I have a table with a date column (DD/MM/YYYY hh:mm:ss) as follows:
02/09/2021 09:50:37
03/09/2021 09:20:27
02/09/2021 14:05:25
03/09/2021 12:50:28
02/09/2021 14:25:26
What's the most efficient way to filter by date?
Note: I tried with
select date
from table
where date = '02/09/2021'
However, this provides results that wrongly lack the time and only contain the date 02/09/2021.
For example, 02/09/2021 09:50:37 would not be returned.
A simple option is
select *
from your_table
where trunc(date_column) = date '2021-09-02'
If there's an index on date_column, it wouldn't be used in that case so performance might suffer so you'll have to either create a function-based index, or use a different where clause:
where date_column >= date '2021-09-02' and date_column < date '2021-03-03'
If you want to return every record that has some date column equal to some date regardless to the time part then you need this:
select date
from table
where trunc(date) = '02/09/2021'
please avoid naming columns date and tables table.
Here is a small demo
Use a date range:
SELECT date_column
FROM table_name
WHERE date_column >= DATE '2021-09-02'
AND date_column < DATE '2021-09-02' + INTERVAL '1' DAY;
This will enable Oracle to use an index on the date_column column.
If you filter using:
TRUNC(date_column) = DATE '2021-09-02' or
TO_CHAR(date_column, 'DD-MM-YYYY') = '02-09-2021'
Then Oracle will not use an index on the date_column and would need a separate function-based index on either TRUNC(date_column) or TO_CHAR(date_column, 'DD-MM-YYYY').
Related
sql table
here in the table above named carpooling contains a column name start_on which has date time as timestamp i have to write a query to select all the rows having date as 25-11-20 using to_char and to_date.
You write a timestamp literal like this:
timestamp '2020-11-25 00:00:00'
so the full filtering condition will be
where start_on >= timestamp '2020-11-25 00:00:00'
and start_on < timestamp '2020-11-26 00:00:00'
Note that dates and timestamps are different in Oracle, and dates include times down to the second (this is for historical reasons - originally there was only the date type, and timestamp was added much later).
Use the TRUNC function, along with date and interval literals:
SELECT *
FROM CARPOOLING
WHERE START_ON BETWEEN DATE '2020-11-25'
AND (DATE '2020-11-26' - INTERVAL '0.000001' SECOND)
You can simply use to_date, but it's recommended to remove the time when comparing the dates. Otherwise, rows having the same date, but a different time will not be selected. Removing the time can be done using TRUNC.
So you can do something like this:
SELECT * FROM carpooling
WHERE TRUNC(start_on) = TO_DATE('2020-11-25','yyyy.mm.dd');
If you don't want to check the 25th of November 2020, but another data, change the date to match your goal.
I'm in PostgreSQL.
I need to print all mailing with creation date strictly more that 2015-04-04. I tried the following queries:
SELECT *
FROM mailing.mailing
WHERE creation_date > '2015-04-04';
and
SELECT *
FROM mailing.mailing
WHERE creation_date >= '2015-04-04';
But they produced the same result set(including '2015-04-04'). Is it possible to write such a query without explicitly saying WHERE creation_date >= '2015-04-05';
UPD: The column's type is timestamp without time zone.
If your creation_date field is of type datetimetry comparing it to '2015-04-04 23:59:59' instead, as '2015-04-04 08:30:00' seems to be greater than '2015-04-04'.
Assuming your default date format for your database is 'YYYY-MM-DD' and creation_date field is a date type, your query will actually be converted automatically to something like:
SELECT *
FROM mailing.mailing
WHERE creation_date > to_date('2015-04-04', 'YYYY-MM-DD');
The date value you have provided represents the first second of that day, that's why you see no difference between your queries. (Your first query would exclude the first second of the day though.)
What you could do to avoid this is:
where creation_date >= to_date('2015-04-05 00:00:00', 'YYYY-MM-DD HH24:MI:SS')
or
where date_trunc(creation_date-1) = '2015-04-04'
I have a column with DATE datatype in a table.
I am trying to retrieve the column values in YYYYMM format. My select query looks like below
select *
from tablename
where date column = to_char(to_date('12/31/4000','MM/DD/YYYY'),'YYYYMM');
I am getting below exception.
ORA-01847: day of month must be between 1 and last day of month
Appreciate any input on this.
I think the simplest method is:
where to_char(datecolumn, 'YYYYMM') = '400012'
Or, if you prefer:
where to_char(datecolumn, 'YYYYMM') = to_char(to_date('12/31/4000', 'MM/DD/YYYY'), 'YYYYMM');
Syntax-wise, the right hand date (to the right of the equals) is OK. But you are doing a character comparison, not a date comparison.
This works for me in multiple databases:
select to_char (to_date('12/31/4000','MM/DD/YYYY'),'YYYYMM')
from dual;
Even though your column is named DATE_COLUMN, you are comparing based on characters in the query.
So, try this instead - this compares based on dates (NOT a character comparison) and truncates off the hour, minute, ETC. so you are only comparing the DAY:
select * from DATE_TAB
where TRUNC(DATE1, 'DDD') = TRUNC(to_date('12/31/4000','MM/DD/YYYY'),'DDD');
NOTE: The DATE1 field above is a DATE field. If you're DATE_COLUMN is not a DATE field, you must
convert it to a DATE datatype first (using TO_DATE, ETC.)
Assuming that "date_column" is actually a date, and that you have an index on date_column, you can do something like this to return the data quickly (without truncating dates in all rows to do a comparison):
with dat as (
select level as id, sysdate - (level*10) as date_column
from dual
connect by level <= 100
)
select id, date_column
from dat
where date_column between to_date('11/1/2013', 'MM/DD/YYYY') and last_day(to_date('11/2013 23:59:59', 'MM/YYYY HH24:MI:SS'))
Here I just dummy up some data with dates going back a few years. This example picks all rows that have a date in the month of November 2013.
If your date_column's data-type is DATE, then use
select *
from tablename
where TO_CHAR(date_column,'YYYYMM') = to_char (to_date('12/31/4000','MM/DD/YYYY'),'YYYYMM');
If your date_column's data-type is VARCHAR, then use:
select *
from tablename
where date_column = to_char (to_date('12/31/4000','MM/DD/YYYY'),'YYYYMM');
I somehow feel your error is because you have a space between date and column as
"date column". If the field name in the table is "COLUMN", then just removing the word "DATE" from your original query would suffice, as:
select *
from tablename
where column = to_char(to_date('12/31/4000','MM/DD/YYYY'),'YYYYMM');
If your column (YYYYMMDD) is in number format, the simplest way to get YYYYMM would be
select floor(DATE/100)
from tablename;
Im trying to do a query where a TIMESTAMP field is = to a specific date but my query is not working:
The field is type TIMESTAMP(6) which I have only ever worked with DATE / DATETIME fields before. Here is example of a value stored here: 04-OCT-13 12.29.53.000000000 PM
Here is my SELECT statement:
SELECT * FROM SomeTable WHERE timestampField = TO_DATE('2013-10-04','yyyy-mm-dd')
I am retrieving no results and I am assuming it has to do with the fact that its not matching the TIME portion of the timestamp
If you want every record that occurs on a given day then this should work:
SELECT * FROM SomeTable
WHERE timestampField >= TO_TIMESTAMP( '2013-03-04', 'yyyy-mm-dd' )
AND timestampField < TO_TIMESTAMP( '2013-03-05', 'yyyy-mm-dd')
That will be likely to take advantage of an index on timestampField if it exists. Another way would be:
SELECT * FROM SomeTable
WHERE TRUNC(timestampField) = TO_DATE( '2013-03-04', 'yyyy-mm-dd' )
in which case you may want a function-based index on TRUNC(timestampField).
(Note that TRUNC applied to a TIMESTAMP returns a DATE.)
I'm trying to retrieve records from table by knowing the date in column contains date and time.
Suppose I have table called t1 which contains only two column name and date respectively.
The data stored in column date like this 8/3/2010 12:34:20 PM.
I want to retrieve this record by this query for example (note I don't put the time):
Select * From t1 Where date="8/3/2010"
This query give me nothing !
How can I retrieve date by knowing only date without the time?
DATE is a reserved keyword in Oracle, so I'm using column-name your_date instead.
If you have an index on your_date, I would use
WHERE your_date >= TO_DATE('2010-08-03', 'YYYY-MM-DD')
AND your_date < TO_DATE('2010-08-04', 'YYYY-MM-DD')
or BETWEEN:
WHERE your_date BETWEEN TO_DATE('2010-08-03', 'YYYY-MM-DD')
AND TO_DATE('2010-08-03 23:59:59', 'YYYY-MM-DD HH24:MI:SS')
If there is no index or if there are not too many records
WHERE TRUNC(your_date) = TO_DATE('2010-08-03', 'YYYY-MM-DD')
should be sufficient. TRUNC without parameter removes hours, minutes and seconds from a DATE.
If performance really matters, consider putting a Function Based Index on that column:
CREATE INDEX trunc_date_idx ON t1(TRUNC(your_date));
Personally, I usually go with:
select *
from t1
where date between trunc( :somedate ) -- 00:00:00
and trunc( :somedate ) + .99999 -- 23:59:59
Convert your date column to the correct format and compare:
SELECT * From my_table WHERE to_char(my_table.my_date_col,'MM/dd/yyyy') = '8/3/2010'
This part
to_char(my_table.my_date_col,'MM/dd/yyyy')
Will result in string '8/3/2010'
You could use the between function to get all records between 2010-08-03 00:00:00:000 AND 2010-08-03 23:59:59:000
trunc(my_date,'DD') will give you just the date and not the time in Oracle.
Simply use this one:
select * from t1 where to_date(date_column)='8/3/2010'
Try the following way.
Select * from t1 where date(col_name)="8/3/2010"