Spark Scala: 将结构数组列转换为字符串列

Question

Spark Scala: 将结构数组列转换为字符串列

6

我有一个列，它是从json文件中推断出的array<Struct>类型。我想将这个array<Struct>转换为字符串，这样我就可以将这个数组列保留在hive中，并将其作为单个列导出到RDBMS中。

temp.json

{"properties":{"items":[{"invoicid":{"value":"923659"},"job_id":
{"value":"296160"},"sku_id":
{"value":"312002"}}],"user_id":"6666","zip_code":"666"}}

处理中：

scala> val temp = spark.read.json("s3://check/1/temp1.json")
temp: org.apache.spark.sql.DataFrame = [properties: struct<items:
array<struct<invoicid:struct<value:string>,job_id:struct<value:string>,sku_id:struct<value:string>>>, user_id: string ... 1 more field>]

    scala> temp.printSchema
    root
     |-- properties: struct (nullable = true)
     |    |-- items: array (nullable = true)
     |    |    |-- element: struct (containsNull = true)
     |    |    |    |-- invoicid: struct (nullable = true)
     |    |    |    |    |-- value: string (nullable = true)
     |    |    |    |-- job_id: struct (nullable = true)
     |    |    |    |    |-- value: string (nullable = true)
     |    |    |    |-- sku_id: struct (nullable = true)
     |    |    |    |    |-- value: string (nullable = true)
     |    |-- user_id: string (nullable = true)
     |    |-- zip_code: string (nullable = true)


scala> temp.select("properties").show
+--------------------+
|          properties|
+--------------------+
|[WrappedArray([[9...|
+--------------------+


scala> temp.select("properties.items").show
+--------------------+
|               items|
+--------------------+
|[[[923659],[29616...|
+--------------------+


scala> temp.createOrReplaceTempView("tempTable")

scala> spark.sql("select properties.items  from tempTable").show
+--------------------+
|               items|
+--------------------+
|[[[923659],[29616...|
+--------------------+

我该如何获得类似于以下结果：

+-----------------------------------------------------------------------------------------+
|               items                                                                     |
+-----------------------------------------------------------------------------------------+
[{"invoicid":{"value":"923659"},"job_id":{"value":"296160"},"sku_id":{"value":"312002"}}] |
+-----------------------------------------------------------------------------------------+

获取数组元素的值而不做任何更改。

- varun r

[{"invoicid":{"value":"923659"},"job_id":{"value":"296160"},"sku_id":{"value":"312002"}}] - Manoj Pandey

1个回答

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- zero323 · Accepted Answer

to_json 是你要寻找的函数。

import org.apache.spark.sql.functions.to_json:

val df = spark.read.json(sc.parallelize(Seq("""
  {"properties":{"items":[{"invoicid":{"value":"923659"},"job_id":
  {"value":"296160"},"sku_id":
  {"value":"312002"}}],"user_id":"6666","zip_code":"666"}}""")))


df
  .select(get_json_object(to_json($"properties"), "$.items").alias("items"))
  .show(false)

+-----------------------------------------------------------------------------------------+
|items                                                                                    |
+-----------------------------------------------------------------------------------------+
|[{"invoicid":{"value":"923659"},"job_id":{"value":"296160"},"sku_id":{"value":"312002"}}]|
+-----------------------------------------------------------------------------------------+