I have a column with date type.
It contains data of this format:
09-02-2015 19:50
08-03-2015 20:26
09-03-2015 18:58
14-03-2015 15:47
I want to find all records which their date is 14-03-2015
How do I remove the time?!
I tried:
select * from tab where strftime('%d-%m-%Y', date)='14-03-2015';
but it doesn't work
You can use the date() function:
where date(date) = '2015-03-14'
Always use ISO/ANSI standard formats for dates, YYYYMMDD or YYYY-MM-DD.
SQLite doesn't have a data type for dates, so your column is probably of type text. Then you can compare the first 10 characters:
where substr(date, 1, 10) = '14-03-2015'
Related
I'm using PostgreSQL, but this question is for any modern dbms
I want to basically convert a datetime column which has yyyy/mm/dd into just yyyy/mm
I tried getting months and year separately and using Concat, but the problem is the month comes as a single digit integers for values < 10 and that messes up ordering
select *,
concat(date_part('year' , date_old), '/', date_part('month' , date_old)) as date_new
from table
date _old
date_new
2010-01-20
2010-1
2010-01-22
2010-1
2010-11-22
2010-11
You can use to_char()
to_char(date_old, 'yyyy/mm')
If you want to display your date in the format YYYY-MM then
In PostgreSQL (db<>fiddle) and Oracle (db<>fiddle), use TO_CHAR:
SELECT TO_CHAR(date_old, 'YYYY/MM') FROM table_name;
In MySQL (db<>fiddle), use DATE_FORMAT:
SELECT DATE_FORMAT(date_old, '%Y/%m') FROM table_name;
In SQL Server (db<>fiddle), use CONVERT or, if you are using SQL Server 12 or later, FORMAT:
SELECT CONVERT(varchar(7), date_old, 111) FROM table_name;
SELECT FORMAT(date_old,'yyyy/MM') FROM table_name;
Don't do this.
If you're able to use the date_part() function, what you have is not actually formatted as the yyyy/mm/dd value you say it is. Instead, it's a binary value that's not human-readable, and what you see is a convenience shown you by your tooling.
You should leave this binary value in place!
If you convert to yyyy/mm, you will lose the ability to directly call functions like date_part(), and you will lose the ability to index the column properly.
What you'll have left is a varchar column that only pretends to be a date value. Schemas that do this are considered BROKEN.
In SQLite, how to compare date without passing timestamp?
The date format is 2018-03-18 08:24:46.101655+00 and I want to compare against only date part as 2018-03-18.
I have tried as where mydate='2018-03-18' but that didn't return any records.
Similarly, tried Date(mydate)='2018-03-18' but that didn't help either.
How can I compare dates ignoring the timestamp part?
select * from mytable
where strftime('%Y-%m-%d', mydate) = '2018-03-18'
This is not one of the supported date formats.
To extract the date part from the string, use substr():
... WHERE substr(mydate, 1, 10) = '2018-03-18'
It might be a better idea to store dates in a correct format in the database to begin with.
It is looking that there is problem with date format.
Sqlite doesn't understand data like '+00' in date.
So date() and strftime() will not work here if data type is 'timestamp with time zone'.
Try by using like clause.
Try using strftime
SELECT strftime('%Y %m %d', 'columnName');
you can find it here strftime.php
In my DB, there is a date field in the format yyyymmdd.
I have to get all the dates in the format dd-mm-yyyy for that particlar date.
ex:
Date
20170130
20170228
20170325
for the above dates, I need the output in the below format with the dates and day of the particular dates
date day
30-01-2017 tuesday
28-02-2017 tuesday
25-03-2017 saturday
If the column is a string, then it can hold invalid date values such as February 31, one way to avoid this is by a small function such as this:
create or replace
function my_to_date( p_str in varchar2 ) return date
is
begin
return to_date( p_str );
exception
when others then
return null;
end;
\\
select to_char(my_to_date('20170231'),'DD-MM-YYYY Day')
from dual
\\
Demo
Try below:
Select to_char(yrdate, 'dd-mm-yyyy'), to_char(yrdate, 'D') from yrtable
It sounds like your dates aren't actually DATE fields but some kind of CHAR field? The best option would be to convert to DATE and then convert back to CHAR:
SELECT TO_CHAR(TO_DATE(mydate, 'YYYYMMDD'), 'DD-MM-YYYY Day')
FROM mytable;
This uses the YYYYMMDD mask to convert your string into a date, then uses the mask DD-MM-YYYY Day to convert it back into a string. Use day if you want the day name in lowercase (as in your OP).
#user2778168 answer will give you the results you want. But why?
Your database does not have dates stored in yyyymmdd format or any other date format for at mater, unless it's defined with a character type definition. Oracle stores all dates in a single internal structure, and with only slight variations timestamps are the same. The format used only tells Oracle how to display the value or to convert a string to a date. Unless a specific format is specified Oracle uses the NLS_DATE_FORMAT for this determination. See here and scan down to "Datetime Format Models" for format specifications.
To see this run the following:
select value
from nls_session_parameters
where parameter = 'NLS_DATE_FORMAT';
Select yrdate default_format
, to_char(yrdate, 'dd-mm-yyyy') specified_format
, dump(yrdate) unformated
from yrtable;
alter session set nls_date_format = 'Month dd,yyyy';
Rerun the above queries.
It seems you hold date column(date1) in character format. Suppose you have a table named days:
SQL> desc days
date1 varchar2(10)
then,
we should convert it into date, and then into char format, with aliases in quotation marks to provide lowercase aliases as you wanted.
perhaps your database's date language is non-english like mine(turkish), then you need to convert it to english.
lastly, it'a appropriate to order the results according to converted date, seen from your output. So we can use the following SQL :
select to_char(to_date(date1,'yyyymmdd'),'dd-mm-yyyy') "date",
to_char(to_date(date1,'yyyymmdd'),'day','nls_date_language=english') "day"
from days
order by to_date(date1,'yyyymmdd');
D e m o
This i got to convert a single binary date to Gregorian date ,
Select to_char(to_date(sum(2415020+42744),'J'),'MMDDYYYY') Last_arrear_date from dual;
But I'm not able to convert a binary date column date into Gregorean in xyz table
For Example
Suppose i have a table borm,That has account_open_date column that stores data in binary format , I want to convert that account_open_date column data into Gregorian date format. Please let me know how to convert that
While you have to change the entire column, Don't use the sum function.
Try the below SQL command it will work.
select acct_no, to_char(to_date((account_open_date + 2415020), 'J'),'DD-MON-YYYY')
from borm;
To Change Binary to Gregorian
select column1 ,...columnN, to_char(to_date((Column_name + 2415020), 'J'),'DD-MON-YYYY')
from Table_Name
WHERE Condition;
To change Gregorian to Binary:
select column1,columnN, to_char(to_date(Column_Name,'YYYYMMDD'),'J')-2415020
from Table_Name
where condition ;
The date column that I have is in varchar2 and I want to convert those values in YYYY-MM-DD
DATE
7/26/2013
7/29/2013
8/1/2013
8/4/2013
7/28/2013
7/31/2013
8/3/2013
7/30/2013
8/5/2013
7/25/2013
8/2/2013
8/6/2013
7/27/2013
You should never store dates in a VARCHAR column
So in order to display it differently now, you need to first convert the string to a date and then back to a string
If you are certain that all "dates" do have the right (and same) format, then the following should work:
select to_char(to_date(date, 'mm/dd/yyyy'), 'yyyy-mm-dd')
from the_table;
But I wouldn't be surprised if that gives you an error because one or more rows have a date which is formatted differently. Something which could not have happened had you defined the column to be of type DATE right from the beginning.
You should really, really consider changing that column to be a real DATE column.
Btw: DATE is a horrible name for such a column. It is a reserved word and it simply doesn't tell you anything about what is actually stored in that column (a "start date", an "end date", a "birth date", a "due date", ...)
You have to convert your string to date and than convert to char with expected format
select TO_CHAR(TO_DATE(datecol, 'MM/DD/YYYY'),'YYYY-MM-DD')
from your_table
Sql Fiddle Demo
you can use REGEXP_REPLACE function to manipulate strings or to_date and to_char to convert string to date and than to string back with different format.
select to_char(TO_DATE("Date", 'MM/DD/YYYY'),'yyyy-mm-dd')
from Table1
SQL FIDDLE