I have some data with the following format:
28/04/2017 00:00:00|20550|22/05/2017 00:00:00|
I am setting | as a separator. For the data in the first and third row in this example, I put timestamp as a data type in HUE, but I get nulls as answer.
I have looked around and it seems to me like Hive supports Unix format timestamp (from places like this )
But, I get a "Invalid date" message (I'm doing it with HUE so far, I'm still new with this technologies). If I try to make this data String, I can see values, but when I try to use to_date() the problem persists.
Is there anything I'm ignoring?
The only supported timestamp format is yyyy-MM-dd HH:mm:ss with optional fraction of seconds.
Anything else should be read as string and converted later.
Demo
with t as (select '28/04/2017 00:00:00' as mydate)
select from_unixtime(to_unix_timestamp (mydate,'dd/MM/yyyy HH:mm:ss'))
from t
2017-04-28 00:00:00
Related
I'm using the following the query to get the date out of string (20191101154559) in Hadoop(hive).
select max(cast(to_date(from_unixtime(unix_timestamp(substr(column3,1,8), 'yyyymmdd'))) as date)) as dt from databasea.table_name_b;
The output i'm getting after running the above script is '31/01/2019' where as the table holds dates of nov 2019. Not sure where i'm going wrong here.
Appreciate if anyone can look into this issue.
from_unixtime you can use it to get in 'yyyymmdd' or 'yyyy-MM-dd'
whatever format you want data in.
For Eg - select from_unixtime(unix_timestamp(substr('20191101154559',1,8), 'yyyymmdd'),'yyyymmdd');
This will return 20191101
I am at a loss as how to insert the current time in a different format than the default. Can somebody help explain?
Here is how my table was created:
CREATE TABLE ACTIVITY_LOG
(
TIME TIMESTAMP NOT NULL
, ACTIVITY VARCHAR2(200) NOT NULL
);
My insert command works:
insert into activity_log
values (localtimestamp,'blah');
But how do i insert the localtimestamp value into my table in a different format using the various MM DD YY HH MM SS tags? I've tried the following, but it gives me the ORA-1830: date format picture ends before converting entire input string error.
insert into activity_log
values (to_timestamp(localtimestamp,'YYYY/MM/DD'),'blah');
You don't insert a timestamp in a particular format. Timestamps (and dates) are stored in the database using an internal representation, which is betwen 7 and 11 bytes depending on the type and precision. There is more about that in this question, among others.
Your client or application decides how to display the value in a human-readable string form.
When you do:
to_timestamp(localtimestamp,'YYYY/MM/DD')
you are implicitly converting the localtimestamp to a string, using your session's NLS settings, and then converting it back to a timestamp. That may incidentally change the value - losing precision - but won't change how the value is stored internally. In your case the mismatch between the NLS setting and the format you are supplying is leading to an ORA-01830 error.
So your first insert is correct (assuming you really want the session time, not the server time). If you want to see the stored values in a particular format then either change your client session's NLS settings, or preferably format it explicitly when you query it, e.g.:
select to_char(time, 'YYYY-MM-DD HH24:MI:SS.FF3') from activity_log
You don't seem to provide any indication of what your 'localtimestamp' is - is that pseudocode? A variable name? A column you haven't shown the definition for?
What data type is 'localtimestamp'? What data does it contain? Pertinent questions as other answers point out, because if it truly is a time stamp then oracle will be converting it to a string for you, before passing that string to to_timestamp() in your final query. Your initial stab at it should just work if the variable is a timestamp, containing a timestamp
Ultimately "date format picture ends" means "you passed me a string looking like '2017-05-17 12:45:59', but claimed it was only 'yyyy-mm-dd'. What was I expected to do with the rest of it?"
Your current final comment on your question "I was hoping to look in the table and see a useful looking time" - that's your query tool's problem. Have a look in the setting of your query tool and change the date format it displays. As has been noted, dates in oracle are stored as a decimal number days since a certain moment in time. If 0 represents 01 Jan 1970, then 1.75 represents 6pm on the 2 Jan 1970. It is up to the end program the user is using, to format the date into something you like.. you cannot "insert a timestamp with a different format" because time stamps don't have a format any more than a number like 1.75 has a format. It is what your query does with it when it gets it out, that gives it the format:
To_char(timestampcol, 'yyyy mm did')
To-char(tomestampcol, 'mon dd yyyy')
These use oracles built in date formatter, that turns that decimal number of the date into a string in the given format; you will see a string.. or you can just write "select * from table" and run it in TOAD and toad will show you the dates according to the format in settings, or you can write a c# program and get a load of date objects out and call my date.ToString("yyyy-MM-dd") on them to format them. The idea I'm trying to get across is that you don't pick the date format on the way in, you pick it on the way out, if you don't like what you're looking at, you have to change it on the way out, not the way in
I'm using Pentaho to datamask some of the information on the oracle DB
I have several transformations of the form:
SELECT -> data mask -> UPDATE rows based on primary key
I have tables where a timestamp is part of the primary key in the update step. Even though I am not masking or updating this field in any way, I get the error ORA-01843: not a valid month when performing the update.
I believe this is because when Pentaho takes in the timestamp from Step 1 it doesn’t actually keep it as a timestamp until I try the update and hence the primary key check. Outputting to excel, I see pentaho giving timestamps in the format
2014-07-30 15:44:31.869033 Europe/London (Pentaho)
But in DB the format is
30-JAN-15 09.21.38.109145000 AM (Oracle - TIMESTAMP(6) WITH LOCAL TIME ZONE)
I have tried to convert the pentaho field to a Timestamp (format: yyyy-MM-dd HH:mm:ss.SSSSSS) before the update step but receive errors if I try and use milliseconds.
2017/03/14 13:19:25 - Select values.0 - AUDIT_CREATE_TS Timestamp : couldn't convert string [2015-01-30 09:21:38.109145 Europe/London] to a timestamp, expecting format [yyyy-mm-dd hh:mm:ss.ffffff]
2017/03/14 13:19:25 - Select values.0 - Timestamp format must be yyyy-mm-dd hh:mm:ss[.fffffffff]
If I replace my formatting to the one suggested by Pentaho I get "Illegal character 'f'" and then I am stuck in a loop.
Ignoring milliseconds seems to succeed but won’t give me any matches because it isn’t precise enough and returns no results from db..
Any help would be appreciated!
Not sure about Pentaho, but if you're looking for a conversion from this string:
'2015-01-30 09:21:38.109145 Europe/London'
to a timestamp with timezone in Oracle, it would be:
select to_timestamp_tz('2015-01-30 09:21:38.109145 Europe/London', 'YYYY-MM-DD HH24:MI:SS.FF6 TZR') from dual;
See Oracle Datetime Format Models document for more.
I want to convert date '11/2/2014' into timestamp using Hive.
This date is in string format.
I tried many ways but I am getting NULL values.
This is the syntax:
CAST(FROM_UNIXTIME(UNIX_TIMESTAMP(<date_column>,'MM/dd/yyyy')) as TIMESTAMP)
You always need to go to unix_timestamp and then cast from unix timestamp to normal timestamp. It doesn't look great, but it works.
Just to check, it might be that your syntax is right but your format is wrong. Perhaps your format is dd/MM/yyyy? That is the reason why you should give us your attempts to get it work next time.
Alternative using only String functions:
SELECT CAST(
CONCAT(
REGEX_REPLACE(
REGEX_REPLACE( dirty_date
, '^([0-9]*)/([0-9]*)/([0-9]*)$','$3-0$1-0$2')
, '-0([0-9][0-9])','-$1')
, ' 00:00:00')
AS TIMESTAMP)
FROM dirty_input
I have the following date which is in varchar2(11) column in database:
select valid_untill from SALES_ORDERS_V where header_id = 7999410;
30-May-2016
Using rtf template and xml source, the report output (PDF) is:
4950-11-19 04:45:49:0
I don't know its equal to "30-May-2016".
Why this is showing this, as I did not do any formating in rtf?
Not familiar with either RTF or XML-Publisher, but whenever you retrieve a date saved in string format, IF you use it as a date in your code and not as a string, you must make sure you retrieve it correctly.
In this case, with your select statement: it shouldn't be select valid-until from... (or is it really misspelled, with two l at the end: valid_until?) If it is meant to be used as a date, it should be
select to_date(valid_until, 'dd-Mon-yyyy') from ...
Really the problem here is that the date is stored as a string and not in the date datatype. Good luck!