【发布时间】:2017-03-06 01:59:32
【问题描述】:
在 Spark SQL 中,我可以使用
val spark = SparkSession
.builder()
.appName("SparkSessionZipsExample")
.master("local")
.config("spark.sql.warehouse.dir", "warehouseLocation-value")
.getOrCreate()
val df = spark.read.json("source/myRecords.json")
df.createOrReplaceTempView("shipment")
val sqlDF = spark.sql("SELECT * FROM shipment")
从“myRecords.json”中获取数据,这个json文件的结构是:
df.printSchema()
root
|-- _id: struct (nullable = true)
| |-- $oid: string (nullable = true)
|-- container: struct (nullable = true)
| |-- barcode: string (nullable = true)
| |-- code: string (nullable = true)
我可以得到这个json的具体列如:
val sqlDF = spark.sql("SELECT container.barcode, container.code FROM shipment")
但是我怎样才能从这个 json 文件中获取 id.$oid 呢?
我试过"SELECT id.$oid FROM shipment_log" 或"SELECT id.\$oid FROM shipment_log",但根本不起作用。
错误信息:
error: invalid escape character
谁能告诉我如何获得id.$oid?
【问题讨论】:
标签: json scala apache-spark apache-spark-sql