在SparkSQL中缓存表中的错误 [英] Error in Caching a Table in SparkSQL
问题描述
我试图在Hive中缓存一个Table(使用 spark-shell
)。以下是我的代码
I am trying to cache a Table available in Hive(using spark-shell
). Given below is my code
scala> val hiveContext = new org.apache.spark.sql.hive.HiveContext(sc)
scala> hiveContext.cacheTable("sparkdb.firsttable")
我得到下面的异常
org.apache.spark.sql.catalyst.analysis.NoSuchTableException
at org.apache.spark.sql.hive.client.ClientInterface$$anonfun$getTable$1.apply(ClientInterface.scala:112)
表 firsttable
可用于数据库 sparkdb
(在Hive中)。看起来问题似乎是提供数据库名称。如何实现这一点?
The table firsttable
is available in database sparkdb
(in Hive). Looks like the issue seems to be in providing database name. How do I achieve this?
PS:像下面显示的那样的HiveQL查询没有任何问题的工作
PS : HiveQL query like the one shown below does work without any issues
阶> hiveContext.sql(select * from sparkdb.firsttable)
从其他几个方法调用中找到下面的结果
Find below results from few other method calls
scala> hiveContext.tables("sparkdb")
res14: org.apache.spark.sql.DataFrame = [tableName: string, isTemporary: boolean]
scala> hiveContext.tables("sparkdb.firsttable")
res15: org.apache.spark.sql.DataFrame = [tableName: string, isTemporary: boolean]
推荐答案
我是对的,这似乎是 SPARK-8105 。所以,现在,你最好的选择是做 select *
并缓存它。
Aha! I was right, this seems to be SPARK-8105. So, for now, your best bet is to do the select *
and cache that.
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