Trying to query an oracle db table having date in format: 2022-06-22T12:25:06.087 (LocalDateTime.now().toString()). Column type for created_time is varchar2.
Trying to query for data between two dates. I have tried the following but it results in error "date format not recognized":
select * from MY_TABLE
where to_date(created_time, 'yyyy-MM-ddTHH:mm:ss.SSS')
between to_date('2022-07-03T10:15:06.091', 'yyyy-MM-ddTHH:mm:ss.SSS')
and to_date('2022-07-03T10:15:06.091', 'yyyy-MM-ddTHH:mm:ss.SSS');
Can anyone help me correct this query?
Format you used looks like data (in CREATED_TIME column) is stored as a timestamp. If that's so, you shouldn't convert it to another datatype (you chose TO_DATE function) but leave it as is. If you stored data as a string (that's usually a huge mistake), then apply the same to_timestamp function with the same format model as the one in between clause.
Apart from that, format model for minutes is mi (not mm; that's month), while fractional seconds is ff3 (not sss).
SELECT *
FROM my_table
WHERE created_time
BETWEEN TO_TIMESTAMP ('2022-07-03T10:15:06.091', 'yyyy-MM-dd"T"HH24:mi:ss.ff3')
AND TO_TIMESTAMP ('2022-07-03T10:15:06.091', 'yyyy-MM-dd"T"HH24:mi:ss.ff3');
I guess column created_time is of data type DATE or TIMESTAMP. Never call TO_DATE() or TO_TIMESTAMP() to a values which is already a DATE
The DATE data type does not support fractional seconds, use TIMESTAMP instead. Format literals have to be enclosed by double quotes.
Format HH is the hour in 12-hour format. I assume you need 24-hour format, which is HH24. mm (or MM) is the Month, for Minute use MI. Format identifiers are not case-sensitive, so SSS is also wrong.
Try this one:
select *
from MY_TABLE
where created_time
between TO_TIMESTAMP('2022-07-03T10:15:06.091', 'yyyy-MM-dd"T"HH24:MI:ss.ff3')
and TO_TIMESTAMP('2022-07-03T10:15:06.091', 'yyyy-MM-dd"T"HH24:MI:ss.ff3');
Related
this is my table:
table name EXAMPLE
column name DATE
this is the output for the following query:
SELECT date
FROM example;
1/23/2010
I want to convert the output to:
23-Jan-10
Is it possible?
Thanks
You can use formatting along with TO_CHAR() conversion such as
SELECT TO_CHAR(dt,'DD-Mon-RR')
FROM example
where
DATE is not a good name for a column as being a reserved keyword. So, I've replaced it with dt
If your column is in varchar/text.. then you can first convert to date then back to char
SELECT
TO_CHAR(
to_date('1/23/2010','mm/dd/yyyy'),'DD-Mon-YY'
)
datec FROM dual;
If it is in date format, to_char only will do
SELECT
TO_CHAR(date_column, 'DD-Mon-YY' )
datec FROM dual;
A DATE data type is a binary format that is stored in 7-bytes that has no format and always contains the components: year (stored as century and year-of-century), month, day, hour, minute and second.
Therefore, you cannot change the format of a DATE data type.
If instead, you ask the question:
How can format a DATE to output it as 23-Jan-10?
Then you can convert the DATE to a formatted string using the TO_CHAR function:
SELECT TO_CHAR(your_date_column, 'DD-MON-YY', 'NLS_DATE_LANGUAGE=English')
AS formatted_date
FROM your_table;
Alternatively, if you want the output as a DATE data type then you can use:
SELECT your_date_column
FROM your_table;
and change the user interface (SQL/Plus, SQL Developer, Toad, PLSQLDeveloper, PHP, Java, etc.) you are using to alter how that that user interface displays dates. The solution is going to depend on which user interface you are using but for SQL/Plus and SQL Developer, you can change Oracle's NLS_DATE_FORMAT session parameter:
ALTER SESSION SET NLS_DATE_FORMAT = 'DD-MON-YY';
Other user interfaces will have different solutions specific to those interfaces.
Attempting to convert a variable character data type field that is time related (eg '2015-Q1') to a timestamp (or time) data type field in order to compare with a different field.
Need to be able to use dateadd() on the field.
Use the TO_DATE function: https://docs.aws.amazon.com/redshift/latest/dg/r_TO_DATE_function.html
https://docs.aws.amazon.com/redshift/latest/dg/r_FORMAT_strings.html
Along with TO_TIMESTAMP
https://docs.aws.amazon.com/redshift/latest/dg/r_TO_TIMESTAMP.html
example:
select to_timestamp(to_date('2015-Q1', 'YYYY-MM-DD'), 'HH24:MI:SS');
In my DB, there is a date field in the format yyyymmdd.
I have to get all the dates in the format dd-mm-yyyy for that particlar date.
ex:
Date
20170130
20170228
20170325
for the above dates, I need the output in the below format with the dates and day of the particular dates
date day
30-01-2017 tuesday
28-02-2017 tuesday
25-03-2017 saturday
If the column is a string, then it can hold invalid date values such as February 31, one way to avoid this is by a small function such as this:
create or replace
function my_to_date( p_str in varchar2 ) return date
is
begin
return to_date( p_str );
exception
when others then
return null;
end;
\\
select to_char(my_to_date('20170231'),'DD-MM-YYYY Day')
from dual
\\
Demo
Try below:
Select to_char(yrdate, 'dd-mm-yyyy'), to_char(yrdate, 'D') from yrtable
It sounds like your dates aren't actually DATE fields but some kind of CHAR field? The best option would be to convert to DATE and then convert back to CHAR:
SELECT TO_CHAR(TO_DATE(mydate, 'YYYYMMDD'), 'DD-MM-YYYY Day')
FROM mytable;
This uses the YYYYMMDD mask to convert your string into a date, then uses the mask DD-MM-YYYY Day to convert it back into a string. Use day if you want the day name in lowercase (as in your OP).
#user2778168 answer will give you the results you want. But why?
Your database does not have dates stored in yyyymmdd format or any other date format for at mater, unless it's defined with a character type definition. Oracle stores all dates in a single internal structure, and with only slight variations timestamps are the same. The format used only tells Oracle how to display the value or to convert a string to a date. Unless a specific format is specified Oracle uses the NLS_DATE_FORMAT for this determination. See here and scan down to "Datetime Format Models" for format specifications.
To see this run the following:
select value
from nls_session_parameters
where parameter = 'NLS_DATE_FORMAT';
Select yrdate default_format
, to_char(yrdate, 'dd-mm-yyyy') specified_format
, dump(yrdate) unformated
from yrtable;
alter session set nls_date_format = 'Month dd,yyyy';
Rerun the above queries.
It seems you hold date column(date1) in character format. Suppose you have a table named days:
SQL> desc days
date1 varchar2(10)
then,
we should convert it into date, and then into char format, with aliases in quotation marks to provide lowercase aliases as you wanted.
perhaps your database's date language is non-english like mine(turkish), then you need to convert it to english.
lastly, it'a appropriate to order the results according to converted date, seen from your output. So we can use the following SQL :
select to_char(to_date(date1,'yyyymmdd'),'dd-mm-yyyy') "date",
to_char(to_date(date1,'yyyymmdd'),'day','nls_date_language=english') "day"
from days
order by to_date(date1,'yyyymmdd');
D e m o
My table in hive has a filed of date in the format of '2016/06/01'. but i find that it is not in harmory with the format of '2016-06-01'.
They can not compare for instance.
Both of them are string .
So I want to know how to make them in harmory and can compare them. Or on the other hand, how to change the '2016/06/01' to '2016-06-01' so that them can compare.
Many thanks.
To convert date string from one format to another you have to use two date function of hive
unix_timestamp(string date, string pattern) convert time string
with given pattern to unix time stamp (in seconds), return 0 if
fail.
from_unixtime(bigint unixtime[, string format]) converts the
number of seconds from unix epoch (1970-01-01 00:00:00 UTC) to a
string representing the timestamp of that moment in the current
system time zone.
Using above two function you can achieve your desired result.
The sample input and output can be seen from below image:
The final query is
select from_unixtime(unix_timestamp('2016/06/01','yyyy/MM/dd'),'yyyy-MM-dd') from table1;
where table1 is the table name present in my hive database.
I hope this help you!!!
Let's say you have a column 'birth_day' in your table which is in your format,
you should use the following query to convert birth_day into the required format.
date_Format(birth_day, 'yyyy-MM-dd')
You can use it in a query in the following way
select * from yourtable
where
date_Format(birth_day, 'yyyy-MM-dd') = '2019-04-16';
Use :
unix_timestamp(DATE_COLUMN, string pattern)
The above command would help convert the date to unix timestamp format which you may format as you want using the Simple Date Function.
Date Function
cast(to_date(from_unixtime(unix_timestamp(yourdate , 'MM-dd-yyyy'))) as date)
here is my solution (for string to real Date type):
select to_date(replace('2000/01/01', '/', '-')) as dt ;
ps:to_date() returns Date type, this feature needs Hive 2.1+; before 2.1, it returns String.
ps2: hive to_date() function or date_format() function , or even cast() function, cannot regonise the 'yyyy/MM/dd' or 'yyyymmdd' format, which I think is so sad, and make me a little crazy.
When ececute the following SQL syntax in Oracle, always not success, please help.
40284.3878935185 represents '2010-04-16 09:18:34', with microsecond.
an epoch date of 01 January 1900 (like Excel).
create table temp1 (date1 number2(5,10));
insert into temp1(date1) values('40284.3878935185');
select to_date(date1, 'yyyy-mm-dd hh24:mi:ssxff') from temp1
Error report: SQL Error: ORA-01861: literal does not match format
string
01861. 00000 - "literal does not match format string"
*Cause: Literals in the input must be the same length as literals in
the format string (with the exception of leading whitespace). If the
"FX" modifier has been toggled on, the literal must match exactly,
with no extra whitespace.
*Action: Correct the format string to match the literal.
Thanks to Mark Bannister
Now the SQL syntax is:
select to_char(to_date('1899-12-30','yyyy-mm-dd') +
date1,'yyyy-mm-dd hh24:mi:ss') from temp1
but can't fetch the date format like 'yyyy-mm-dd hh24:mi:ss.ff'. Continue look for help.
Using an epoch date of 30 December 1899, try:
select to_date('1899-12-30','yyyy-mm-dd') + date1
Simple date addition doesn't work with timestamps, at least if you need to preserve the fractional seconds. When you do to_timestamp('1899-12-30','yyyy-mm-dd')+ date1 (in a comment on Mark's answer) the TIMESTAMP is implicitly converted to a DATE before the addition, to the overall answer is a DATE, and so doesn't have any fractional seconds; then you use to_char(..., '... .FF') it complains with ORA-01821.
You need to convert the number of days held by your date1 column into an interval. Fortunately Oracle provides a function to do exactly that, NUMTODSINTERVAL:
select to_timestamp('1899-12-30','YYYY-MM-DD')
+ numtodsinterval(date1, 'DAY') from temp3;
16-APR-10 09.18.33.999998400
You can then display that in your desired format, e.g. (using a CTE to provide your date1 value):
with temp3 as ( select 40284.3878935185 as date1 from dual)
select to_char(to_timestamp('1899-12-30','YYYY-MM-DD')
+ numtodsinterval(date1, 'DAY'), 'YYYY-MM-DD HH24:MI:SSXFF') from temp3;
2010-04-16 09:18:33.999998400
Or to restrict to thousandths of a second:
with temp3 as ( select 40284.3878935185 as date1 from dual)
select to_char(to_timestamp('1899-12-30','YYYY-MM-DD')+
+ numtodsinterval(date1, 'DAY'), 'YYYY-MM-DD HH24:MI:SS.FF3') from temp3;
2010-04-16 09:18:33.999
An epoch of 1899-12-30 sounds odd though, and doesn't correspond to Excel as you stated. It seems more likely that your expected result is wrong and it should be 2010-04-18, so I'd check your assumptions. Andrew also makes some good points, and you should be storing your value in the table in a TIMESTAMP column. If you receive data like this though, you still need something along these lines to convert it for storage at some point.
Don't know the epoch date exactly, but try something like:
select to_date('19700101','YYYYMMDD')+ :secs_since_epoch/86400 from dual;
Or, cast to timestamp like:
select cast(to_date('19700101', 'YYYYMMDD') + :secs_since_epoch/86400 as timestamp with local time zone) from dual;
I hope this doesn't come across too harshly, but you've got to totally rethink your approach here.
You're not keeping data types straight at all. Each line of your example misuses a data type.
TEMP1.DATE1 is not a date or a varchar2, but a NUMBER
you insert not the number 40284.3878935185, but the STRING >> '40284.3878935185' <<
your SELECT TO_DATE(...) uses the NUMBER Temp1.Date1 value, but treats it as a VARCHAR2 using the format block
I'm about 95% certain that you think Oracle transfers this data using simple block data copies. "Since each Oracle date is stored as a number anyway, why not just insert that number into the table?" Well, because when you're defining a column as a NUMBER you're telling Oracle "this is not a date." Oracle therefore does not manage it as a date.
Each of these type conversions is calculated by Oracle based on your current session variables. If you were in France, where the '.' is a thousands separator rather than a radix, the INSERT would completely fail.
All of these conversions with strings are modified by the locale in which Oracle thinks your running. Check dictionary view V$NLS_PARAMETERS.
This gets worse with date/time values. Date/time values can go all over the map - mostly because of time zone. What time zone is your database server in? What time zone does it think you're running from? And if that doesn't spin your head quite enough, check out what happens if you change Oracle's default calendar from Gregorian to Thai Buddha.
I strongly suggest you get rid of the numbers ENTIRELY.
To create date or date time values, use strings with completely invariant and unambiguous formats. Then assign, compare and calculate date values exclusively, e.g.:
GOODFMT constant VARCHAR2 = 'YYYY-MM-DD HH24:MI:SS.FFF ZZZ'
Good_Time DATE = TO_DATE ('2012-02-17 08:07:55.000 EST', GOODFMT);