我想访问存储在表(实木复合地板)中的json文件的嵌套属性。我可以通过 select * from test来访问表中的所有记录;如何在火花中访问嵌套的属性sql
然而,我无法通过写入查询来作为访问嵌套属性: VAL标签= sqlContext.sql( “选择文本,user.screen_name从测试LIMIT 1”)
Schema是如下:
|-- text: string (nullable = true)
| |-- truncated: boolean (nullable = true)
| |-- user: struct (nullable = true)
| | |-- created_at: string (nullable = true)
| | |-- id: long (nullable = true)
| | |-- id_str: string (nullable = true)
| | |-- is_translator: boolean (nullable = true)
| | |-- lang: string (nullable = true)
| | |-- location: string (nullable = true)
| | |-- name: string (nullable = true)
| | |-- screen_name: string (nullable = true)
下面是我的代码:
import scala.tools.nsc.doc.model.Object
import scala.tools.nsc.interactive.Main
import org.apache.spark.SparkConf
import org.apache.spark.SparkContext
import org.apache.spark.SparkContext._
import org.apache.spark.sql._
import org.apache.spark.sql.SQLContext
object SimpleSparkSQL {
def main(args:Array[String]) {
val path = args(0);
val conf = new SparkConf().setAppName("Simple Application").setMaster("local[2]").set("spark.executor.memory", "1g")
val sc = new SparkContext(conf)
val data = sc.textFile(path)
val sqlContext = new org.apache.spark.sql.SQLContext(sc)
import sqlContext.implicits._
val sqlc = new SQLContext(sc)
val df = sqlc.read.json(data)
df.select("text", "user.screen_name").write.format("parquet").save("staging.parquet")
val parquetFile = sqlContext.read.parquet("staging.parquet")
parquetFile.registerTempTable("test")
//= 514621627494322176 where user.screen_name='abyschan'
val tab= sqlContext.sql("select * from test LIMIT 1")
df.printSchema()
tab.collect().foreach{println}
}
}
注:SELECT * FROM测试工作正常,但当我尝试选择user.screen_name(嵌套属性)我GETT出现“无法解析user.screen_name”的错误
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