How to extract data between 2 days for every month. For example, say the date range is between 2nd and 10th. Then I need to extract data for 2nd and 10th of every month from the table.
Use EXTRACT
SELECT *
FROM table_name
WHERE EXTRACT( DAY FROM date_column ) BETWEEN 2 AND 10;
or TO_CHAR then TO_NUMBER:
SELECT *
FROM table_name
WHERE TO_NUMBER( TO_CHAR( date_column, 'DD' ) ) BETWEEN 2 AND 10;
Related
I'm attempting to create a breakdown of requests per hour. As such the date part of the TimeStamp object needs to be ignored. This is what I've tried to far:
select
trunc(request_time, 'HH'),
count(*)
from
service_request
group by
trunc(request_time, 'HH');
This seems to group all data into one hour sections but also groups by the day. How do I go about removing the date part so I end up with results like:
Time Count
--------------
00:00 1
01:00 4
02:00 2
You should be able to do
select to_number( to_char( request_time, 'HH24' ) ),
count(*)
from service_request
group by to_number( to_char( request_time, 'HH24' ) );
The to_number is probably not strictly necessary but it makes more sense to return a numeric hour than a string hour for things like sorting.
If the data type is actually a timestamp, then you could improve this
select extract( hour from request_time ),
count(*)
from service_request
group by extract( hour from request_time );
I have a table with data and columns containing a date but separated into several columns :
a column with the year,
a column with the month,
a column with the day.
I would like to keep only the data with a date that is less than 2 months.
I have tried to concat the date but the month and the day does not always have 2 characters. Sometimes it is one number : 1 for january for example.
Could you give some tips to make this request?
Thanks in advance,
select *
from etude
where concat(year,month,day) > NOW()
It is not working as expected
The TIMESTAMP_FORMAT() function should be robust to months/days being either one or two digits:
SELECT *
FROM etude
WHERE TIMESTAMP_FORMAT(CONCAT(year, month, day), 'YYYYMMDD') > NOW();
But note that while this may fix your immediate problem, a much better long term solution would be to stop storing the various components of your date in separate columns. Instead, just maintain a single bona-fide date/timestamp column.
Use lpad function along with timestamp_format function
select *
from etude
where timestamp_format(year || lpad(month,2,'0') || lpad(day,2,'0'), 'YYYYMMDD') > now();
If keep only the data with a date that is less than 2 months means keep only the data with a date that is not earlier than 2 months from the current date, then try this as is:
SELECT dt
FROM
(
SELECT date(digits(dec(year, 4)) || '-' || digits(dec(month, 2))|| '-' || digits(dec(day, 2))) dt
FROM table
(
values
(2019, 1, 1)
, (2019, 7, 13)
) etude (year, month, day)
)
WHERE dt > current date - 2 month;
to combine your 2 questions into 1 answer and using Satya's answer
select *
from etude
where timestamp_format(year || lpad(month,2,'0') || lpad(day,2,'0'), 'YYYYMMDD') > now() - 2 months;
This is another option
select *
from etude
where DATE('0001-12-31') + (year -2) YEARS + month MONTHS + day DAYS > CURRENT DATE
I have table as below:
Table Temp:
ID MAX MIN DATE_C
1 34 24 21-APR-17 02.41.38.520000 PM
2 32 26 20-APR-17 02.42.44.569000 PM
I execute the below SQL query to get temperature details on respective date:
SELECT *
FROM Temp t
WHERE t.date_c = TO_DATE( '2017-04-21', 'YYYY-MM-DD')
order by t.id
But it's returning empty records. Whats wrong with my query?
You need to remove the time component on the column. Here is one way:
SELECT *
FROM Temp t
WHERE TRUNC(t.date_c) = DATE '2017-04-21'
ORDER BY t.id;
However, I usually recommend using inequalities, rather than a function on the column:
SELECT *
FROM Temp t
WHERE t.date_c >= DATE '2017-04-21' AND
t.date_c < DATE '2017-04-22'
ORDER BY t.id;
This allows the query to use an index on date_c. I should add that the original version can use an index on (trunc(date_c, id).
21-APR-17 02.41.38.520000 PM is not a DATE; it has a fractional seconds component so it is a TIMESTAMP.
So, if you want to find items that are on a particular day (inputting the TIMESTAMP using an ISO/ANSI timestamp literal):
SELECT *
FROM Temp
WHERE date_c >= TIMESTAMP '2017-04-21 00:00:00' AND
date_c < TIMESTAMP '2017-04-21 00:00:00' + INTERVAL '1' DAY;
or
SELECT *
FROM Temp
WHERE date_c >= TO_TIMESTAMP( :your_date_string, 'YYYY-MM-DD' ) AND
date_c < TO_TIMESTAMP( :your_date_string, 'YYYY-MM-DD' ) + INTERVAL '1' DAY;
it's returning empty records. Whats wrong with my query?
date_c = TO_DATE( '2017-04-21', 'YYYY-MM-DD') matches all rows where the date_c value is exactly 2017-04-21 00:00:00.000000 (including the time component); if you do not have any rows with exactly that date and time then, as you noticed, it will return nothing. If you want to get records matching that day then you need to get values within a range of times between the start and end of the day.
You need to pass date on the column. Here is a way...
SELECT *
FROM Temp t
WHERE CAST(t.CREATED_ON as date)= N'2017-04-22'
ORDER BY t.id
I have dates stored like String in database.
The format is 'yyyy-ww' (example: '2015-43').
I need to get the first day of the week.
I tried to convert this string into date but there is no 'ww' option for the function "to_date".
Do you have an idea to perform this convertion?
EDIT
Test results based on the answers -
Thanks for your anwsers, but I have many problems to apply your solutions to my context:
select
TRUNC ( 2015 + ((43 - 1) * 7), 'IW' )
from dual
==> Error : ORA-01722: invalid number
select
TRUNC(to_date('2015','YYYY')+ to_number('01') *7, 'IW')
from dual
==> 2015-02-02 00:00:00
I waited for a date in january
select
trunc(to_date(regexp_substr('2015-01', '\d+',1,2), 'YYYY') + regexp_substr('2015-01', '\d+') * 7, 'IW') dt2
from dual
==> 0039-09-14 00:00:00
select
regexp_substr('2015-01', '\d+',1,2) as res1,
regexp_substr('2015-01', '\d+') * 7 as res2
from dual
==> res1 = 01
==> res2 = 14105
try to use by truncate
with t as (
select '16-2010' dt from dual
)
--
--
select dt,
trunc(to_date(regexp_substr(dt, '\d+',1,2), 'YYYY') + regexp_substr(dt, '\d+') * 7, 'IW') dt2
from t
I have dates stored like String in database.
You should never do that. It is a bad design. you should store date as DATE and not as a string. For all kinds of requirements for date manipulations Oracle provides the required DATE functions and format models. As and when needed, you could extract/display the way you want.
I need to get the first day of the week.
TRUNC (dt, 'IW') returns the Monday on or before the given date.
Anyway, in your case, you have the literal as YYYY-WW format. You could first extract the year and week number and combine them together to get the date using TRUNC.
TRUNC ( year + ((week_number - 1) * 7)
, 'IW
)
So, the above should give you the Monday of the week number passed for that year.
SQL> WITH DATA AS
2 ( SELECT '2015-43' str FROM dual
3 )
4 SELECT TRUNC(to_date(SUBSTR(str, 1, 4),'YYYY')+ to_number(SUBSTR(str, instr(str, '-',1)+1))*7, 'IW')
5 FROM DATA
6 /
TRUNC(TO_
---------
23-NOV-15
SQL>
Similar to Lalit's, however, I think I've corrected the math (his seemed to be off a bit when I tested .. )
with w_data as (
select sysdate + level +200 d from dual connect by level <= 10
),
w_weeks as (
select d, to_char(d,'yyyy-iw') c
from w_data
)
SELECT d, c, trunc(d,'iw'),
TRUNC(
to_date(SUBSTR(c, 1, 4)||'0101','yyyymmdd')-8+to_char(to_date(SUBSTR(c, 1, 4)||'0101','yyyymmdd'),'d')
+to_number(SUBSTR(c, instr(c, '-',1)+1)-1)*7 ,'IW')
FROM w_weeks;
The extra columns help show the dates before, and after.
I would do the following:
WITH d1 AS (
SELECT '2015-43' AS mydate FROM dual
)
SELECT TRUNC(TRUNC(TO_DATE(REGEXP_SUBSTR(mydate, '^\d{4}'), 'YYYY'), 'YEAR') + (COALESCE(TO_NUMBER(REGEXP_SUBSTR(mydate, '\d+$')), 0)-1) * 7, 'IW')
FROM d1
The first thing the above query does is get the first four digits of the string 2015-43 and truncates that to the closest year (if you convert convert 2015 using TO_DATE() it returns a date within the current month; that is SELECT TO_DATE('2015', 'YYYY') FROM dual returns 01-FEB-2015; we need to truncate this value to the YEAR in order to get 01-JAN-2015). I then add the number of weeks minus one times seven and truncate the whole thing by IW. This returns a date of 01-OCT-2015 (see SQL Fiddle here).
According ISO the 4th of January is always in week 1, so your query should look like
Select
TRUNC(TO_DATE(REGEXP_SUBSTR(your_column, '^\d{4}')||'-01-04', 'YYYY-MM-DD')
+ 7*(REGEXP_SUBSTR(your_column, '\d$')-1), 'IW')
from your_table;
However, there is a problem. ISO year used for Week number can be different than actual year. For example, 1st Jan 2008 was in ISO week number 53 of 2007.
I think a proper working solution you get only when you generate ISO weeks from date value.
WITH w AS
(SELECT TO_CHAR(DATE '2010-01-04' + LEVEL * INTERVAL '7' DAY, 'IYYY-IW') AS week_number,
TRUNC(DATE '2010-01-04' + LEVEL * INTERVAL '7' DAY, 'IW') AS first_day
FROM dual
CONNECT BY DATE '2010-01-04' + LEVEL * INTERVAL '7' DAY < SYSDATE)
SELECT your_Column, first_day
FROM w your_table
JOIN w ON week_number = your_Column;
Your date range must bigger than 2010-01-04 and not bigger than current day.
This is what I used:
select
to_date(substr('2017/01',1,4)||'/'||to_char(to_number(substr('2017/01',6,2)*7)-5),'yyyy/ddd') from dual;
I need to select recods from oracle table for the current calendar week based on a date datatype field. Currently I am doing like this:
select * from my_table where enter_date > sysdate -7
If today is Thursday, this query will select records seven days from today which infiltrates to last week on Thursday. Is there a way to select records for the current calendar week dynamically?
If your week starts on a Monday then:
select ...
from ...
where dates >= trunc(sysdate,'IW')
For alternative definitions of the first day of the week, trunc(sysdate,'W') truncates to the day of the week with which the current month began, and trunc(sysdate,'WW') truncates to the day of the week with which the current year began.
See other available truncations here: http://docs.oracle.com/cd/E11882_01/server.112/e26088/functions255.htm#i1002084
to_char(sysdate, 'd') returns day of week in [1..7] range; so try using
select *
from my_table
where enter_date >= trunc(sysdate) - to_char(sysdate, 'd') + 1
Here is the SQLFiddel Demo
Below is the query which you can try
select Table1.*
from Table1
where dates > sysdate - TO_CHAR(SYSDATE,'D')