首页 文章

Spark 2.2非法模式组件：XXX java.lang.IllegalArgumentException：非法模式组件：XXX

提问于
浏览 次
5

我正在尝试从Spark 2.1升级到2.2 . 当我尝试将数据帧读取或写入某个位置（CSV或JSON）时，我收到此错误：

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)

我没有为dateFormat设置默认值，所以我不知道它来自哪里 .

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)

我仍然得到这个错误：

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")

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

scala apache-spark spark-dataframe

2 回答

  • 6

    我找到了答案 .

    timestampFormat的默认值是 yyyy-MM-dd'T'HH:mm:ss.SSSXXX ，这是一个非法参数 . 在编写数据帧时需要设置它 .

    修复是将其更改为ZZ，其中包括时区 .

    df.write
.option("timestampFormat", "yyyy/MM/dd HH:mm:ss ZZ")
.mode(SaveMode.Overwrite)
.csv("my.csv")
    回复于
  • 21

    确保您使用的是commons-lang3的正确版本

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

相关问题