i want to display in output as : today is wednesday
my code:
select to_char(sysdate,'today is' 'day' ) from dual ;
it is not working.
You need to include the raw text you want in the output in double quotes in the format specification:
select to_char(sysdate, '"today is" day') from dual ;
Output (as of 2020-03-04)
today is wednesday
Demo on dbfiddle
Use concat() or the concatenation operator || to prepend the 'today is ' string.
SELECT concat('today is ', to_char(sysdate, 'day'))
FROM dual;
or
SELECT 'today is ' || to_char(sysdate, 'day')
FROM dual;
Related
When I do...
Select TO_CHAR (date_field, 'Month DD, YYYY')
from...
I get the following:
July 01, 2011
April 01, 2011
January 01, 2011
Why are there extra spaces between my month and day? Why doesn't it just put them next to each other?
Why are there extra spaces between my month and day? Why does't it just put them next to each other?
So your output will be aligned.
If you don't want padding use the format modifier FM:
SELECT TO_CHAR (date_field, 'fmMonth DD, YYYY')
FROM ...;
Reference: Format Model Modifiers
if you use 'Month' in to_char it right pads to 9 characters; you have to use the abbreviated 'MON', or to_char then trim and concatenate it to avoid this. See, http://www.techonthenet.com/oracle/functions/to_char.php
select trim(to_char(date_field, 'month')) || ' ' || to_char(date_field,'dd, yyyy')
from ...
or
select to_char(date_field,'mon dd, yyyy')
from ...
You should use fm element to delete blank spaces.
SELECT TO_CHAR(sysdate, 'fmDAY DD "de" MONTH "de" YYYY') CURRENT_DATE
FROM dual;
SQL> -- original . . .
SQL> select
2 to_char( sysdate, 'Day "the" Ddth "of" Month, yyyy' ) dt
3 from dual;
DT
----------------------------------------
Friday the 13th of May , 2016
SQL>
SQL> -- collapse repeated spaces . . .
SQL> select
2 regexp_replace(
3 to_char( sysdate, 'Day "the" Ddth "of" Month, yyyy' ),
4 ' * *', ' ') datesp
5 from dual;
DATESP
----------------------------------------
Friday the 13th of May , 2016
SQL>
SQL> -- and space before commma . . .
SQL> select
2 regexp_replace(
3 to_char( sysdate, 'Day "the" Ddth "of" Month, yyyy' ),
4 ' *(,*) *', '\1 ') datesp
5 from dual;
DATESP
----------------------------------------
Friday the 13th of May, 2016
SQL>
SQL> -- space before punctuation . . .
SQL> select
2 regexp_replace(
3 to_char( sysdate, 'Day "the" Ddth "of" Month, yyyy' ),
4 ' *([.,/:;]*) *', '\1 ') datesp
5 from dual;
DATESP
----------------------------------------
Friday the 13th of May, 2016
try this:-
select to_char(to_date('01/10/2017','dd/mm/yyyy'),'fmMonth fmDD,YYYY') from dual;
select to_char(sysdate,'fmMonth fmDD,YYYY') from dual;
select to_char(sysdate, 'DD-fmMONTH-YYYY') "Date" from Dual;
The above query result will be as given below.
Date
01-APRIL-2019
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
I would like to add 10 minutes to sysdate,
select to_char(SYSDATE,'dd-Mon-yyyy hh:mi:ss') + 10/1440 from dual
when I tried the above I got the error
ORA-01722: invalid number
The error appears because you add 10/1440 to char not to date.
Try this:
select SYSDATE + 10/1440 from dual;
or
select to_char(SYSDATE+ 10/1440,'dd-Mon-yyyy hh:mi:ss') from dual;
or
select to_char( sysdate + interval '10' minute,'dd-Mon-yyyy hh:mi:ss')
from dual;
Here you can find more information.
Here you can find similar problem on SO, there are more solutions.
You don't need to "to_char" the date first.
Either
select SYSDATE + 10/1440 from dual;
or
select to_char(SYSDATE + 10/1440,'dd-Mon-yyyy hh:mi:ss') from dual;
depending on whether you just want a date or a string representation of the date formatted in a certain way.
When I do...
Select TO_CHAR (date_field, 'Month DD, YYYY')
from...
I get the following:
July 01, 2011
April 01, 2011
January 01, 2011
Why are there extra spaces between my month and day? Why doesn't it just put them next to each other?
Why are there extra spaces between my month and day? Why does't it just put them next to each other?
So your output will be aligned.
If you don't want padding use the format modifier FM:
SELECT TO_CHAR (date_field, 'fmMonth DD, YYYY')
FROM ...;
Reference: Format Model Modifiers
if you use 'Month' in to_char it right pads to 9 characters; you have to use the abbreviated 'MON', or to_char then trim and concatenate it to avoid this. See, http://www.techonthenet.com/oracle/functions/to_char.php
select trim(to_char(date_field, 'month')) || ' ' || to_char(date_field,'dd, yyyy')
from ...
or
select to_char(date_field,'mon dd, yyyy')
from ...
You should use fm element to delete blank spaces.
SELECT TO_CHAR(sysdate, 'fmDAY DD "de" MONTH "de" YYYY') CURRENT_DATE
FROM dual;
SQL> -- original . . .
SQL> select
2 to_char( sysdate, 'Day "the" Ddth "of" Month, yyyy' ) dt
3 from dual;
DT
----------------------------------------
Friday the 13th of May , 2016
SQL>
SQL> -- collapse repeated spaces . . .
SQL> select
2 regexp_replace(
3 to_char( sysdate, 'Day "the" Ddth "of" Month, yyyy' ),
4 ' * *', ' ') datesp
5 from dual;
DATESP
----------------------------------------
Friday the 13th of May , 2016
SQL>
SQL> -- and space before commma . . .
SQL> select
2 regexp_replace(
3 to_char( sysdate, 'Day "the" Ddth "of" Month, yyyy' ),
4 ' *(,*) *', '\1 ') datesp
5 from dual;
DATESP
----------------------------------------
Friday the 13th of May, 2016
SQL>
SQL> -- space before punctuation . . .
SQL> select
2 regexp_replace(
3 to_char( sysdate, 'Day "the" Ddth "of" Month, yyyy' ),
4 ' *([.,/:;]*) *', '\1 ') datesp
5 from dual;
DATESP
----------------------------------------
Friday the 13th of May, 2016
try this:-
select to_char(to_date('01/10/2017','dd/mm/yyyy'),'fmMonth fmDD,YYYY') from dual;
select to_char(sysdate,'fmMonth fmDD,YYYY') from dual;
select to_char(sysdate, 'DD-fmMONTH-YYYY') "Date" from Dual;
The above query result will be as given below.
Date
01-APRIL-2019
I'd like to format a number as "1st", "2nd", "4th", "9th", etc. Is there an Oracle function that will do this for me?
Assuming the value supplied is numeric, rather than DATE, you can use TO_CHAR but you have to convert the numeric value to a string, then a DATE (Julian) before ultimately formatting it:
SELECT TO_CHAR(TO_DATE('1', 'dd'), 'ddth')
FROM DUAL
Result:
01st
When testing, using 'd' for the format didn't return expected results because the value is interpreted as a Julian date. Either substring the output to remove the leading zero, or provide a full date string (doesn't matter to the TO_CHAR because it's only interested in the day of the month):
SELECT TO_CHAR(TO_DATE('1900-01-01', 'YYYY-MM-dd'), 'dth')
FROM DUAL
Because calendar days end at 31, use the year value instead to handle numbers greater than 31:
SELECT TO_CHAR(TO_DATE('32-01-01', 'YYYY-MM-dd'), 'yyth')
FROM DUAL
Result:
32nd
Maybe I'm oversimplifying, but it seems like the following should work just fine (for integers) and is a lot more readable than converting to a date and back:
select case
when initial_extent is null then null
when substr(initial_extent,-2,1) = '1'
then initial_extent || 'th'
else case substr(initial_extent,-1,1)
when '1' then initial_extent || 'st'
when '2' then initial_extent || 'nd'
when '3' then initial_extent || 'rd'
else initial_extent || 'th'
end
end as formatted_number
from user_tables
select substr( to_char( to_date( abs( decode( mod( l_value, 10 ), 0, 4, mod( l_value , 10 ) ) ), 'YYYY' ), 'YTH' ), 2 ) as value
from dual
Replace l_value with appropriate, hmmm, value. Should cover any numbers.