使用 spark xml 读取值 xml 标记值，想要获取值但给我列表

Posted 2023-04-15

【中文标题】使用 spark xml 读取值 xml 标记值，想要获取值但给我列表

【英文标题】：Reading value xml tag value using spark xml , want to get the value but give me the list

【发布时间】：2018-12-18 00:56:53

【问题描述】：

<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>

SPARK 与 SPARK XML

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()


  

printSchema

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   |
+--------------------+------+

请指导我如何更正代码以产生预期结果

【参考方案1】：

以您拥有数据的方式，很难产生您想要的输出。

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

您有两个标签，并且 spark 结构将其视为一个数组。你想要 185685445477437.02 来自第一个 c26 _value 和 3 来自第二个 c26 _attribute 这可能不正确。

如果您正在寻找以下输出。用下面的语句替换你的最后一行

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，我通过展开 cr26 数组并从展开的新列中选择值创建了一个新列 c26Struct。

希望这会有所帮助！

谢谢， 纳文