How to compare day of week to string in IF statements? - date

I need to use IF to compare the date(day format) to the string 'SUNDAY' which is today.
If today is Sunday, the code prints still prints the ELSE statement
Today is Sunday, but it still says it's not Sunday.
Expecting the IF statement and not ELSE statement

The output of to_char(datetime) is blank-padded by default:
SQL> select '"'||to_char(sysdate,'DAY')||'"' as actual_value from dual;
ACTUAL_VALUE
--------------------------------------
"SUNDAY "
You can avoid this with the fm modifier:
SQL> select '"'||to_char(sysdate,'fmDAY')||'"' as actual_value from dual;
ACTUAL_VALUE
--------------------------------------
"SUNDAY"
However, it is safer to specify a date language, in case the session is not set to default to English (and fractionally more efficient to specify a shorter format):
SQL> alter session set nls_date_language = German;
SQL> select '"'||to_char(sysdate,'fmDAY')||'"' as actual_value from dual;
ACTUAL_VALUE
------------------------------------------
"SONNTAG"
Shorter ('DY' instead of 'DAY') and safer (ensures 'SUN' not 'SON', 'DIM' etc):
SQL> select '"'||to_char(sysdate,'DY','nls_date_language = English')||'"' as actual_value from dual;
ACTUAL_VALUE
--------------
"SUN"

The problem is TO_CHAR(..., 'DAY'). It returns a string of fixed length regardless of the day of the week. The length depends on NLS_DATE_LANGUAGE. For example, for English, the longest day-of-the-week name is WEDNESDAY which is nine characters. Then, for any other day of the week, TO_CHAR(..., 'DAY') returns a string of the same length (nine characters for English), by right-padding with spaces as needed.
The simplest way to fix your code is to wrap the left-hand side of your comparisons within TRIM().
There are a bunch of odd things in your code (for example, in the ELSE branch you should say "today is not SUNDAY", not "today is not ... whatever the TO_CHAR returned"), but that's not what you asked. The answer to your question, and no more, is given above.

Related

Add_months function error based on postgres database

I tried to ruh this query in postgres :
Select to_char((select add_months (to_date ('10/10/2019', 'dd/mm/yyyy'), '11/11/2019') ) , 'dd/mm/yyyy') as temp_date
I got an error :
Function add_months (date, unknown) does not exist
Hint: no function matches the given name and argument types. You might need to add explicit type casts.
Please help
As documented in the manual there is no add_months function in Postgres
But you can simply add an interval:
select to_date('10/10/2019', 'dd/mm/yyyy') + interval '10 months'
If you need to format that date value to something:
select to_char(to_date('10/10/2019', 'dd/mm/yyyy') + interval '10 months', 'yyyy-mm-dd')
No one, even running on Oracle, has run the original query- at least not successfully. It appears that query is expecting to add two months together (in this case Oct and Nov). That is not what the function does. It adds an integer number of months to the specified date and returns the resulting date. As indicated in Postgres just adding the desired interval. However, if you have many occurrences ( like converting) of this the following implements a Postgres version.
create or replace function add_months(
date_in date
, n_months_in integer)
returns date
language sql immutable strict
as
$$
-- given a date and an integer for number of months return the calendar date for the specified number of months away.
select (date_in + n_months_in * interval '1 month')::date
$$ ;
-- test
-- +/- 6 months from today.
select current_date "today"
, add_months(current_date,6) "6 months from now"
, add_months(current_date,-6) "6 months ago"
;

Leading Zero when extract Day and Month

Hello i'm trying to extract current day and month in a query. This query is filling table from csv file.
the file is named like this:
LOG_01_01_2018.csv
My query search for file:
LOG_1_1_2018.csv
With no Zero in front day and month. How to add Zero numbers?
Here is the code:
execute format ($f$COPY tmp_x FROM 'D:\Programs\PS\download_files_from_ftp_avtomat\files\LOG_%s_%s_%s.csv'
(header, FORMAT CSV, DELIMITER ',', NULL ' ', ENCODING 'WIN1251');
$f$,extract(day from now()),extract(month from now()),extract(year from now()));
One option uses LPAD:
execute format ($f$COPY tmp_x FROM 'D:\Programs\PS\download_files_from_ftp_avtomat\files\LOG_%s_%s_%s.csv'
(header, FORMAT CSV, DELIMITER ',', NULL ' ', ENCODING 'WIN1251');
$f$,
lpad(extract(day from now())::text, 2, '0'),
lpad(extract(month from now())::text, 2, '0'),
extract(year from now()));
The year would always be a fixed width four digit number, unless you plan to work with data which existed before computers were around.
https://www.postgresql.org/docs/current/static/functions-datetime.html#FUNCTIONS-DATETIME-EXTRACT
extract(field from timestamp) returns double precision, so you either lpad zero to it (making the value text), or just use to_char with data mask, eg:
t=# select extract(day from now()), to_char(now(),'MM'),extract(year from now());
date_part | to_char | date_part
-----------+---------+-----------
19 | 01 | 2018
(1 row)
Looks shorter to use to_char():
SELECT to_char(now(),'DD')||'/'||to_char(now(),'MM')||'/'||to_char(now(),'YYYY')

Oracle:Update date column in a table

Following is my oracle update query that`s throwing error--
Update table Set col1 = To_date(:dateFill, 'mm/dd/yyyy hh24:mi:ss') Where Fil1 = :ID;
dateFill = 01/05/2012,
ID= 15
this statement is executing in a procedure,
error -:ORA-01722: invalid number(date field)
Can someone tell me why is 'select To_date('01/05/2012 00:00:00', 'mm/dd/yyyy hh24:mi:ss') from dual;' giving me result like '05-JAN-2012 00:00:00'.???
Please suggest me some answers.
If dateFill = 01/05/2012, why are you specifying a date format that includes "hh24:mi:ss"?
An ORA-01722 error occurs when an attempt is made to convert a character string into a number, and the string cannot be converted into a valid number. Valid numbers contain the digits '0' through '9', with possibly one decimal point, a sign (+ or -) at the beginning or end of the string, or an 'E' or 'e' (if it is a floating point number in scientific notation).
to_date always work with 'character' ie
to_date(char[,'format'[,nls_lang])
Your variable 'dateFill' is of Number datatype. Cast or convert this 'dateFill'
field into varchar and things will work.
Can someone tell me why is 'select To_date('01/05/2012 00:00:00',
'mm/dd/yyyy hh24:mi:ss') from dual;' giving me result like
'05-JAN-2012 00:00:00'.???
how a date displays is down to your client / Nls date format setting:
SQL> select To_date('01/05/2012 00:00:00','mm/dd/yyyy hh24:mi:ss') from dual;
TO_DATE('01/05/20120
--------------------
05-jan-2012 00:00:00
SQL> alter session set nls_date_format='mm/dd/yyyy hh24:mi:ss';
Session altered.
SQL> select To_date('01/05/2012 00:00:00','mm/dd/yyyy hh24:mi:ss') from dual;
TO_DATE('01/05/2012
-------------------
01/05/2012 00:00:00
SQL>

T-sql IF Condition date evaluation

I have a simple question regarding T-SQL. I have a stored procedure which calls a Function which returns a date. I want to use an IF condition to compare todays date with the Functions returned date. IF true to return data.
Any ideas on the best way to handle this. I am learning t-sql at the moment and I am more familar with logical conditions from using C#.
ALTER FUNCTION [dbo].[monday_new_period](#p_date as datetime) -- Parameter to find current date
RETURNS datetime
BEGIN
-- 1 find the year and period given the current date
-- create parameters to store period and year of given date
declare #p_date_period int, #p_date_period_year int
-- assign the values to the period and year parameters
select
#p_date_period=period,
#p_date_period_year = [year]
from client_week_uk where #p_date between start_dt and end_dt
-- 2 determine the first monday given the period and year, by adding days to the first day of the period
-- this only works on the assumption a period lasts a least one week
-- create parameter to store the first day of the period
declare #p_start_date_for_period_x datetime
select #p_start_date_for_period_x = min(start_dt)
from client_week_uk where period = #p_date_period and [year] = #p_date_period_year
-- create parameter to store result
declare #p_result datetime
-- add x days to the first day to get a monday
select #p_result = dateadd(d,
case datename(dw, #p_start_date_for_period_x)
when 'Monday' then 0
when 'Tuesday' then 6
when 'Wednesday' then 5
when 'Thursday' then 4
when 'Friday' then 3
when 'Saturday' then 2
when 'Sunday' then 1 end,
#p_start_date_for_period_x)
Return #p_result
END
ALTER PROCEDURE [dbo].[usp_data_to_retrieve]
-- Add the parameters for the stored procedure here
AS
BEGIN
-- SET NOCOUNT ON added to prevent extra result sets from
-- interfering with SELECT statements.
SET NOCOUNT ON;
IF monday_new_period(dbo.trimdate(getutcdate()) = getutcdate()
BEGIN
-- SQL GOES HERE --
END
Thanks!!
I assume you are working on Sql2008. See documentation of IF and CASE keywords for more details.
CREATE FUNCTION dbo.GetSomeDate()
RETURNS datetime
AS
BEGIN
RETURN '2012-03-05 13:12:14'
END
GO
IF CAST(GETDATE() AS DATE) = CAST(dbo.GetSomeDate() AS DATE)
BEGIN
PRINT 'The same date'
END
ELSE
BEGIN
PRINT 'Different dates'
END
-- in the select query
SELECT CASE WHEN CAST(GETDATE() AS DATE) = CAST(dbo.GetSomeDate() AS DATE) THEN 1 ELSE 0 END AS IsTheSame
This is the basic syntax for a T-SQL IF and a date compare.
If you are comparing just the date portion for equality you will need to use:
select dateadd(dd,0, datediff(dd,0, getDate()))
This snippet will effectively set the time portion to 00:00:00 so you can compare just dates. So in use it will look something like this.
IF dateadd(dd,0, datediff(dd,0, fn_yourFunction())) = dateadd(dd,0, datediff(dd,0, GETDATE()))
BEGIN
RETURN SELECT * FROM SOMEDATA
END
Hope that helps!

Problem when extracting year and week number from string in PSQL

Let's say that I have a range of SQL tables that are named name_YYYY_WW where YYYY = year and WW = week number. If I call upon a function that guides a user defined date to the right table.
If the date entered is "20110101":
SELECT EXTRACT (WEEK FROM DATE '20110101') returns 52 and
SELECT EXTRACT (YEAR FROM DATE '20110101') returns 2011.
While is nothing wrong with these results I want "20110101" to either point to table name_2010_52 or name_2011_01, not name_2011_52 as it does now when I concanate the results to form the query for the table.
Any elegant solutions to this problem?
The function to_char() will allow you to format a date or timestamp to output correct the iso week and iso year.
SELECT to_char('2011-01-01'::date, 'IYYY_IW') as iso_year_week;
will produce:
iso_year_week
---------------
2010_52
(1 row)
You could use a CASE:
WITH sub(field) AS (
SELECT CAST('20110101' AS date) -- just to test
)
SELECT
CASE
WHEN EXTRACT (WEEK FROM field ) > 1 AND EXTRACT (MONTH FROM field) = 1 AND EXTRACT (DAY FROM field) < 3 THEN 1
ELSE
EXTRACT (WEEK FROM field)
END
FROM
sub;