I have date format (TIMESTAMP) like 2018-03-26-08.30.00.000000 and i want to get it as 2018-03-26-08 how i can do it in sql in DB2(just i want year-month-day-hour )'
Read DB2's manual:
https://www.ibm.com/docs/en/db2-for-zos/12?topic=sf-char
Using that docu:
I need to convert the non-standard in-format with a dash between day and hour, and dots as hour/minute and minute/second separators to a timestamp, using TO_TIMESTAMP().
From the obtained timestamp, I can use TO_CHAR() with a format string to give me back a string with the format I desire.
WITH
indata(ts) AS (
SELECT
TO_TIMESTAMP(
'2018-03-26-08.30.00.000000','YYYY-MM-DD-HH24.MI.SS.US'
)
FROM sysibm.sysdummy1
)
SELECT TO_CHAR(ts,'YYYY-MM-DD-HH24') AS new_format FROM indata;
new_format
---------------
2018-03-26-08
VARCHAR_FORMAT(<timestamp>, '<desired format>')
So:
SELECT VARCHAR_FORMAT(CURRENT TIMESTAMP, 'YYYY-MM-DD-hh24')
FROM SYSIBM.SYSDUMMY1
Related
I want yo get only the 'date hours:minutes:seconds' from the Date column
Date
10/11/22 12:14:01,807000000
11/12/22 13:15:46,650000000
29/12/22 14:30:46,501000000
and I want to get a string column with date hours:minutes:seconds
Date_string
10/11/22 12:14:01
11/12/22 13:15:46
29/12/22 14:30:46
I tried this code but it doesn't work:
select*, TO_CHAR(extract(hour from (Date)))||':'||TO_CHAR(extract(minute from (Date)))||':'||TO_CHAR(extract(second from (Date))) as Date_string
from table;
If this is a date column, you could use to_char directly:
SELECT m.*, TO_CHAR(my_date_column, 'dd/mm/yy hh24:mi:ss')
FROM mytable m
You can use REGEX SUBSTRING function to get the date string on the left.
SELECT REGEXP_SUBSTR (Date_string, '[^,]+', 1, 1)
AS left_part
FROM Table1;
where ^, means look for chars that are NOT comma on 1st position
and get the first occurrence (on the left)
Result:
LEFT_PART
10/11/22 12:14:01
11/12/22 13:15:46
29/12/22 14:30:46
reference:
https://docs.oracle.com/cd/B12037_01/server.101/b10759/functions116.htm
Just do it with the TO_DATE() and TO_CHAR() function pair, both operating on the Oracle date format strings:
Building the scenario:
-- your input ..
WITH indata(dt) AS (
SELECT '10/11/22 12:14:01,807000000' FROM dual UNION ALL
SELECT '11/12/22 13:15:46,650000000' FROM dual UNION ALL
SELECT '29/12/22 14:30:46,501000000' FROM dual
)
-- end of your input. Real query starts here.
-- Change following comma to "WITH" ..
,
-- Now convert to TIMESTAMP(9) ...
as_ts AS (
SELECT
TO_TIMESTAMP(dt ,'DD/MM/YY HH24:MI:SS,FF9') AS ts
FROM indata
)
SELECT
ts
, CAST(ts AS TIMESTAMP(0)) AS recast -- note: this is rounded
, TO_CHAR(ts,'DD/MM/YY HH24:MI:SS') AS reformatted -- this is truncated
FROM as_ts
Result:
TS
RECAST
REFORMATTED
10-NOV-22 12.14.01.807000000
10-NOV-22 12.14.02
10/11/22 12:14:01
11-DEC-22 13.15.46.650000000
11-DEC-22 13.15.47
11/12/22 13:15:46
29-DEC-22 14.30.46.501000000
29-DEC-22 14.30.47
29/12/22 14:30:46
Going by what you have in your question, it appears that the data in the field Date is a timestamp. This isn't a problem, but the names of the table (TABLE) and field (Date) present some challenges.
In Oracle, TABLE is a reserved word - so to use it as the name of a table it must be quoted by putting it inside double-quotes, as "TABLE". Similarly, Date is a mixed-case identifier and must likewise be quoted (e.g. "Date") every time it's used.
Given the above your query becomes:
SELECT TO_CHAR("Date", 'DD/MM/YY HH24:MI:SS') AS FORMATTED_DATE
FROM "TABLE"
and produces the desired results. db<>fiddle here
Generally, it's best in Oracle to avoid using reserved words as identifiers, and to allow the database to convert all names to upper case - if you do that you don't have to quote the names, and you can refer to them by upper or lower case as the database automatically converts all unquoted names to upper case internally.
I have Time column in BigQuery, the values of which look like this: 2020-09-01-07:53:19 it is a STRING format. I need to extract just the date. Desired output: 2020-09-01.
My query:
SELECT
CAST(a.Time AS date) as Date
from `table_a`
The error message is: Invalid datetime string "2020-09-02-02:17:49"
You could also use the parse_datetime(), then convert to a date.
with temp as (select '2020-09-02-02:17:49' as Time)
select
date(parse_datetime('%Y-%m-%d-%T',Time)) as new_date
from temp
How about just taking the left-most 10 characters?
select substr(a.time, 1, 10)
If you want this as a date, then:
select parse_date('%Y-%m-%d', substr(a.time, 1, 10))
select STR_TO_DATE('2020-09-08 00:58:09','%Y-%m-%d') from DUAL;
or to be more specific as your column do as:
select STR_TO_DATE(a.Time,'%Y-%m-%d') from `table_a`;
Note: this format is applicable where mysql is supported
Got a Export_date field with YYYY-MM-DD. I'd like to get YYYYMM as output.
input : 2020-06-02
Expected output : 202006
I'm using SQL Standard in GCP Bigquery
How to Convert a date YYYY-MM-DD to a string YYYYMM, using Big Query SQL Standard ?
I've tried this function
substr(cast(export_date as string), 0,7) as date,
But I got YYYY-DD as output (2020-06)
Thank you
You can use format_timestamp:
SELECT format_timestamp('%Y%m', '2020-06-02')
it gives back '202006'.
A simple method is:
select extract(year from export_date) * 100 + extract(month from export_date) as yyyymm
You could also format this as a string using date_format().
If your value is stored as a string and not a date, then just use string functions:
select concat(substr(export_date, 1, 4), substr(export_date, 6, 2))
i am having problems with converting a varchar(yyyymmdd) to date(yyyymmdd).
i have a procedure with a parameter (pdate varchar2, yyyymmdd format) which needed to be converted to date type in yyyymmdd format as well.
so far i tried.
vdate date;
vdate := (to_date(SUBSTR(pdate,1,4)+SUBSTR(pdate,5,2)+SUBSTR(pdate,7,2), 'yyyymmdd'));
this threw a error of ORA-01840: input value not long enough for date format.
any help would be appreciated.
Just use a to_date
select to_date(MyDate, 'yyyymmdd')
from mytable
Test with:
select to_date('20170831','yyyymmdd')
from dual
Also, to concatenate in Oracle, use a double pipe ||
select 'Chicken'||'Food'
from dual
If pdate has other characters after yyyymmdd, and yyyymmdd is in the beginning of the whole text, you can just use
SELECT TO_DATE(SUBSTR(pdate,1,8), 'yyyymmdd')
FROM yourtable;
Example
SELECT TO_DATE(SUBSTR('20170831 10:30am',1,8), 'yyyymmdd')
FROM dual;
Otherwise, you can directly use TO_DATE() as suggested by most that replied
I'm storing Dates as string in the database with this format DD-MM-YYYY.
When I tried to make a select query with an orderby on the date column. I didn't get the expected result.
example of result :
28/02/2013
27/02/2013
01/03/2013
My sql query :
SELECT * FROM data ORDER BY strftime('%s', date_column)
Thank you.
The problem is that you store dates as DD-MM-YYYY strings, which does not only prevent natural ordering of dates as strings, but also parsing them with SQLite's date and time functions. Click the link and scroll down to 'Time Strings' section.
SQLite expects date/time strings in the natural order, most significant digit to least significant, that is, YYYY-MM-DD. You can use string operations to transform your DD-MM-YYYY strings into that form. For instance:
select
substr(reversed_date, 7,4) || '-' ||
substr(reversed_date, 4, 2)|| '-' ||
substr(reversed_date, 1, 2) as proper_date
from (
select '12-03-2000' as reversed_date
)
;
You can either transform your date column into this format (as #peterm suggests) or just use the value of proper_date for sorting. You don't need to use strftime for that, but date-related functions will work with such values.
IMHO you need to change the format you store dates in from
DD-MM-YYYY
to
YYYY-MM-DD
From docs
Time Strings
A time string can be in any of the following formats:
YYYY-MM-DD
YYYY-MM-DD HH:MM
YYYY-MM-DD HH:MM:SS
YYYY-MM-DD HH:MM:SS.SSS
YYYY-MM-DDTHH:MM
YYYY-MM-DDTHH:MM:SS
YYYY-MM-DDTHH:MM:SS.SSS
...
Then your original query and this one will work as expected
SELECT * FROM Table1 ORDER BY date(date_column);
SELECT * FROM Table1 ORDER BY strftime('%s', date_column);
Output:
| date_column |
---------------
| 2013-02-27 |
| 2013-02-28 |
| 2013-03-01 |
sqlfiddle
According to the documentation the following should work
SELECT *
FROM data
ORDER BY strftime('%Y-%m-%d', date_column)
try:
SELECT * FROM data ORDER BY to_date(date_column)
probably this might solve you problem as it is going for string comparison rather than date comparison so
01/03/2013 appears smaller than 28/02/2013 or 27/02/2013
thus output is :
01/03/2013, 27/02/2013, 28/02/2013
This query worked for me filter dates
SELECT inst.*,
Substr(columnname, 4, 2) AS newdate,
Substr(columnname, 0, 3) AS newday,
Substr(columnname, 12, 5) AS newtime
FROM table_name AS inst
WHERE child_id = id
ORDER BY newdate DESC, newday DESC, newtime DESC