, true))
}
asDataFrame.write
.options(Map("path" -> myTablePath))
.mode(SaveMode.Append)
.partitionBy(partitionColumns: _*)
.saveAsTable(tableName)
}
//Now test
tableOverwrite(asDataFrame, partitionColumns, tableName)
spark.sqlContext.sql(s"select * from $tableName").show(1000)
tableOverwrite(asDataFrame, partitionColumns, tableName)
import spark.implicits._
val asLocalSet = spark.sqlContext.sql(s"select * from $tableName").as[DataExample].collect().toSet
if (asLocalSet == listData.toSet) {
println("Overwrite is working !!!")
}
}
If you use DataFrame, possibly you want to use Hive table over data.
In this case you need just call method
df.write.mode(SaveMode.Overwrite).partitionBy("partition_col").insertInto(table_name)
It'll overwrite partitions that DataFrame contains.
There's not necessity to specify format (orc), because Spark will use Hive table format.
It works fine in Spark version 1.6
Instead of writing to the target table directly, i would suggest you create a temporary table like the target table and insert your data there.
CREATE TABLE tmpTbl LIKE trgtTbl LOCATION '<tmpLocation';
Once the table is created, you would write your data to the tmpLocation
df.write.mode("overwrite").partitionBy("p_col").orc(tmpLocation)
Then you would recover the table partition paths by executing:
MSCK REPAIR TABLE tmpTbl;
Get the partition paths by querying the Hive metadata like:
SHOW PARTITONS tmpTbl;
Delete these partitions from the trgtTbl and move the directories from tmpTbl to trgtTbl
I would suggest you doing clean-up and then writing new partitions with Append mode:
import scala.sys.process._
def deletePath(path: String): Unit = {
s"hdfs dfs -rm -r -skipTrash $path".!
}
df.select(partitionColumn).distinct.collect().foreach(p => {
val partition = p.getAs[String](partitionColumn)
deletePath(s"$path/$partitionColumn=$partition")
})
df.write.partitionBy(partitionColumn).mode(SaveMode.Append).orc(path)
This will delete only new partitions. After writing data run this command if you need to update metastore:
sparkSession.sql(s"MSCK REPAIR TABLE $db.$table")
Note: deletePath assumes that hfds command is available on your system.
My solution implies overwriting each specific partition starting from a spark dataframe. It skips the dropping partition part. I'm using pyspark>=3 and I'm writing on AWS s3:
def write_df_on_s3(df, s3_path, field, mode):
# get the list of unique field values
list_partitions = [x.asDict()[field] for x in df.select(field).distinct().collect()]
df_repartitioned = df.repartition(1,field)
for p in list_partitions:
# create dataframes by partition and send it to s3
df_to_send = df_repartitioned.where("{}='{}'".format(field,p))
df_to_send.write.mode(mode).parquet(s3_path+"/"+field+"={}/".format(p))
The arguments of this simple function are the df, the s3_path, the partition field, and the mode (overwrite or append). The first part gets the unique field values: it means that if I'm partitioning the df by daily, I get a list of all the dailies in the df. Then I'm repartition the df. Finally, I'm selecting the repartitioned df by each daily and I'm writing it on its specific partition path.
You can change the repartition integer by your needs.
You could do something like this to make the job reentrant (idempotent):
(tried this on spark 2.2)
# drop the partition
drop_query = "ALTER TABLE table_name DROP IF EXISTS PARTITION (partition_col='{val}')".format(val=target_partition)
print drop_query
spark.sql(drop_query)
# delete directory
dbutils.fs.rm(<partition_directoy>,recurse=True)
# Load the partition
df.write\
.partitionBy("partition_col")\
.saveAsTable(table_name, format = "parquet", mode = "append", path = <path to parquet>)
For >= Spark 2.3.0 :
spark.conf.set("spark.sql.sources.partitionOverwriteMode","dynamic")
data.write.insertInto("partitioned_table", overwrite=True)

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