Just use column number in your group by.
select datediff(day,now()) from test_table where day>='2018-06-01' group by 1
-Original Message-
From: skyyws [mailto:sky...@163.com]
Sent: Tuesday, June 05, 2018 9:49 AM
To: dev
Subject: Re: A question about AnalysisException
Here is the corret
Hello all,
Recently, I found a probelm when I used impala to do ad-hoc analysis. When I
executed the sql below:
select datediff(day,now()) from test_table where day>=(now() - interval 5 days)
group by datediff(day,now());
I got an exception like this:
Things are starting to look healthier now.
I went through the broken-build JIRAs and downgraded some of the infrequent
infrastructure issues to critical so we have a clearer idea of what's
actually breaking the build now versus what's an occasional infra issue:
I develop a UDA function as follow:
input:string
Output:string
I want to caculate every name' md5 code and then caculate the md5 code with
XOR. I will get one value which is all md5 code caculate.
md5 code is saved as unsigned char [16], as I know StringVal has a member
called ptr(uint8_t *)
Hi 周胜为,
I notice you are replying to other threads about different subjects when
you ask your questions. I think you will be more likely to get help if you
start new threads with relevant subjects and if you be as specific as
possible with your questions.
The Impala wiki has some advice for
We're happy to give you pointers. If you could share your uda code and
"create function" that would help us help you
On Tue., 5 Jun. 2018, 19:31 Jim Apple, wrote:
> Hi 周胜为,
>
> I notice you are replying to other threads about different subjects when
> you ask your questions. I think you will be
Thanks for your reply, and I knew that both column number and alias worked like
this:
--
select datediff(day,now()) from test_table where day>='2018-06-01' group by 1
select datediff(day,now()) d from test_table where day>='2018-06-01' group by d
--
I just wonder why built-in function in
