removing milliseconds from a oracle tmstmp field

removing milliseconds from a oracle tmstmp field - sql

I have a TIMESTAMP(6) field in Oracle and I need to remove the millisecond component from the time.
For example I have
10/20/2014 10:34:06.356000 AM
and I would like to remove the milliseconds so that I have
10/20/2014 10:34:06 AM
Do you know the best way to do this?
Thank you!

How about this?
select cast(col as timestamp(0))
EDIT:
The easiest way to avoid rounding is to use trunc() or to subtract half a second:
select cast(col - 0.5/(24*60*60) as timestamp(0))

try this
SELECT TO_CHAR(SYSDATE, 'MM-DD-YYYY HH24:MI:SS') "NOW"
FROM DUAL;
if you need 12-hour date format
SELECT TO_CHAR(SYSDATE, 'MM-DD-YYYY HH:MI:SS AM') "NOW"
FROM DUAL;
SQL FIDDLE

You can either cast it to a timestamp with no fractional seconds (this will round to the nearest second):
CAST( your_timestamp AS TIMESTAMP(0) )
Or to a DATE data type (this will truncate to the nearest second):
CAST( your_timestamp AS DATE )
If you want it as a TIMESTAMP(0) data type then cast it back:
CAST( CAST( your_timestamp AS DATE ) AS TIMESTAMP(0) )
Or you can convert it to a formatted string and specify the format model you want to use (this will truncate to the nearest second):
TO_CHAR( your_timestamp, 'YYYY-MM-DD HH24:MI:SS' )
Like this:
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE your_table ( your_timestamp ) AS
SELECT TIMESTAMP '2017-10-25 12:53:12.10076' FROM DUAL;
Query 1:
SELECT CAST( your_timestamp AS TIMESTAMP(0) ) AS "Timestamp",
CAST( your_timestamp AS DATE ) AS "Date",
TO_CHAR( your_timestamp, 'DD-MM-YYYY HH24:MI:SS' ) AS "String"
FROM your_table
Results:
| Timestamp | Date | String |
|-----------------------|----------------------|---------------------|
| 2017-10-25 12:53:12.0 | 2017-10-25T12:53:12Z | 25-10-2017 12:53:12 |
note: How the TIMESTAMP and DATE are formatted in the output will depend on your NLS_TIMESTAMP_FORMAT and NLS_DATE_FORMAT session parameters but you can directly control the formatting of TO_CHAR when you specify a format model.

This might help!
select substr(to_char('10/20/2014 10:34:06.356000 AM'),1,instr(to_char('10/20/2014 10:34:06.356000 AM'),'.')-1)||' '||
substr(to_char('10/20/2014 10:34:06.356000 AM'),-2,instr(to_char('10/20/2014 10:34:06.356000 AM'),'.')-1) "Date"
from dual;

Related

Convert Oracle Date into RFC 3339 Format

I have a date field in oracle which returns
17-APR-19 12:00:00 AM
I also have a time column (VARCHAR) which returns HHMM in Military
1810
I'd like to combine these two fields to create a timestamp that is formatted to RFC 3339 standards. Preferable like this.
2019-04-17T18:10:00Z
I can convert a timestamp into the correct time using this:
SELECT
TO_CHAR(
SYSTIMESTAMP AT TIME ZONE 'UTC',
'yyyy-mm-dd"T"hh24:mi:ss"Z"'
)
FROM dual;
Is there a way to convert my date and time field into this timestamp format? The time on the date field is incorrect and needs to be replaced by the time field.

You can TRUNCate your date back to midnight and then use NUMTODSINTERVAL to add hours and minutes to it to get the correct time component:
Oracle Setup:
CREATE TABLE your_table ( your_date_column, your_time_column ) AS
SELECT DATE '2019-04-17', '1810' FROM DUAL
Query:
SELECT TO_CHAR(
TRUNC( your_date_column )
+ NUMTODSINTERVAL( SUBSTR( your_time_column, 1, 2 ), 'HOUR' )
+ NUMTODSINTERVAL( SUBSTR( your_time_column, 3, 2 ), 'MINUTE' ),
'YYYY-MM-DD"T"HH24:MI:SS"Z"'
) AS combined_date_time
FROM your_table
Output:
| COMBINED_DATE_TIME |
| :------------------- |
| 2019-04-17T18:10:00Z |
db<>fiddle here
If you want the value as a TIMESTAMP WITH TIME ZONE then:
SELECT CAST(
TRUNC( your_date_column )
+ NUMTODSINTERVAL( SUBSTR( your_time_column, 1, 2 ), 'HOUR' )
+ NUMTODSINTERVAL( SUBSTR( your_time_column, 3, 2 ), 'MINUTE' )
AS TIMESTAMP
) AT TIME ZONE 'UTC' AS combined_date_time
FROM your_table

Just do a bit of string concatenation
to_char( your_date, 'yyyy-mm-dd' ) ||
'T' ||
substr( your_time, 1, 2 ) ||
':' ||
substr( your_time, 3, 2 ) ||
':00Z'
assuming that your_time is always 4 characters long (i.e. 2 AM is represented as the string '0200' rather than '200'). This also assumes that the seconds will always be '00'.

You can achieve this by converting your_number into minutes and add it to your date, then cast it to timestamp as following:
SELECT CAST(
your_date +
(FLOOR(YOUR_TIME/100)*60 + MOD(YOUR_TIME,100)) / 1440
AS TIMESTAMP
) AT TIME ZONE 'UTC' AS YOUR_TIME_STAMP
FROM your_table;
Cheers!!

Inject character to date column

I have a date column.
select RETAIL_ACQUISITION_DTTM from XXXXX.TABLE_2348
13/07/2018
I want to select the value as such it returns as below,
2018-07-13T00:00:00
so it has to display the date as YYYY-MM-DD with T and then HH24:MI:SS

You need to convert your string (why is it stored as a string?) to a date using a format model that matches the actual string value:
to_date(RETAIL_ACQUISITION_DTTM, 'DD/MM/YYYY')
Having that format model wrong is why you get the 0013 year in your result.
Then you can convert that date back to a string with to_char(), and you can embed the fixed T as a character literal with double quotes, using a format model like 'YYYY-MM-DD"T"HH24:MI:SS':
with TABLE_2348 (RETAIL_ACQUISITION_DTTM) as (
select '13/07/2018' from dual
)
select to_char(to_date(RETAIL_ACQUISITION_DTTM, 'DD/MM/YYYY'),
'YYYY-MM-DD"T"HH24:MI:SS') as RETAIL_ACQUISITION_DTTM
from XXXXX.TABLE_2348;
RETAIL_ACQUISITION_
-------------------
2018-07-13T00:00:00
You could also just ignore that it is a date and use string manipulation:
with TABLE_2348 (RETAIL_ACQUISITION_DTTM) as (
select '13/07/2018' from dual
)
select substr(RETAIL_ACQUISITION_DTTM, 7, 4)
|| '-' || substr(RETAIL_ACQUISITION_DTTM, 4, 2)
|| '-' ||substr(RETAIL_ACQUISITION_DTTM, 1, 2)
|| 'T00:00:00' as RETAIL_ACQUISITION_DTTM
from XXXXX.TABLE_2348;
RETAIL_ACQUISITION_
-------------------
2018-07-13T00:00:00
If the column is actually a date rather than a string then you are doing unnecessary conversions, including implicit ones which rely on your NLS settings, and you are losing the original time from the value if it was not midnight anyway:
alter session set nls_date_format = 'DD-MON-RR';
with TABLE_2348 (RETAIL_ACQUISITION_DTTM) as (
select to_date('2018-07-13 12:34:56', 'YYYY-MM-DD HH24:MI:SS') from dual
)
select to_char(to_date(RETAIL_ACQUISITION_DTTM, 'YYYY-MM-DD-HH24:MI:SS'),
'YYYY-MM-DD HH24:MI:SS') as RETAIL_ACQUISITION_DTTM from XXXXX.TABLE_2348;
RETAIL_ACQUISITION_
-------------------
0013-07-20 18:00:00
That is really doing:
to_char(
to_date(
to_char(
RETAIL_ACQUISITION_DTTM,
'DD-MON-RR'), ---- from your session NLS_DATE_FORMAT setting
'YYYY-MM-DD-HH24:MI:SS'),
'YYYY-MM-DD HH24:MI:SS')
If you skip to extra steps you can just format the date directly:
with TABLE_2348 (RETAIL_ACQUISITION_DTTM) as (
select to_date('2018-07-13 12:34:56', 'YYYY-MM-DD HH24:MI:SS') from dual
)
select to_char(RETAIL_ACQUISITION_DTTM,
'YYYY-MM-DD"T"HH24:MI:SS') as RETAIL_ACQUISITION_DTTM
from XXXXX.TABLE_2348;
RETAIL_ACQUISITION_
-------------------
2018-07-13T12:34:56
which also doesn't rely on your NLS settings, so won't break in interesting ways if it's run in another session with different settings.

replace sysdate with your column name if its date type.
Using Single to_char:
select to_char(sysdate,'--YYYY-MM-DD"T"hh24:mi:ss') from dual;
Using two to_char and concatenation for simplification.
select to_char(sysdate,'--'||'YYYY-MM-DD')||'T'||to_char(sysdate,'hh24:mi:ss') from dual; -- 13/07/2018
select '--'||to_char(sysdate,'YYYY-MM-DD')||'T'||to_char(sysdate,'hh24:mi:ss') from dual; -- 13/07/2018

Oracle how to convert time in UTC to the local time (offset information is missing)

I have a table which contains a date column. I think that the date in that column is saved in UTC.
I would like when the date is retrieved that it is printed in local time. That means when I call the date from Germany, the result should be something like this:
2015-04-29 11:24:06 +0200UTC EUROPE/BERLIN
I tried the following sql:
SELECT TO_CHAR(CAST(dateColum as TIMESTAMP WITH LOCAL TIME ZONE), 'YYYY-MM-DD HH24:MI:SS TZR') from myTable;
the result looks like this:
2015-04-29 11:24:06 EUROPE/BERLIN
+/- offset is missing.
Any idea?

Oracle Setup:
CREATE TABLE table_name ( value ) AS
SELECT DATE '2016-07-13' FROM DUAL;
Query:
SELECT TO_CHAR(
FROM_TZ( CAST( value AS TIMESTAMP ), 'UTC' )
AT TIME ZONE 'EUROPE/BERLIN',
'YYYY-MM-DD HH24:MI:SS TZH:TZM TZR'
) AS berlin_time
FROM table_name;
Output:
BERLIN_TIME
----------------------------------------
2016-07-13 02:00:00 +02:00 EUROPE/BERLIN
Query 2:
SELECT TO_CHAR(
FROM_TZ( CAST( value AS TIMESTAMP ), 'UTC' ) AT LOCAL,
'YYYY-MM-DD HH24:MI:SS TZH:TZM TZR'
) AS local_time
FROM table_name;
Output:
LOCAL_TIME
----------------------------------------
2016-07-13 02:00:00 +02:00 EUROPE/BERLIN

SQL filter query output based on difference of two timestamp field [duplicate]

I have a table as follows:
Filename - varchar
Creation Date - Date format dd/mm/yyyy hh24:mi:ss
Oldest cdr date - Date format dd/mm/yyyy hh24:mi:ss
How can I calcuate the difference in hours minutes and seconds (and possibly days) between the two dates in Oracle SQL?
Thanks

You can substract dates in Oracle. This will give you the difference in days. Multiply by 24 to get hours, and so on.
SQL> select oldest - creation from my_table;
If your date is stored as character data, you have to convert it to a date type first.
SQL> select 24 * (to_date('2009-07-07 22:00', 'YYYY-MM-DD hh24:mi')
- to_date('2009-07-07 19:30', 'YYYY-MM-DD hh24:mi')) diff_hours
from dual;
DIFF_HOURS
----------
2.5
Note:
This answer applies to dates represented by the Oracle data type DATE.
Oracle also has a data type TIMESTAMP, which can also represent a date (with time). If you subtract TIMESTAMP values, you get an INTERVAL; to extract numeric values, use the EXTRACT function.

To get result in seconds:
select (END_DT - START_DT)*60*60*24 from MY_TABLE;
Check [https://community.oracle.com/thread/2145099?tstart=0][1]

select
extract( day from diff ) Days,
extract( hour from diff ) Hours,
extract( minute from diff ) Minutes
from (
select (CAST(creationdate as timestamp) - CAST(oldcreationdate as timestamp)) diff
from [TableName]
);
This will give you three columns as Days, Hours and Minutes.

declare
strTime1 varchar2(50) := '02/08/2013 01:09:42 PM';
strTime2 varchar2(50) := '02/08/2013 11:09:00 PM';
v_date1 date := to_date(strTime1,'DD/MM/YYYY HH:MI:SS PM');
v_date2 date := to_date(strTime2,'DD/MM/YYYY HH:MI:SS PM');
difrence_In_Hours number;
difrence_In_minutes number;
difrence_In_seconds number;
begin
difrence_In_Hours := (v_date2 - v_date1) * 24;
difrence_In_minutes := difrence_In_Hours * 60;
difrence_In_seconds := difrence_In_minutes * 60;
dbms_output.put_line(strTime1);
dbms_output.put_line(strTime2);
dbms_output.put_line('*******');
dbms_output.put_line('difrence_In_Hours : ' || difrence_In_Hours);
dbms_output.put_line('difrence_In_minutes: ' || difrence_In_minutes);
dbms_output.put_line('difrence_In_seconds: ' || difrence_In_seconds);
end ;
Hope this helps.

You may also try this:
select to_char(to_date('1970-01-01 00:00:00', 'yyyy-mm-dd hh24:mi:ss')+(end_date - start_date),'hh24:mi:ss')
as run_time from some_table;
It displays time in more human readable form, like: 00:01:34.
If you need also days you may simply add DD to last formatting string.

Calculate age from HIREDATE to system date of your computer
SELECT HIREDATE||' '||SYSDATE||' ' ||
TRUNC(MONTHS_BETWEEN(SYSDATE,HIREDATE)/12) ||' YEARS '||
TRUNC((MONTHS_BETWEEN(SYSDATE,HIREDATE))-(TRUNC(MONTHS_BETWEEN(SYSDATE,HIREDATE)/12)*12))||
'MONTHS' AS "AGE " FROM EMP;

You could use to_timestamp function to convert the dates to timestamps and perform a substract operation.
Something like:
SELECT
TO_TIMESTAMP ('13.10.1990 00:00:00','DD.MM.YYYY HH24:MI:SS') -
TO_TIMESTAMP ('01.01.1990:00:10:00','DD.MM.YYYY:HH24:MI:SS')
FROM DUAL

In oracle 11g
SELECT end_date - start_date AS day_diff FROM tablexxx
suppose the starT_date end_date is define in the tablexxx

select days||' '|| time from (
SELECT to_number( to_char(to_date('1','J') +
(CLOSED_DATE - CREATED_DATE), 'J') - 1) days,
to_char(to_date('00:00:00','HH24:MI:SS') +
(CLOSED_DATE - CREATED_DATE), 'HH24:MI:SS') time
FROM request where REQUEST_ID=158761088 );

If you want something that looks a bit simpler, try this for finding events in a table which occurred in the past 1 minute:
With this entry you can fiddle with the decimal values till you get the minute value that you want. The value .0007 happens to be 1 minute as far as the sysdate significant digits are concerned. You can use multiples of that to get any other value that you want:
select (sysdate - (sysdate - .0007)) * 1440 from dual;
Result is 1 (minute)
Then it is a simple matter to check for
select * from my_table where (sysdate - transdate) < .00071;

If you select two dates from 'your_table' and want too see the result as a single column output (eg. 'days - hh:mm:ss') you could use something like this.
First you could calculate the interval between these two dates and after that export all the data you need from that interval:
select extract (day from numtodsinterval (second_date
- add_months (created_date,
floor (months_between (second_date,created_date))),
'day'))
|| ' days - '
|| extract (hour from numtodsinterval (second_date
- add_months (created_date,
floor (months_between (second_date,created_date))),
'day'))
|| ':'
|| extract (minute from numtodsinterval (second_date
- add_months (created_date,
floor (months_between (second_date, created_date))),
'day'))
|| ':'
|| extract (second from numtodsinterval (second_date
- add_months (created_date,
floor (months_between (second_date, created_date))),
'day'))
from your_table
And that should give you result like this:
0 days - 1:14:55

select (floor(((DATE2-DATE1)*24*60*60)/3600)|| ' : ' ||floor((((DATE2-DATE1)*24*60*60) -floor(((DATE2-DATE1)*24*60*60)/3600)*3600)/60)|| ' ' ) as time_difference from TABLE1

(TO_DATE(:P_comapre_date_1, 'dd-mm-yyyy hh24:mi') - TO_DATE(:P_comapre_date_2, 'dd-mm-yyyy hh24:mi'))*60*60*24 sum_seconds,
(TO_DATE(:P_comapre_date_1, 'dd-mm-yyyy hh24:mi') - TO_DATE(:P_comapre_date_2, 'dd-mm-yyyy hh24:mi'))*60*24 sum_minutes,
(TO_DATE(:P_comapre_date_1, 'dd-mm-yyyy hh24:mi') - TO_DATE(:P_comapre_date_2, 'dd-mm-yyyy hh24:mi'))*24 sum_hours,
(TO_DATE(:P_comapre_date_1, 'dd-mm-yyyy hh24:mi') - TO_DATE(:P_comapre_date_2, 'dd-mm-yyyy hh24:mi')) sum_days

select to_char(actual_start_date,'DD-MON-YYYY hh24:mi:ss') start_time,
to_char(actual_completion_date,'DD-MON-YYYY hh24:mi:ss') end_time,
floor((actual_completion_date-actual_start_date)*24*60)||'.'||round(mod((actual_completion_date-actual_start_date)*24*60*60,60)) diff_time
from fnd_concurrent_requests
order by request_id desc;

If You want get date defer from using table and column.
SELECT TO_DATE( TO_CHAR(COLUMN_NAME_1, 'YYYY-MM-DD'), 'YYYY-MM-DD') -
TO_DATE(TO_CHAR(COLUMN_NAME_2, 'YYYY-MM-DD') , 'YYYY-MM-DD') AS DATEDIFF
FROM TABLE_NAME;

This will count time between to dates:
SELECT
(TO_CHAR( TRUNC (ROUND(((sysdate+1) - sysdate)*24,2))*60,'999999')
+
TO_CHAR(((((sysdate+1)-sysdate)*24)- TRUNC(ROUND(((sysdate+1) - sysdate)*24,2)))/100*60 *100, '09'))/60
FROM dual

Here's another option:
with tbl_demo AS
(SELECT TO_DATE('11/26/2013 13:18:50', 'MM/DD/YYYY HH24:MI:SS') dt1
, TO_DATE('11/28/2013 21:59:12', 'MM/DD/YYYY HH24:MI:SS') dt2
FROM dual)
SELECT dt1
, dt2
, round(dt2 - dt1,2) diff_days
, round(dt2 - dt1,2)*24 diff_hrs
, numtodsinterval((dt2 - dt1),'day') diff_dd_hh_mm_ss
from tbl_demo;

Single query that will return time difference of two timestamp columns:
select INS_TS, MAIL_SENT_TS, extract( hour from (INS_TS - MAIL_SENT_TS) ) timeDiff
from MAIL_NOTIFICATIONS;

select round( (tbl.Todate - tbl.fromDate) * 24 * 60 * 60 )
from table tbl

for oracle sql I justbn did this and works perfect :
SELECT trunc(date_col_1) - trunc(date_col_2)
FROM TABLE;

$sql="select bsp_bp,user_name,status,
to_char(ins_date,'dd/mm/yyyy hh12:mi:ss AM'),
to_char(pickup_date,'dd/mm/yyyy hh12:mi:ss AM'),
trunc((pickup_date-ins_date)*24*60*60,2),message,status_message
from valid_bsp_req where id >= '$id'";

Subtracting the difference between two time intervals in oracle sql [duplicate]

I have a table as follows:
Filename - varchar
Creation Date - Date format dd/mm/yyyy hh24:mi:ss
Oldest cdr date - Date format dd/mm/yyyy hh24:mi:ss
How can I calcuate the difference in hours minutes and seconds (and possibly days) between the two dates in Oracle SQL?
Thanks

You can substract dates in Oracle. This will give you the difference in days. Multiply by 24 to get hours, and so on.
SQL> select oldest - creation from my_table;
If your date is stored as character data, you have to convert it to a date type first.
SQL> select 24 * (to_date('2009-07-07 22:00', 'YYYY-MM-DD hh24:mi')
- to_date('2009-07-07 19:30', 'YYYY-MM-DD hh24:mi')) diff_hours
from dual;
DIFF_HOURS
----------
2.5
Note:
This answer applies to dates represented by the Oracle data type DATE.
Oracle also has a data type TIMESTAMP, which can also represent a date (with time). If you subtract TIMESTAMP values, you get an INTERVAL; to extract numeric values, use the EXTRACT function.

To get result in seconds:
select (END_DT - START_DT)*60*60*24 from MY_TABLE;
Check [https://community.oracle.com/thread/2145099?tstart=0][1]

select
extract( day from diff ) Days,
extract( hour from diff ) Hours,
extract( minute from diff ) Minutes
from (
select (CAST(creationdate as timestamp) - CAST(oldcreationdate as timestamp)) diff
from [TableName]
);
This will give you three columns as Days, Hours and Minutes.

declare
strTime1 varchar2(50) := '02/08/2013 01:09:42 PM';
strTime2 varchar2(50) := '02/08/2013 11:09:00 PM';
v_date1 date := to_date(strTime1,'DD/MM/YYYY HH:MI:SS PM');
v_date2 date := to_date(strTime2,'DD/MM/YYYY HH:MI:SS PM');
difrence_In_Hours number;
difrence_In_minutes number;
difrence_In_seconds number;
begin
difrence_In_Hours := (v_date2 - v_date1) * 24;
difrence_In_minutes := difrence_In_Hours * 60;
difrence_In_seconds := difrence_In_minutes * 60;
dbms_output.put_line(strTime1);
dbms_output.put_line(strTime2);
dbms_output.put_line('*******');
dbms_output.put_line('difrence_In_Hours : ' || difrence_In_Hours);
dbms_output.put_line('difrence_In_minutes: ' || difrence_In_minutes);
dbms_output.put_line('difrence_In_seconds: ' || difrence_In_seconds);
end ;
Hope this helps.

You may also try this:
select to_char(to_date('1970-01-01 00:00:00', 'yyyy-mm-dd hh24:mi:ss')+(end_date - start_date),'hh24:mi:ss')
as run_time from some_table;
It displays time in more human readable form, like: 00:01:34.
If you need also days you may simply add DD to last formatting string.

Calculate age from HIREDATE to system date of your computer
SELECT HIREDATE||' '||SYSDATE||' ' ||
TRUNC(MONTHS_BETWEEN(SYSDATE,HIREDATE)/12) ||' YEARS '||
TRUNC((MONTHS_BETWEEN(SYSDATE,HIREDATE))-(TRUNC(MONTHS_BETWEEN(SYSDATE,HIREDATE)/12)*12))||
'MONTHS' AS "AGE " FROM EMP;

You could use to_timestamp function to convert the dates to timestamps and perform a substract operation.
Something like:
SELECT
TO_TIMESTAMP ('13.10.1990 00:00:00','DD.MM.YYYY HH24:MI:SS') -
TO_TIMESTAMP ('01.01.1990:00:10:00','DD.MM.YYYY:HH24:MI:SS')
FROM DUAL

In oracle 11g
SELECT end_date - start_date AS day_diff FROM tablexxx
suppose the starT_date end_date is define in the tablexxx

select days||' '|| time from (
SELECT to_number( to_char(to_date('1','J') +
(CLOSED_DATE - CREATED_DATE), 'J') - 1) days,
to_char(to_date('00:00:00','HH24:MI:SS') +
(CLOSED_DATE - CREATED_DATE), 'HH24:MI:SS') time
FROM request where REQUEST_ID=158761088 );

If you want something that looks a bit simpler, try this for finding events in a table which occurred in the past 1 minute:
With this entry you can fiddle with the decimal values till you get the minute value that you want. The value .0007 happens to be 1 minute as far as the sysdate significant digits are concerned. You can use multiples of that to get any other value that you want:
select (sysdate - (sysdate - .0007)) * 1440 from dual;
Result is 1 (minute)
Then it is a simple matter to check for
select * from my_table where (sysdate - transdate) < .00071;

If you select two dates from 'your_table' and want too see the result as a single column output (eg. 'days - hh:mm:ss') you could use something like this.
First you could calculate the interval between these two dates and after that export all the data you need from that interval:
select extract (day from numtodsinterval (second_date
- add_months (created_date,
floor (months_between (second_date,created_date))),
'day'))
|| ' days - '
|| extract (hour from numtodsinterval (second_date
- add_months (created_date,
floor (months_between (second_date,created_date))),
'day'))
|| ':'
|| extract (minute from numtodsinterval (second_date
- add_months (created_date,
floor (months_between (second_date, created_date))),
'day'))
|| ':'
|| extract (second from numtodsinterval (second_date
- add_months (created_date,
floor (months_between (second_date, created_date))),
'day'))
from your_table
And that should give you result like this:
0 days - 1:14:55

select (floor(((DATE2-DATE1)*24*60*60)/3600)|| ' : ' ||floor((((DATE2-DATE1)*24*60*60) -floor(((DATE2-DATE1)*24*60*60)/3600)*3600)/60)|| ' ' ) as time_difference from TABLE1

(TO_DATE(:P_comapre_date_1, 'dd-mm-yyyy hh24:mi') - TO_DATE(:P_comapre_date_2, 'dd-mm-yyyy hh24:mi'))*60*60*24 sum_seconds,
(TO_DATE(:P_comapre_date_1, 'dd-mm-yyyy hh24:mi') - TO_DATE(:P_comapre_date_2, 'dd-mm-yyyy hh24:mi'))*60*24 sum_minutes,
(TO_DATE(:P_comapre_date_1, 'dd-mm-yyyy hh24:mi') - TO_DATE(:P_comapre_date_2, 'dd-mm-yyyy hh24:mi'))*24 sum_hours,
(TO_DATE(:P_comapre_date_1, 'dd-mm-yyyy hh24:mi') - TO_DATE(:P_comapre_date_2, 'dd-mm-yyyy hh24:mi')) sum_days

select to_char(actual_start_date,'DD-MON-YYYY hh24:mi:ss') start_time,
to_char(actual_completion_date,'DD-MON-YYYY hh24:mi:ss') end_time,
floor((actual_completion_date-actual_start_date)*24*60)||'.'||round(mod((actual_completion_date-actual_start_date)*24*60*60,60)) diff_time
from fnd_concurrent_requests
order by request_id desc;

If You want get date defer from using table and column.
SELECT TO_DATE( TO_CHAR(COLUMN_NAME_1, 'YYYY-MM-DD'), 'YYYY-MM-DD') -
TO_DATE(TO_CHAR(COLUMN_NAME_2, 'YYYY-MM-DD') , 'YYYY-MM-DD') AS DATEDIFF
FROM TABLE_NAME;

This will count time between to dates:
SELECT
(TO_CHAR( TRUNC (ROUND(((sysdate+1) - sysdate)*24,2))*60,'999999')
+
TO_CHAR(((((sysdate+1)-sysdate)*24)- TRUNC(ROUND(((sysdate+1) - sysdate)*24,2)))/100*60 *100, '09'))/60
FROM dual

Here's another option:
with tbl_demo AS
(SELECT TO_DATE('11/26/2013 13:18:50', 'MM/DD/YYYY HH24:MI:SS') dt1
, TO_DATE('11/28/2013 21:59:12', 'MM/DD/YYYY HH24:MI:SS') dt2
FROM dual)
SELECT dt1
, dt2
, round(dt2 - dt1,2) diff_days
, round(dt2 - dt1,2)*24 diff_hrs
, numtodsinterval((dt2 - dt1),'day') diff_dd_hh_mm_ss
from tbl_demo;

Single query that will return time difference of two timestamp columns:
select INS_TS, MAIL_SENT_TS, extract( hour from (INS_TS - MAIL_SENT_TS) ) timeDiff
from MAIL_NOTIFICATIONS;

select round( (tbl.Todate - tbl.fromDate) * 24 * 60 * 60 )
from table tbl

for oracle sql I justbn did this and works perfect :
SELECT trunc(date_col_1) - trunc(date_col_2)
FROM TABLE;

$sql="select bsp_bp,user_name,status,
to_char(ins_date,'dd/mm/yyyy hh12:mi:ss AM'),
to_char(pickup_date,'dd/mm/yyyy hh12:mi:ss AM'),
trunc((pickup_date-ins_date)*24*60*60,2),message,status_message
from valid_bsp_req where id >= '$id'";

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removing milliseconds from a oracle tmstmp field - sql

I have a TIMESTAMP(6) field in Oracle and I need to remove the millisecond component from the time. For example I have 10/20/2014 10:34:06.356000 AM and I would like to remove the milliseconds so that I have 10/20/2014 10:34:06 AM Do you know the best way to do this? Thank you!

How about this? select cast(col as timestamp(0)) EDIT: The easiest way to avoid rounding is to use trunc() or to subtract half a second: select cast(col - 0.5/(246060) as timestamp(0))

try this SELECT TO_CHAR(SYSDATE, 'MM-DD-YYYY HH24:MI:SS') "NOW" FROM DUAL; if you need 12-hour date format SELECT TO_CHAR(SYSDATE, 'MM-DD-YYYY HH:MI:SS AM') "NOW" FROM DUAL; SQL FIDDLE

This might help! select substr(to_char('10/20/2014 10:34:06.356000 AM'),1,instr(to_char('10/20/2014 10:34:06.356000 AM'),'.')-1)||' '|| substr(to_char('10/20/2014 10:34:06.356000 AM'),-2,instr(to_char('10/20/2014 10:34:06.356000 AM'),'.')-1) "Date" from dual;

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sql objective-c vba vb.net react-native apache vue.js tensorflow api pandas

removing milliseconds from a oracle tmstmp field - sql

I have a TIMESTAMP(6) field in Oracle and I need to remove the millisecond component from the time. For example I have 10/20/2014 10:34:06.356000 AM and I would like to remove the milliseconds so that I have 10/20/2014 10:34:06 AM Do you know the best way to do this? Thank you!

How about this? select cast(col as timestamp(0)) EDIT: The easiest way to avoid rounding is to use trunc() or to subtract half a second: select cast(col - 0.5/(24*60*60) as timestamp(0))

try this SELECT TO_CHAR(SYSDATE, 'MM-DD-YYYY HH24:MI:SS') "NOW" FROM DUAL; if you need 12-hour date format SELECT TO_CHAR(SYSDATE, 'MM-DD-YYYY HH:MI:SS AM') "NOW" FROM DUAL; SQL FIDDLE

This might help! select substr(to_char('10/20/2014 10:34:06.356000 AM'),1,instr(to_char('10/20/2014 10:34:06.356000 AM'),'.')-1)||' '|| substr(to_char('10/20/2014 10:34:06.356000 AM'),-2,instr(to_char('10/20/2014 10:34:06.356000 AM'),'.')-1) "Date" from dual;

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Convert Oracle Date into RFC 3339 Format

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SQL filter query output based on difference of two timestamp field [duplicate]

Subtracting the difference between two time intervals in oracle sql [duplicate]

Categories

Resources

How about this? select cast(col as timestamp(0)) EDIT: The easiest way to avoid rounding is to use trunc() or to subtract half a second: select cast(col - 0.5/(246060) as timestamp(0))