在Spark SQL中无法基于TimeUUID进行查询 [英] Cannot query based on TimeUUID in Spark SQL

问题描述

我想使用SparkSQL终端查询Cassandra数据库。
查询：

  select * from keyspace.tablename 
其中user_id = e3a119e0-8744-11e5-a557 -e789fe3b4cc1; 
  

错误：java.lang.RuntimeException：[1.88] failure： 'union''expected but identifier e5 found

也尝试过：

  user_id = UUID.fromString（\`e3a119e0-8744-11e5-a557-e789fe3b4cc1\'））
 
 user_id = \'e3a119e0-8744- 
 
 
 
 
 
 
 
 
 
 
 
 $ e $ pre> 
 
 
我不知道如何查询 timeuuid 上的数据。
解决方案
 TimeUUID不支持作为SparkSQL中的类型，所以你只能进行直接字符串比较。以字符串形式表示TIMEUUID 
  select * from keyspace.tablename where user_id =e3a119e0-8744-11e5-a557-e789fe3b4cc1 
  
 
I am trying to query the Cassandra database using SparkSQL terminal.
Query: 
select * from keyspace.tablename 
where user_id = e3a119e0-8744-11e5-a557-e789fe3b4cc1;



  
Error: java.lang.RuntimeException: [1.88] failure: ``union'' expected but identifier e5 found
Also tried: 
user_id= UUID.fromString(\`e3a119e0-8744-11e5-a557-e789fe3b4cc1\`)")

user_id= \'e3a119e0-8744-11e5-a557-e789fe3b4cc1\'")

token(user_id)= token(`e3a119e0-8744-11e5-a557-e789fe3b4cc1`)
I am not sure how can I query data on timeuuid.
解决方案
TimeUUIDs are not supported as a type in SparkSQL so you are only allowed to do direct string comparisons. Represent the TIMEUUID as a string
select * from keyspace.tablename where user_id = "e3a119e0-8744-11e5-a557-e789fe3b4cc1"


                        
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