I have the following values in 1 column in Oracle:
20123103
20113112
20103006
The data type is varchar.
I need to go back 1 year and find the dates in the same format in Oracle.
So, the output should be:
20123103 -> 20110104
20113112 -> 20110101
20103006 -> 20090107
Please advise.
You could convert it to a date, subtract a year, and then format it back to a string:
SELECT TO_CHAR(TO_DATE(date_column, 'YYYYDDMM')
- INTERVAL '1' YEAR
+ INTERVAL '1' DAY,
'YYYYDDMM')
FROM my_table
Related
I have the following query that gets the week of a date:
SELECT pdm.serie, rta.matricula_ant, TO_CHAR (fecha, 'ww') semana,
SUM (rta.kms_acumulados) kms,
COUNT
(DISTINCT (CASE
WHEN v.secuencia BETWEEN rta.sec_origen AND rta.sec_destino
THEN v.cod_inc
ELSE '0'
END
)
)
- 1 numincidencias
FROM (SELECT ms.tren, ms.fecha_origen_tren, ms.secuencia, ri.cod_inc
FROM r_incidencias ri, mer_sitra ms
WHERE ri.cod_serv = ms.tren
AND ri.fecha_origen_tren = ms.fecha_origen_tren
AND ri.cod_tipoin IN (SELECT cod_tipo_iincidencia
FROM v_tipos_incidencias
WHERE grupo = '45')
AND ri.punto_desde = ms.cod_estacion) v,
r_trenes_asignar rta,
r_maquinas rm,
planificador.pl_dh_material pdm
WHERE rta.fecha BETWEEN TO_DATE ('21/09/2018', 'dd/mm/yyyy') AND TO_DATE ('21/09/2018',
'dd/mm/yyyy'
)
AND rta.serie >= 4000
AND rta.matricula_ant IS NOT NULL
AND rm.matricula_maq = rta.matricula_ant
AND rm.cod_serie = pdm.id_material
AND rta.grafico BETWEEN pdm.desde AND pdm.hasta
AND v.tren(+) = rta.tren
AND v.fecha_origen_tren(+) = rta.fecha
GROUP BY pdm.serie, rta.matricula_ant, TO_CHAR (fecha, 'ww')
ORDER BY pdm.serie, rta.matricula_ant, TO_CHAR (fecha, 'ww')
For example week 1
I want to display
week 1 : 1 january - 7 january
How can I get this?
Oracle offers the TRUNC(datestamp, format) function to manipulate dates this way. You may use a variety of format strings to get the first day of a quarter, year, or even the top of the hour.
Given a particular datestamp value, Oracle returns midnight on the first day of the present week with this expression:
TRUNC(datestamp,'DY')
You can add days to a datestamp. Therefore this expression gives you midnight on the last day of the week
TRUNC(datestamp,'DY') + 6
A WHERE-clause selector for all rows in the present week might be this.
WHERE datestamp >= TRUNC(SYSDATE,'DY')
AND datestamp < TRUNC(SYSDATE,'DY') + 7
Notice that the end of the range is just before (<) midnight on the first day of the next week. You need that because you may have datestamps after midnight on the last day of the week. (Beware using BETWEEN for datestamp ranges.)
And,
SELECT TO_CHAR(TRUNC(SYSDATE,'DY'),'YYYY-MM-DD'),
TO_CHAR(TRUNC(SYSDATE,'DY')+6,'YYYY-MM-DD')
FROM DUAL;
displays the first and last dates of the present week in ISO-like format.
Date arithmetic is cool. It's worth your trouble to study the date-arithmetic functions in your DBMS at least once a year.
I am trying to add a day with a specific date using add_months in oracle database.
I wrote this line:
SELECT ADD_MONTHS('01-JAN-2018', MONTHS_BETWEEN('02-JAN-2018', '01-JAN-2018')) FROM DUAL;
this returns:
01-JAN-18
Why doesn't it return 02-JAN-18?? Can I add one day to the date using this function?
Why doesn't it return 02-JAN-18??
According to MONTHS_BETWEEN documentation,
The MONTHS_BETWEEN function calculates the number of months between
two dates. When the two dates have the same day component or are both
the last day of the month, then the return value is a whole number.
Otherwise, the return value includes a fraction that considers the
difference in the days based on a 31-day month
So,
select MONTHS_BETWEEN('02-JAN-2018', '01-JAN-2018') FROM DUAL ;
yields
.0322580645161290322580645161290322580645
ADD_MONTHS returns the date date plus integer months.
So, .0322.. is considered as integer 0 and your query is equivalent to
SELECT ADD_MONTHS('01-JAN-2018', 0) FROM DUAL;
In order to add 1 months, simply take the difference of two dates.
SELECT ADD_MONTHS(DATE '2018-01-01', DATE '2018-01-02' - DATE '2018-01-01') FROM DUAL;
Or better, add an INTERVAL of 1 month
SELECT DATE '2018-01-01' + INTERVAL '1' MONTH FROM DUAL;
To answer your question, add 1 day, simply use
SELECT DATE '2018-01-01' + 1 FROM DUAL;
Am getting the below issue when am using 'mon-d-yyyy' to convert date to char, as i need a single day digit for values from 1 to 9 days in a month.
When i use the 'mon-d-yyyy' format, am losing out on 5 days and getting a wrong date. Any help on this would be great.
select to_char(sysdate-22,'mon-d-yyyy') from dual;--aug-2-2017
select to_char(sysdate-22,'mon-dd-yyyy') from dual;--aug-07-2017
select sysdate-22 from dual;--07-AUG-17 11.06.43
In Oracle date formats, d gets the day of week. The 2 in your output means monday, not august the 2nd.
Try using Fill Mode as Format Model Modifier
select to_char(sysdate-22,'mon-fmdd-yyyy') from dual;
One option might be to piece together the date output you want:
SELECT
TO_CHAR(sysdate-22, 'mon-') ||
TRIM(LEADING '0' FROM TO_CHAR(sysdate-22, 'dd-')) ||
TO_CHAR(sysdate-22, 'yyyy')
FROM dual;
The middle term involving TRIM strips off the leading zeroes, if present, from the date.
Output:
Demo here:
Rextester
SQL>SELECT TO_CHAR(TO_DATE('29-AUG-2017','DD-MON-YYYY') - 22,'"WEEKDAY :"D, MON-FMDD-YYYY') "Before22Days" FROM DUAL;
D- Gives you a numeric weekday(2nd weekday in a week) on AUG-07-2017.
DD-Gives a Numeric Month Day i.e,07th
FMDD-Gives 7th
Before22Days
----------------------
WEEKDAY :2, AUG-7-2017
I need to query 2 tables, one contains a TIMESTAMP(6) column, other contains a DATE column. I want to write a select statement that prints both values and diff between these two in third column.
SB_BATCH.B_CREATE_DT - timestamp
SB_MESSAGE.M_START_TIME - date
SELECT SB_BATCH.B_UID, SB_BATCH.B_CREATE_DT, SB_MESSAGE.M_START_TIME,
to_date(to_char(SB_BATCH.B_CREATE_DT), 'DD-MON-RR HH24:MI:SS') as time_in_minutes
FROM SB_BATCH, SB_MESSAGE
WHERE
SB_BATCH.B_UID = SB_MESSAGE.M_B_UID;
Result:
Error report -
SQL Error: ORA-01830: date format picture ends before converting entire input string
01830. 00000 - "date format picture ends before converting entire input string"
You can subtract two timestamps to get an INTERVAL DAY TO SECOND, from which you calculate how many minutes elapsed between the two timestamps. In order to convert SB_MESSAGE.M_START_TIME to a timestamp you can use CAST.
Note that I have also removed your implicit table join with an explicit INNER JOIN, moving the join condition to the ON clause.
SELECT t.B_UID,
t.B_CREATE_DT,
t.M_START_TIME,
EXTRACT(DAY FROM t.diff)*24*60 +
EXTRACT(HOUR FROM t.diff)*60 +
EXTRACT(MINUTE FROM t.diff) +
ROUND(EXTRACT(SECOND FROM t.diff) / 60.0) AS diff_in_minutes
FROM
(
SELECT SB_BATCH.B_UID,
SB_BATCH.B_CREATE_DT,
SB_MESSAGE.M_START_TIME,
SB_BATCH.B_CREATE_DT - CAST(SB_MESSAGE.M_START_TIME AS TIMESTAMP) AS diff
FROM SB_BATCH
INNER JOIN SB_MESSAGE
ON SB_BATCH.B_UID = SB_MESSAGE.M_B_UID
) t
Convert the timestamp to a date using cast(... as date). Then take the difference between the dates, which is a number - expressed in days, so if you want it in minutes, multiply by 24*60. Then round the result as needed. I made up a small example below to isolate just the steps needed to answer your question. (Note that your query has many other problems, for example you didn't actually take a difference of anything anywhere. If you need help with your query in general, please post it as a separate question.)
select ts, dt, round( (sysdate - cast(ts as date))*24*60, 2) as time_diff_in_minutes
from (select to_timestamp('2016-08-23 03:22:44.734000', 'yyyy-mm-dd hh24:mi:ss.ff') as ts,
sysdate as dt from dual )
;
TS DT TIME_DIFF_IN_MINUTES
-------------------------------- ------------------- --------------------
2016-08-23 03:22:44.734000000 2016-08-23 08:09:15 286.52
I tried the below query in oracle
select cast(TO_DATE (cal.MONTH,'MM') AS varchar2(30)) as result
FROM JOBCONTROL_USER.ods_calendar_weeks cal
WHERE cal.YEAR NOT IN (0, 9999)
it gives result in dd-mon-yy format. Now I want only mon from the result, how can I achieve this without using to_char()?
If you're avoiding Oracle functions and the month number is stored on its own as a varchar2 field, then you could brute-force it:
select case cast(month as number)
when 1 then 'Jan'
when 2 then 'Feb'
when 3 then 'Mar'
when 4 then 'Apr'
when 5 then 'May'
when 6 then 'Jun'
when 7 then 'Jul'
when 8 then 'Aug'
when 9 then 'Sep'
when 10 then 'Oct'
when 11 then 'Nov'
when 12 then 'Dec'
end as mon
from ods_calendar_weeks cal
where cal.year not in (0, 9999);
But you're having to specify the language; you don't get Oracle's conversion of the month to the NLS date language. Which might be a bonus or a problem depending on your context.
I'd be tempted to put the conversions into a look-up table instead and join to that; or to add the month name as a separate column on the table, as a virtual column in 11g.
You can try somthing like this:-
SELECT EXTRACT(MONTH FROM CAST(TO_DATE (cal.MONTH,'MM') AS varchar2(30))) as RESULT
FROM JOBCONTROL_USER.ods_calendar_weeks cal
WHERE cal.YEAR NOT IN (0, 9999)
Hope this will help you.
select to_char(cal.Month,'month')
) AS result
FROM JOBCONTROL_USER.ods_calendar_weeks cal
WHERE cal.YEAR NOT IN (0, 9999);
This will gives month. the to_char() is a function which has two arguments 1. Column name ans 2. Month. Column name is of date data type so we have to convert the date into character data type and we required only the month so the second column will describes what will be extracted from the date. Finally the result is displayed as a character datatype. This query will returns the months name if year neither 0 nor 9999.