I have a table with date column in which date is updated in this format - 11/21/2022.
How can I get the results for the last 15 days using this date column in Teradata? Looks like need to change the date format in where clause.
I was using below query which does not work with this format
select *
from table_A
WHERE date BETWEEN (CURRENT_DATE - INTERVAL '30' DAY)
AND CURRENT_DATE
This does not give any results.
Assuming your date column be text, then you must first convert it to a bona fide date using the TO_DATE() function:
SELECT *
FROM table_A
WHERE TO_DATE("date", 'MM/DD/YYYY') BETWEEN (CURRENT_DATE - INTERVAL '30' DAY) AND CURRENT_DATE;
Note that ideally you should be storing your dates in actual date columns, not as text. This would avoid the need to call the clunky TO_DATE() function.
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 have these varchar : 20211026231735.
So I would like a query to substract actual sysdate to that date and convert the substraction to DAY HOURS AND SECONDS.
select TO_CHAR(SYSDATE,'YYYYMMDDHH24MISS') - start_time from TABLEA where job_name='jOB_AA_BB';
I get 4220.
Any help please? Thanks
When you do datetime arithmetic with the DATE datatype, you get back a NUMBER of days. To get an INTERVAL you can subtract two TIMESTAMPs. You don't say what the data type is for start_time, but you might get away with this:
select localtimestamp - start_time
from tablea where job_name='jOB_AA_BB';
LOCALTIMESTAMP gives you a TIMESTAMP value in the current session time zone. There's also CURRENT_TIMESTAMP, which give you the same thing in a TIMESTAMP WITH TIME ZONE and SYSTIMESTAMP that gives you the database time in TIMESTAMP WITH TIME ZONE. You may need to convert your start_time to avoid time zone differences, if any.
You can us the function numtodsinterval to convert the results of date arithmetic to an interval. If necessary then use extract to pull out the needed components.
with tablea(job_name, start_time) as
(select 'jOB_AA_BB','20211026231735' from dual)
select numtodsinterval((SYSDATE - to_date( start_time,'yyyymmddhh24miss')),'hour') date_diff
from tablea where job_name='jOB_AA_BB' ;
with tablea(job_name, start_time) as
(select 'jOB_AA_BB','20211026231735' from dual)
select extract (hour from date_diff) || ':' || extract (minute from date_diff)
from (
select numtodsinterval((sysdate - to_date( start_time,'yyyymmddhh24miss')),'day') date_diff
from tablea where job_name='jOB_AA_BB'
);
NOTE: I am not sure how you got any result, other than an error, as your query winds up as a string - a string. You should not convert sysdate to a string but your string to a date (better yet store it as the proper data type - date).
You can convert the value to a date (rather than converting SYSDATE to a string) and then subtract and explicitly return the value as an INTERVAL DAY TO SECOND type:
SELECT (SYSDATE - TO_DATE('20211026231735', 'YYYYMMDDHH24MISS')) DAY TO SECOND
FROM DUAL;
Or, for your table:
SELECT (SYSDATE - TO_DATE(start_time,'YYYYMMDDHH24MISS')) DAY(5) TO SECOND
FROM TABLEA
WHERE job_name='jOB_AA_BB';
db<>fiddle here
I'm having issues in my WHERE clause selecting data from a specific day to today's date. The day/time format in my date column is '7/2/2020 3:12:08 PM'.
I've tested a couple options but keep getting this error - 'literal does not match format string'.
Any idea's of how I can select all data from March 1, 2020 to current date?
Thanks!
In Oracle date columns are not strings, they are exactly in date datatype, so you don't need to convert/cast it. Just use simple date literals:
https://docs.oracle.com/en/database/oracle/oracle-database/19/sqlrf/Literals.html#GUID-8F4B3F82-8821-4071-84D6-FBBA21C05AC1
select * from table where your_date_columg >= date'2015-12-31'
or with to_date function for your string:
select * from table
where
your_date_columg >= to_date('2019-11-25 13:57:52',
'yyyy-mm-dd hh24:mi:ss')
I'm querying a table to get some date, like so:
SELECT date - INTERVAL '10 day' AS date
FROM example_table
WHERE username = 'Bob'
LIMIT 1;
The date column in the example_table does not have a timestamp. All dates in the column are stored in the following manner:
YYYY-MM-DD
The query above will return a result like so:
2016-11-20 00:00:00.000000
It takes the date found, goes back 10 days, and returns that date. But I want it to return the date without adding the timestamp, like so:
2016-11-20
If I use INTERVAL it always seems to add a timestamp. Is there a way to only get the date?
Your query is fine (but can be simplified, as demonstrated by a_horse_with_no_name). What you are seeing is a display issue. You can format your date to a string in the relevant format using to_char():
SELECT to_char("date" - INTERVAL '10 day', 'yyyy-mm-dd') AS "date"
FROM example_table
WHERE username = 'Bob'
LIMIT 1;
Note: LIMIT without an ORDER BY does not make sense: if there is more than one record in the resultset, you actually get a random record out of them.
You can use the interval notation and convert back to a date:
SELECT (date - INTERVAL '10 day')::date AS date
You can subtract (or add) an integer from a date. That integer represents the number of days:
SELECT "date" - 10 AS "date"
FROM example_table
WHERE username = 'Bob'
LIMIT 1;
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;