CODEWITHSUNDEEP
scalaapache-sparkapache-spark-sql
Lee
Lee

Reputation: 658

Spark 2.2 Illegal pattern component: XXX java.lang.IllegalArgumentException: Illegal pattern component: XXX

I'm trying to upgrade from Spark 2.1 to 2.2. When I try to read or write a dataframe to a location (CSV or JSON) I am receiving this error:

Illegal pattern component: XXX
java.lang.IllegalArgumentException: Illegal pattern component: XXX
at org.apache.commons.lang3.time.FastDatePrinter.parsePattern(FastDatePrinter.java:282)
at org.apache.commons.lang3.time.FastDatePrinter.init(FastDatePrinter.java:149)
at org.apache.commons.lang3.time.FastDatePrinter.<init>(FastDatePrinter.java:142)
at org.apache.commons.lang3.time.FastDateFormat.<init>(FastDateFormat.java:384)
at org.apache.commons.lang3.time.FastDateFormat.<init>(FastDateFormat.java:369)
at org.apache.commons.lang3.time.FastDateFormat$1.createInstance(FastDateFormat.java:91)
at org.apache.commons.lang3.time.FastDateFormat$1.createInstance(FastDateFormat.java:88)
at org.apache.commons.lang3.time.FormatCache.getInstance(FormatCache.java:82)
at org.apache.commons.lang3.time.FastDateFormat.getInstance(FastDateFormat.java:165)
at org.apache.spark.sql.catalyst.json.JSONOptions.<init>(JSONOptions.scala:81)
at org.apache.spark.sql.catalyst.json.JSONOptions.<init>(JSONOptions.scala:43)
at org.apache.spark.sql.execution.datasources.json.JsonFileFormat.inferSchema(JsonFileFormat.scala:53)
at org.apache.spark.sql.execution.datasources.DataSource$$anonfun$7.apply(DataSource.scala:177)
at org.apache.spark.sql.execution.datasources.DataSource$$anonfun$7.apply(DataSource.scala:177)
at scala.Option.orElse(Option.scala:289)
at org.apache.spark.sql.execution.datasources.DataSource.getOrInferFileFormatSchema(DataSource.scala:176)
at org.apache.spark.sql.execution.datasources.DataSource.resolveRelation(DataSource.scala:366)
at org.apache.spark.sql.DataFrameReader.load(DataFrameReader.scala:178)
at org.apache.spark.sql.DataFrameReader.json(DataFrameReader.scala:333)
at org.apache.spark.sql.DataFrameReader.json(DataFrameReader.scala:279)

I am not setting a default value for dateFormat, so I'm not understanding where it is coming from.

spark.createDataFrame(objects.map((o) => MyObject(t.source, t.table, o.partition, o.offset, d)))
    .coalesce(1)
    .write
    .mode(SaveMode.Append)
    .partitionBy("source", "table")
    .json(path)

I still get the error with this:

import org.apache.spark.sql.{SaveMode, SparkSession}
val spark = SparkSession.builder.appName("Spark2.2Test").master("local").getOrCreate()
import spark.implicits._
val agesRows = List(Person("alice", 35), Person("bob", 10), Person("jill", 24))
val df = spark.createDataFrame(agesRows).toDF();

df.printSchema
df.show

df.write.mode(SaveMode.Overwrite).csv("my.csv")

Here is the schema:

root
 |-- name: string (nullable = true)
 |-- age: long (nullable = false)

Upvotes: 17

Views: 22661

Answers (4)

Zhang Xujie
Zhang Xujie

Reputation: 13

I also met this problem, and my solution(reason) is: Because I put a wrong format json file to hdfs. After I put a correct text or json file, it can go correctly.

Upvotes: -2

danzhi
danzhi

Reputation: 51

Use commons-lang3-3.5.jar fixed the original error. I didn't check the source code to tell why but it is no surprising as the original exception happens at org.apache.commons.lang3.time.FastDatePrinter.parsePattern(FastDatePrinter.java:282). I also noticed the file /usr/lib/spark/jars/commons-lang3-3.5.jar (on an EMR cluster instance) which also suggest 3.5 is the consistent version to use.

Upvotes: 4

Lee
Lee

Reputation: 658

I found the answer.

The default for the timestampFormat is yyyy-MM-dd'T'HH:mm:ss.SSSXXX which is an illegal argument. It needs to be set when you are writing the dataframe out.

The fix is to change that to ZZ which will include the timezone.

df.write
.option("timestampFormat", "yyyy/MM/dd HH:mm:ss ZZ")
.mode(SaveMode.Overwrite)
.csv("my.csv")

Upvotes: 36

Mauro Pirrone
Mauro Pirrone

Reputation: 401

Ensure you are using the correct version of commons-lang3

<dependency>
  <groupId>org.apache.commons</groupId>
  <artifactId>commons-lang3</artifactId>
  <version>3.5</version>
</dependency>

Upvotes: 26

Related Questions