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使用spark xml读取值xml标记值,希望获取该值但给出列表

apache-spark-xml apache-spark scala

tree em · 技术社区 · 6 年前

<row id='185685445477437.020001' xml:space='preserve'>
    <c2>KH0013001</c2>
    <c3>-2271164.00</c3>
    <c4>9</c4>
    <c7>65395</c7>
    <c9>1</c9>
    <c12>KHR</c12>
    <c16>TR</c16>
    <c17>6-71-10-1-001-030</c17>
    <c20>1</c20>
    <c22>1</c22>
    <c23>DC183050001030071</c23>
    <c24>DC</c24>
    <c25>20181101</c25>
    <c26>185685445477437.02</c26>
    <c26 m='3'>1</c26>
    <c29>1</c29>
    <c30>5011_DMUSER__OFS_DM.OFS.SRC.VAL</c30>
    <c31>1811012130</c31>
    <c32>6010_DMUSER</c32>
    <c56>PL.65395.......1.....KH0013001</c56>
    <c98></c98>
</row>

用Spark XML激发

import org.apache.spark.sql.{SQLContext, SparkSession}

object sparkXml {
  def main(args: Array[String]): Unit = {

    val spark = SparkSession.
      builder.master("local[*]")
      //.config("spark.debug.maxToStringFields", "100")
      .appName("Insight Application Big Data")
      .getOrCreate()

    val df = spark.read
      .format("com.databricks.spark.xml")
      .option("rowTag", "row")
      .load("src/main/resources/in/FBNK_CATEG_ENTRY.xml")
    df.createOrReplaceTempView("categ_entry")

   df.printSchema()
  spark.sql("Select c26['_VALUE'] as value, c26['_m'] as option from categ_entry").show()


  }
}

印刷模式

root
 |-- _id: double (nullable = true)
 |-- _space: string (nullable = true)
 |-- c12: string (nullable = true)
 |-- c16: string (nullable = true)
 |-- c17: string (nullable = true)
 |-- c2: string (nullable = true)
 |-- c20: long (nullable = true)
 |-- c22: long (nullable = true)
 |-- c23: string (nullable = true)
 |-- c24: string (nullable = true)
 |-- c25: long (nullable = true)
 |-- c26: array (nullable = true)
 |    |-- element: struct (containsNull = true)
 |    |    |-- _VALUE: double (nullable = true)
 |    |    |-- _m: long (nullable = true)
 |-- c29: long (nullable = true)
 |-- c3: double (nullable = true)
 |-- c30: string (nullable = true)
 |-- c31: long (nullable = true)
 |-- c32: string (nullable = true)
 |-- c4: long (nullable = true)
 |-- c56: string (nullable = true)
 |-- c7: long (nullable = true)
 |-- c9: long (nullable = true)
 |-- c98: string (nullable = true)

运行后的结果

+--------------------+------+
|[1.85685445477437...| [, 3]|
+--------------------+------+

我希望结果是这样的。

+--------------------+------+
| 185685445477437.02  | 3   |
+--------------------+------+

任何人请指导我应该如何更正代码以产生预期的结果

1 回复 | 直到 6 年前

NNK 6 年前

数据的获取方式很难生成所需的输出。

<c26>185685445477437.02</c26>   
<c26 m='3'>1</c26>

您有两个标签,Spark结构将其视为一个数组。您需要从第一个c26_值得到185685445477437.02,从第二个c26_属性得到3,这可能不正确。

如果你在寻找下面的输出。用下面的语句替换最后一行

val df2 = df.withColumn("c26Struct",explode(col("c26")))
df2.select(col("c26Struct._VALUE").alias("value"),col("c26Struct._m").alias("option") ).show(false)


+---------------------+------+
|value                |option|
+---------------------+------+
|1.8568544547743703E14|null  |
|1.0                  |3     |
+---------------------+------+

在这里,我通过分解CR26数组和从分解的新列中选择的值创建了一个新的列C26结构。

希望这有帮助!!

谢谢, 纳文