How to convert current time to date without time in sql? - sql

I want to select data where the operation_date between '01-Jan-2016' and yesterday. I used code as follows:
select *
where operation_date between '01-Jan-2016' and sysdate-1
from TABLE
But sysdate returns both date and time. Therefore, the above output includes all the data between '01-Jan-2016' and say, 14 Nov-2017 09:50:51. I only want the data before today. How to convert current time to date without time? Thanks.

Seems that you are looking for trunc().
Example
SELECT sysdate - 1 AS current_Date_Time
,trunc(sysdate) - 1 AS CURRENT_DATE1
,trunc(sysdate - 1) AS CURRENT_DATE2
FROM dual
Result
CURRENT_DATE_TIME CURRENT_DATE1 CURRENT_DATE2
----------------------------------------------------------
13.11.2017 18:08:41 13.11.2017 00:00:00 13.11.2017 00:00:00
DEMO
So the correct query will be as below.
SELECT *
WHERE operation_date BETWEEN DATE '2016-01-01' -- ANSI Date Literal
AND trunc(sysdate - 1)
FROM TABLE
OR
SELECT *
WHERE operation_date BETWEEN DATE '2016-01-01' -- ANSI Date Literal
AND trunc(sysdate) - 1
FROM TABLE

Related

How to find data of a period when the date format in is american date style

TABLE NAME = CONTRACT_DATA
CONTRACT DATE
A 10/25/2019
B 06/20/2019
C 10/01/2019
D 10/05/2019
Query:
select *
from CONTRACT_DATA
WHERE DATE(VALUE_DATE,'MM/DD/YYYY') > '10/01/2019' AND
DATE(VALUE_DATE,'MM/DD/YYYY') < '10/31/2019'
Tried to get the data in Oct month but failed
Required output A,C,D
In Oracle a DATE data type always has a year, month, day, hour, minute and second components. Your user interface may not show the time component but it is always there. If its not specified then it will default to 00:00:00 (midnight) but you should not rely on this.
Also, dates do not have a format. They are stored internally as 7 bytes (2 for year and 1 for each other month-second component). It is the user interface you are using that will implicitly apply a format to your data when you want to display it so that it is meaningful to the user.
You can specify dates using:
ANSI date literals: DATE '2019-10-01'
ANSI timestamp literals: TIMESTAMP '2019-10-01 12:34:56' (and Oracle will cast it to a date as needed)
The TO_DATE function: TO_DATE( '10/01/2019', 'MM/DD/YYYY' )
If you have a DATE column then do not use TO_DATE on it as it is already a DATE; instead just compare it to another DATE.
If you want the dates for October 2019 then you need to find the values between 2019-10-01 00:00:00 and 2019-10-31 23:59:59 and the simplest way to do this is to find values that are greater than or equal to midnight of the first day of the month and less than midnight of the first day of the next month:
Test Data:
CREATE TABLE contract_data (
CONTRACT CHAR(1),
VALUE_DATE DATE
);
INSERT INTO contract_data
SELECT 'A', DATE '2019-10-25' FROM DUAL UNION ALL
SELECT 'B', DATE '2019-06-20' FROM DUAL UNION ALL
SELECT 'C', DATE '2019-10-01' FROM DUAL UNION ALL
SELECT 'D', DATE '2019-10-05' FROM DUAL UNION ALL
SELECT 'E', TIMESTAMP '2019-10-31 23:59:59' FROM DUAL;
Query:
SELECT *
FROM contract_data
WHERE VALUE_DATE >= DATE '2019-10-01'
AND VALUE_DATE < DATE '2019-11-01'
Output:
CONTRACT | VALUE_DATE
:------- | :---------
A | 10/25/2019
C | 10/01/2019
D | 10/05/2019
E | 10/31/2019
db<>fiddle here
You would need to convert your string to date for comparison, using TO_DATE(); on the other side of the operator, for fixed values, you can use a DATE litteral:
SELECT *
FROM contract_data
WHERE
TO_DATE(value_date, 'MM/DD/YYY') > DATE'2019-10-01'
AND TO_DATE(value_date, 'MM/DD/YYY') < DATE'2019-10-31'
Note: if you want the entire month, then you need to ajust the boundaries
SELECT *
FROM contract_data
WHERE
TO_DATE(value_date, 'MM/DD/YYY') >= DATE'2019-10-01'
AND TO_DATE(value_date, 'MM/DD/YYY') < DATE'2019-11-01'
If your date is stored as string in the table then you can simply use the to_date function and DATE literal as following:
select *
from CONTRACT_DATA
WHERE TO_DATE(VALUE_DATE,'MM/DD/YYYY')
BETWEEN DATE '2019-10-01' AND DATE '2019-10-31'
Cheers!!

How to extract No. of Days between 2 dates in oracle sql?

I want No. of days between these 2 dates using Oracle SQL
Dates:
BETWEEN "1/1/2018" AND "6/11/2018"
How to write SQL Query?
between date '2018-01-01' and date '2018-11-06'
where DATE literal looks exactly like that: DATE 'YYYY-MM-DD'
In your example:
double quote's can't be used
even if you used single quotes, that would be a string, not DATE so you'd depend on whether Oracle is capable of converting it (implicitly) to date or not
therefore, always use dates, not strings
[EDIT]
This is how you select the whole calendar between those two dates:
select date '2018-01-01' + level - 1
from dual
connect by level <= date '2018-11-06' - date '2018-01-01' + 1;
As other answers have pointed out you can simply divide two dates, but there is also no need for any additional arithmetic.
The code:
select to_date('6/11/2018', 'DD/MM/YYYY') - to_date('1/1/2018', 'DD/MM/YYYY')
from dual;
The result: 309
you can simple do:
select date1-date2 form dual;
or
select (sysdate-to_date('01-jan-2018'))-(sysdate-to_date('10-jan-2018'))from dual;
Just use
select date'2018-11-06' - date'2018-01-01' + 1 as days_difference
from dual;
DAYS_DIFFERENCE
---------------
310
or
with t( myDate ) as
(
select date'2018-11-06' from dual union all
select date'2018-01-01' from dual
)
select max(myDate) - min(myDate) + 1 as days_difference
from t;
DAYS_DIFFERENCE
---------------
310

How to search by to_date function?

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

How to use Oracle SQL Date and Time

I have the following example :
select * from my_table
where date between ('19-06-2014 00:00:00,000000000 EUROPE/BUCHAREST') and ('19-06-2014 23:59:59,000000000 EUROPE/BUCHAREST'
If I run this query it returns the correct values . What I need is something like this :
select * from my_table
where date between ('today - 2DAYS ') and ('today - 1DAY')
Do you have any idea how I can achieve this?
Try following solution:
--for one day results
select * from my_table where date between sysdate - 2 and sysdate - 1
--If you wish to start from the beginning of the day:
select * from my_table where date between trunc(sysdate) - 2 and trunc(sysdate) - 1
FYI - sysdate referes the current date and substrating value will be counted in terms of day. If you wish to substract 12 hrs, you should substract 0.5.

Select rows where day is less than today

How do I select rows in the past starting from yesterday in Oracle DB where a field like created_date is a timestamp(6)?
I don't want to compare time, just date.
If you want exactly one day prior to the current time:
select *
from table t
where created_date < sysdate - 1;
If you want times before today:
select *
from table t
where created_date <= trunc(sysdate);
From the Oracle documentation on SELECT :
SELECT * FROM orders
WHERE created_date < TO_DATE('2014-04-28', 'YYYY-MM-DD');
I can pass this date format from my application, worked like a charm.
As you want to compare just date:
select *
from table t
where date(created_date) < DATE_SUB(CURRENT_DATE(), INTERVAL 1 DAY);
you can use cast function to deal with timestamp as date:
SELECT cast(SYSTIMESTAMP(6) as date)
FROM dual;
so you can select rows with "yesterdate" date by:
select ....
where cast(SYSTIMESTAMP(6) as date) like sysdate - 1
note: replace SYSTIMESTAMP(6) with column name which has timestamp type.