I have an Oracle DB with this data (extract - there are more columns - totally)
The first is the name, the second the department.
Then there is at CJDate the JDE Julian date - Format.
And cStart and cEnds inludes the timestamp.
I would like to get the Date in the Format "DD.MM.YYYY" and the duration between cStart and cEnd in "D.HH:MM:SS".
But there are some exotics. If there are values in cEnd between 0 and 62000, the cJDate should be -1.
Is it possible to get the correct duration, if cEnd has the value 240000?Or sould it be transformed in 23:59:59 or in cJDate -1 and 00:00:00?
I use already this, to get the correct date:
DATE '1900-01-01' + FLOOR(ZZ."cJDate" / 1000) * INTERVAL'1' YEAR+
(MOD(ZZ."cJDate", 1000) -1) * INTERVAL'1' DAY AS "Date",
Does someone have any idea?
Best regards
Until now I've been using this:
select to_date(to_char(cjDate), 'YYYDDD')
from dual;
Is not julian date it's more like JDE Julian date
Example output:
118316 → 12/11/18
Related
how can I convert the unix date into a specified date mentioned below in snowflake
unix date = 96227 expected output = 1996-08-14
I have tried using below query in snowflake
select cast(to_timestamp(1900000+96227) as date) as "effDate"
but both the above query and the unix converter online gives me the below result--
how can i construct a query in snowflake that gives me 1996-08-14, what number should I add while converting it ?
Adding to answer from Gordon, and maybe simpler to manage use the following:
Set UNIX_DATE='96227';
Select DATEADD(day,RIGHT($UNIX_DATE,3)-1,TRY_TO_DATE(LEFT($UNIX_DATE,2),'YY'));
Note: Two-digit year, controlled by the TWO_DIGIT_CENTURY_START session parameter, default is 1970, which matches Unix epoch.
Your data format appears to be in the format YYDDD, where YY is the year minus 1900 and DDD is the day of the year.
So:
select date_from_parts(1900 + d / 1000, 1, 1) + (d % 1000 - 1) * interval '1 day'
from (values (96227)) v(d);
I'm querying a table to get some date, like so:
SELECT date - INTERVAL '10 day' AS date
FROM example_table
WHERE username = 'Bob'
LIMIT 1;
The date column in the example_table does not have a timestamp. All dates in the column are stored in the following manner:
YYYY-MM-DD
The query above will return a result like so:
2016-11-20 00:00:00.000000
It takes the date found, goes back 10 days, and returns that date. But I want it to return the date without adding the timestamp, like so:
2016-11-20
If I use INTERVAL it always seems to add a timestamp. Is there a way to only get the date?
Your query is fine (but can be simplified, as demonstrated by a_horse_with_no_name). What you are seeing is a display issue. You can format your date to a string in the relevant format using to_char():
SELECT to_char("date" - INTERVAL '10 day', 'yyyy-mm-dd') AS "date"
FROM example_table
WHERE username = 'Bob'
LIMIT 1;
Note: LIMIT without an ORDER BY does not make sense: if there is more than one record in the resultset, you actually get a random record out of them.
You can use the interval notation and convert back to a date:
SELECT (date - INTERVAL '10 day')::date AS date
You can subtract (or add) an integer from a date. That integer represents the number of days:
SELECT "date" - 10 AS "date"
FROM example_table
WHERE username = 'Bob'
LIMIT 1;
Unfortunately, this is my first approach with SQL!
I am creating with the following code a query between an oracle DB and Excel (Power Query).
select "$Table"."Order" as "Order",
"$Table"."NR" as "Nr",
"$Table"."JDDATE" as "JDDATE"
from "POOLDB2"."3112" "$Table"
WHERE "Key" >118001
AND "CodeAA" = 1
This code works!
Now I want to format the Julian Date (CYYDDD) - for example 118001 for the 01.01.2019 - to a normal date format.
Does anyone know, how to implement this into the code above?
Maybe something like :
select "$Table"."Order" as "Order",
"$Table"."NR" as "Nr",
DATEADD(DAY, JDDATE % 1000 - 1, DATEADD(year, JDDATE/1000, 0))
"$Table"."JDDATE" as "JDDATE"
from "POOLDB2"."3112" "$Table"
WHERE "Key" >118001
AND "CodeAA" = 1
Best regards
There are many different formats for Julian Date... In your use case, this should do it :
with t as (select 118001 jd from dual)
select to_char( to_date(to_char(1901 + floor(jd / 1000)),'YYYY') + mod(jd,1000) - 1, 'dd.mm.yyyy' ) from t
Yields : 01.01.2019
For Oracle,
select to_char(sysdate,'J') from dual; --To Julian Date
select to_date(2456143,'J') from dual; --To Normal Date
must work.
Edit: Sorry I didn't see oracle tag.
Edit: For the requested behavior by OP
select to_date(to_char(1901 + floor(118001 / 1000)),'YYYY') from dual;
You can use the 118001 value you have, split into separate year and day sections, by adding to the nominal starting date 1900-01-01 (based on your comment that 118001 is actually 2018-01-01, not 2019-01-01):
select date '1900-01-01'
+ floor(118001 / 1000) * interval '1' year
+ (mod(118001, 1000) - 1) * interval '1' day
from dual;
DATE'1900-
----------
2018-01-01
or by startng the fixed date a day earlier you can remove the explicit -1:
select date '1899-12-31'
+ floor(118019 / 1000) * interval '1' year
+ mod(118019, 1000) * interval '1' day
from dual;
DATE'1899-
----------
2018-01-19
This avoids having to build up a longer string to convert to a date, though you could do that (modifying #GMB's approach) as:
select to_date(to_char(1900 + floor(118001 / 1000)) || '-01-01', 'YYYY-MM-DD')
+ (mod(118001, 1000) - 1)
from dual;
You need to specify the month, at least, in the to_date() call as Oracle defaults to the current month if that is not supplied. That behaviour is tucked away in the documentation:
If you specify a date value without a time component, then the default time is midnight. If you specify a date value without a date, then the default date is the first day of the current month.
The first part of that is fairly well known and makes sense ; the second part is a bit less obvious, and doesn't make it clear that it applies to partial dates too - so ifyou don't supply a year then the current year is used; if you don't supply a month then the current month is used; but if you don't supply a day then the 1st is used.
You can see what it's doing with some test conversions:
select to_date('2018-12-25', 'YYYY-MM-DD') as demo_a,
to_date('12:34:56', 'HH24:MI:SS') as demo_b,
to_date('2019', 'YYYY') as demo_c,
to_date('07-04', 'MM-DD') as demo_d,
to_date('2019-01', 'YYYY-MM') as demo_e
from dual;
DEMO_A DEMO_B DEMO_C DEMO_D DEMO_E
------------------- ------------------- ------------------- ------------------- -------------------
2018-12-25 00:00:00 2018-12-01 12:34:56 2019-12-01 00:00:00 2018-07-04 00:00:00 2019-01-01 00:00:00
I have a unix timstamp with millsecond precision like below:
1523572200000
I need to convert it to timestamp(6). This is the format I need:
05-NOV-14 09.45.00.000000000 AM
(Fyi examples above are not matching dates, just using as example.)
What's the best way to go about this?
Thanks!
The following might work for you (where myunixtimestamp is the name of the column in which your Unix timestamps are stored):
SELECT TIMESTAMP'1970-01-01 00:00:00.000' + NUMTODSINTERVAL(myunixtimestamp/1000, 'SECOND')
FROM mytable;
For example,
SELECT TIMESTAMP'1970-01-01 00:00:00.000' + NUMTODSINTERVAL(1523572200000/1000, 'SECOND')
FROM dual;
gives a result of 2018-04-12 10:30:00.000000000 PM.
Hope this helps.
Assuming that current timestamp is: 1523572200000, try following:
select cast (to_date('1970-01-01', 'YYYY-MM-DD') + 1523572200000/1000/60/60/24 as timestamp) from dual;
where:
to_date('1970-01-01', 'YYYY-MM-DD') is epoch time
<unix_timestamp>/60/60/24 was divided by 1000 miliseconds 60 second and 60 minutes and 24 hours because in oracle we are adding days
By using normal minus '-' function between two timestamps, the answer given from oracle is incorrect.
This is what i want to do:
ALTER SESSION SET NLS_TIMESTAMP_TZ_FORMAT='DD-MON-RR HH24:MI TZR';
Created table:
CREATE TABLE TEST (
StartTime timestamp with time zone
,EndTime timestamp with time zone
,Science varchar2(7)
);
I create the column data type as timestamp with time zone. This is value I have inserted:
INSERT INTO TEST
VALUES('05-OCT-2013 01:00 +08:00'
,'05-OCT-2013 23:00 +06:00'
,'SCIENCE');
INSERT INTO TEST
VALUES('05-OCT-2013 12:00 +08:00'
,'05-OCT-2013 15:00 -12:00'
,'Maths');
Attempted for rounding time:
CREATE VIEW TESTRECRDS AS
SELECT (Extract(hour FROM(ENDTIME- STARTTIME)) || 'Hours' ||
Extract(minute FROM(ENDTIME- STARTTIME))>=60 Then (Extract(hour FROM(ENDTIME- STARTTIME)) + Extract(minute FROM(ENDTIME- STARTTIME))/60 ELSE 0 END || 'Minutes' AS DURATION,
Science
FROM Test;
Now i have two questions regarding on the calculation and rounding off the minutes to nearest hours.
First let's say the endtime is 1535 +0600 and starttime is 01:50 +0800
So when i deduct endtime - starttime:
the formula should be:
2135 - 0950 = 2085 - 0950
= 1135
But if i use my successful attempt answer to calculate, it is not the correct exact answer. The oracle answer would be 15 hours 45 minutes.
In your last CREATE VIEW statement you try to multiply text, which cannot work:
SELECT To_Char(STARTTIME - ENDTIME, 'HH24:MI TZR')*24 AS DURATION
*24 is operating on the text to_char() returns.
You have to multiply the interval before converting to text.
You define the column Science varchar2(6), then you insert 'SCIENCE', a 7-letter word?
I also fixed a syntax error in your INSERT statement: missing '.
About your comment:
"I would like to insert timestamp with timezone during creation of my tables. Can DATE data type do that too?
Read about data types in the manual.
The data type date does not include time zone information.
If by "timezone difference" you mean the difference between the timezone modifiers, use this to calculate:
SELECT EXTRACT(timezone_hour FROM STARTTIME) AS tz_modifier FROM tbl
Keywords here are timezone_hour and is timezone_minute. Read more in the manual.
But be aware that these numbers depend on the daylight saving hours and such shenanigans. Very uncertain territory!
Get it in pretty format - example:
SELECT to_char((EXTRACT (timezone_hour FROM STARTTIME) * 60
+ EXTRACT (timezone_minutes FROM STARTTIME))
* interval '1 min', 'HH:MI')
In PostgreSQL you would have the simpler EXTRACT (timezone FROM STARTTIME), but I don't think Oracle supports that. Can't test now.
Here is a simple demo how you could round minutes to hours:
SELECT EXTRACT(hour FROM (ENDTIME - STARTTIME))
+ CASE WHEN EXTRACT(minute FROM (ENDTIME - STARTTIME)) >= 30 THEN 1 ELSE 0 END
FROM Test;
I'm not sure what number you're trying to calculate, but when you subtract two dates in Oracle, you get the difference between the dates in units of days, not a DATE datatype
SELECT TO_DATE('2011-01-01 09:00', 'yyyy-mm-dd hh24:mi') -
TO_DATE('2011-01-01 08:00', 'yyyy-mm-dd hh24:mi') AS diff
FROM dual
DIFF
----------
.041666667
In this case 8am and 9am are 0.41667 days apart. This is not a date object, this is a scalar number, so formatting it as HH24:MI doesn't make any sense.
To round you will need to do a bit of more math. Try something like:
TO_DATE(ROUND((ENDTIME - STARTTIME) * 96) / 96, 'HH24:MI')
The difference between dates is in days. Multiplying by 96 changes the measure to quarter hours. Round, then convert back to days, and format. It might be better to use a numeric format want to format, in which case you would divide by 4 instead of 96.
Timezone is not particularly relevant to a time difference. You will have to adjust the difference from UTC to that timezone to get the right result with Timezone included.