I have column named datenum which is int type. For ex. 20220208
Need to convert datenum column type to date with YYYY/MM/DD format.
Tried to_date("datenum", 'YYYY/MM/DD') but isn't working.
Dates in Redshift (or any SQL database) do not really have any internal format. To convert your input text dates to bona fide Redshift dates, use TO_DATE with the appropriate date mask:
SELECT TO_DATE(datenum, 'YYYYMMDD')
FROM yourTable;
If you really want to view your input dates in some other text format, then do a full roundtrip back to string, using TO_CHAR:
SELECT TO_CHAR(TO_DATE(datenum, 'YYYYMMDD'), 'YYYY/MM/DD')
FROM yourTable;
Related
I would like to replace a date timestamp variable 'date' with the format yyyy-mm-dd hh:mm:ss e.g, 2021-12-28 00:00:00
with two other string variables; one named date with the format: 'yyyymmdd' e.g, 20211228, and one named month with the format: 'yyyymm' e.g, 202112.
Can you give me some suggestions using a SELECT statement?
Thanks
Just use TO_CHAR (or whatever the equivalent is for your DBMS) and the appropriate format string
so I'm trying to learn how to deal with this.
the key partition column is "date" as a string (YYYYMMDD).
so its a character string, not a numeric date format.
I'm having trouble understanding how I can WHERE filter off this by normal date ranges like
where date > CURRENT_DATE -30
or
WHERE date BETWEEN '20210512' AND '20210517'
How does this work? Should the date column be converted first?
This is SQL in a HIVE database.
The best way is to convert the comparison date to a string. Many databases support to_char():
where date > to_char(current_date - 30, 'YYYYMMDD')
Those that do not would have some other function to convert the date to a string.
Note that this formulation allows the use of indexes and table partitions to optimize the query.
I'm looking to convert (using Oracle) a date timestamp
[2018-01-25T00:00:00.000+00:00] to a date [2018-01-24]. I've tried several formats however I can't seem to find the right one to convert it. I'm unsure of how to handle the +00:00.
Thanks in advance
It depends on what you really ask.
It you have a real Oracle timestamp and you want a string in format 'YYYY-MM-DD', you can use to_char():
select to_char(col, 'YYYY-MM-DD') as res from mytable
If you have a string in ISO timestamp format, and you want a string as a result:
select substr(col, 1, 10) as res from mytable
If you have a timestamp column and you want to set the time portion to 00:00:00:
select trunc(col) as res from mytable;
This returns a value of datatype date.
I have that field in my table:
2020-01-16T10:55:16..296518000
How to convert this Varchar into a date in format 'YYYY-MM-DD' ?
I tried:
select TRUNC(to_date(DATE_UPDATED ,'YYYY-MM-DD hh24:mi:ss')) from JOB_SCHEDULE_TBL
but I'm getting an error:
ORA-01830: date format picture ends before converting entire input string
Just use substr():
select to_date(SUBSTR(DATE_UPDATED, 1, 10) ,'YYYY-MM-DD')
The trunc() is unnecessary.
You are confusing the format of DATE with what you are seeing on the screen. A DATE data type has no "format". It's Oracle's internal, binary format. Even when you SELECT TO_DATE ..., the result is going to get displayed by the client, and to do so the client will have to peform (under the covers) a TO_CHAR on the resulting DATE. And that implied to_char will use the current system/session/ settings for TNS_DATE_FORMAT.
I have a requirement to convert MM/DD/YYYY to YYYYMMDD in amazon redshift database.
My result of this query gives me some weird result. Can some one please help me.
select to_date ('07/17/2017','YYYYMMDD');
0007-07-20
If you just wish to convert the hard-coded string into a DATE:
select to_date('07/17/2017', 'MM/DD/YYYY')
If you have a column already formatted as DATE, then use:
to_char(fieldname, 'YYYYMMDD')
Combining the two concepts:
select to_char(to_date('07/17/2017', 'MM/DD/YYYY'), 'YYYYMMDD')
TO_DATE - converts a date represented in a character string to a DATE data type.
TO_CHAR - converts a time stamp or numeric expression to a character-string data format.
select to_char(sysdate,'YYYYMMDD');
If I’ve made a bad assumption please comment and I’ll refocus my answer.