failure:
``STRING'' expected but identifier timestamp found*
Thanks
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s,
>
> Alessandro
>
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Can you please post how did you overcome this issue.
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as “timestamp” is the keyword of data type in
> Hive/Spark SQL.)
>
> <>
> From: Alessandro Panebianco [mailto:[hidden email]
> ]
> Sent: Monday, November 24, 2014 11:12 AM
> To: Wang, Daoyuan
> Cc: [hidden email]
> Subject: Re: SparkSQL Timestamp query failure
the keyword of data type in
Hive/Spark SQL.)
From: Alessandro Panebianco [mailto:ale.panebia...@me.com]
Sent: Monday, November 24, 2014 11:12 AM
To: Wang, Daoyuan
Cc: u...@spark.incubator.apache.org
Subject: Re: SparkSQL Timestamp query failure
Hey Daoyuan,
following your suggestion I obtain the
4 12:11 AM
> To: u...@spark.incubator.apache.org
> Subject: Re: SparkSQL Timestamp query failure
>
> Thanks for your answer Akhil,
>
> I have already tried that and the query actually doesn't fail but it doesn't
> return anything either as it should.
> Using single quo
Hi,
I think you can try
cast(l.timestamp as string)='2012-10-08 16:10:36.0'
Thanks,
Daoyuan
-Original Message-
From: whitebread [mailto:ale.panebia...@me.com]
Sent: Sunday, November 23, 2014 12:11 AM
To: u...@spark.incubator.apache.org
Subject: Re: SparkSQL Timestamp que
chance?
Thanks,
Alessandro
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-
a:902)
> at org.apache.spark.repl.SparkILoop.process(SparkILoop.scala:997)
> at org.apache.spark.repl.Main$.main(Main.scala:31)
> at org.apache.spark.repl.Main.main(Main.scala)
> at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
> a
ache.spark.deploy.SparkSubmit$.launch(SparkSubmit.scala:328)
at org.apache.spark.deploy.SparkSubmit$.main(SparkSubmit.scala:75)
at org.apache.spark.deploy.SparkSubmit.main(SparkSubmit.scala)
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