如何在 scala 中将嵌套 JSON 转换为映射对象

我有以下 JSON 对象：

{
    "user_id": "123",
    "data": {
        "city": "New York"
    },
    "timestamp": "1563188698.31",
    "session_id": "6a793439-6535-4162-b333-647a6761636b"
}
{
    "user_id": "123",
    "data": {
        "name": "some_name",
        "age": "23",
        "occupation": "teacher"
    },
    "timestamp": "1563188698.31",
    "session_id": "6a793439-6535-4162-b333-647a6761636b"
}

我在用着val df = sqlContext.read.json("json")将文件读取到数据框

它将所有数据属性组合成数据结构，如下所示：

root
 |-- data: struct (nullable = true)
 |    |-- age: string (nullable = true)
 |    |-- city: string (nullable = true)
 |    |-- name: string (nullable = true)
 |    |-- occupation: string (nullable = true)
 |-- session_id: string (nullable = true)
 |-- timestamp: string (nullable = true)
 |-- user_id: string (nullable = true)

是否可以将数据字段转换为 MAP[String, String] 数据类型？那么它只具有与原始 json 相同的属性？

是的，您可以通过从 JSON 数据导出 Map[String, String] 来实现这一点，如下所示：

import org.apache.spark.sql.types.{MapType, StringType}
import org.apache.spark.sql.functions.{to_json, from_json}

val jsonStr = """{
    "user_id": "123",
    "data": {
        "name": "some_name",
        "age": "23",
        "occupation": "teacher"
    },
    "timestamp": "1563188698.31",
    "session_id": "6a793439-6535-4162-b333-647a6761636b"
}"""

val df = spark.read.json(Seq(jsonStr).toDS)

val mappingSchema = MapType(StringType, StringType)

df.select(from_json(to_json($"data"), mappingSchema).as("map_data"))

//Output
// +-----------------------------------------------------+
// |map_data                                             |
// +-----------------------------------------------------+
// |[age -> 23, name -> some_name, occupation -> teacher]|
// +-----------------------------------------------------+

首先我们提取出内容data字段转换为字符串to_json($"data")，然后我们解析并提取 Mapfrom_json(to_json($"data"), schema).