CODEWITHSUNDEEP
apache-sparkapache-spark-sqldatabricks
BdEngineer
BdEngineer

Reputation: 3199

How to set jdbc/partitionColumn type to Date in spark 2.4.1

I am trying to retrieve data from oracle using spark-sql-2.4.1 version. I tried to set the JdbcOptions as below :

    .option("lowerBound", "31-MAR-02");
    .option("upperBound", "01-MAY-19");
    .option("partitionColumn", "data_date");
    .option("numPartitions", 240);

But gives error :

    java.lang.IllegalArgumentException: Timestamp format must be yyyy-mm-dd hh:mm:ss[.fffffffff]
        at java.sql.Timestamp.valueOf(Timestamp.java:204)
        at org.apache.spark.sql.execution.datasources.jdbc.JDBCRelation$.toInternalBoundValue(JDBCRelation.scala:179)

Then tried as below

    .option("lowerBound", "2002-03-31"); //changed the date format
    .option("upperBound", "2019-05-02");
    .option("partitionColumn", "data_date"); 
    .option("numPartitions", 240);

Still no luck. So what is the correct way to pass the date as "lower/upperBound"? Is there a way to specify/set option parameter data Type ?

Part-2 Checked the options properly. they were overwritten in between before executing the query. So corrected it. ... now that error resolved.

But for below options:

.option("lowerBound", "2002-03-31 00:00:00"); 
.option("upperBound", "2019-05-01 23:59:59");
.option("timestampFormat", "yyyy-mm-dd hh:mm:ss");

Query String : 

query ->  ( SELECT * FROM MODEL_VALS ) T

It is throwing another error :

java.sql.SQLException: ORA-12801: error signaled in parallel query server P022, instance nj0005

ORA-01861: literal does not match format string

Upvotes: 10

Views: 13842

Answers (4)

alexm
alexm

Reputation: 530

If you are using Oracle, see: apache/spark/.../OracleIntegrationSuite.scala

val df1 = spark.read.format("jdbc")
      .option("url", jdbcUrl)
      .option("dbtable", "datetimePartitionTest")
      .option("partitionColumn", "d")
      .option("lowerBound", "2018-07-06")
      .option("upperBound", "2018-07-20")
      .option("numPartitions", 3)
      // oracle.jdbc.mapDateToTimestamp defaults to true. If this flag is not disabled, column d
      // (Oracle DATE) will be resolved as Catalyst Timestamp, which will fail bound evaluation of
      // the partition column. E.g. 2018-07-06 cannot be evaluated as Timestamp, and the error
      // message says: Timestamp format must be yyyy-mm-dd hh:mm:ss[.fffffffff].
      .option("oracle.jdbc.mapDateToTimestamp", "false")
      .option("sessionInitStatement", "ALTER SESSION SET NLS_DATE_FORMAT = 'YYYY-MM-DD'")
      .load()

Upvotes: 15

Ankush
Ankush

Reputation: 519

I stumbled on this question as I am solving a similar problem. But in this case Spark 2.4.2 is sending date in format 'yyyy-MM-dd HH:mm:ss.ssss' to Oracle and it returned "Not a valid month" as it expects 'dd-MMM-yy HH:mm:ss.ssss'. To solve that I followed: Spark GitHub Link , it says:

Override beforeFetch method in OracleDialect to finish the following two things:

Set Oracle's NLS_TIMESTAMP_FORMAT to "YYYY-MM-DD HH24:MI:SS.FF" to match java.sql.Timestamp format. Set Oracle's NLS_DATE_FORMAT to "YYYY-MM-DD" to match java.sql.Date format.

And it solved the issue. Hope it helps.

Upvotes: 3

raam86
raam86

Reputation: 6871

all of the following options must be set in this way in order for it to work:

spark.read
      .option("header", true)
      .option("inferSchema", true)
      .option("timestampFormat", "MM/dd/yyyy h:mm:ss a")
      .csv("PATH_TO_CSV")

Upvotes: 1

Alex Ott
Alex Ott

Reputation: 87249

The given parameters have type timestamp, but you're providing the only date. Timestamp has format as yyyy-mm-dd hh:mm:ss, so you need to provide your dates as 2002-03-31 00:00:00 and 2019-05-01 23:59:59 correspondingly...

Upvotes: 2

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