I would like to know if we can got values from date if the date is in the current month
For example, If I have :
date value
28/06/2021 50
02/07/2021 100
05/07/2021 18
Then I search to have :
02/07/2021 100
05/07/2021 18
because 28/06/2021 is not in the current month
thanks in advance !
In Mysql you can use now() and month() functions to get the current date ang extract the month.
Select *
from table
where <your filters>
having MONTH(NOW()) = MONTH( table.date_field)
Different databases like oracle , postrest change the name of this functions , so you must find it
TSQL
select * from tbl
where MONTH(date)=month(getdate())
You'll want to match the YEAR along with the MONTH if you will have data that spans more than a year.
SELECT date, value
FROM table
WHERE YEAR(date) = YEAR(GETDATE()) AND MONTH(date) = MONTH(GETDATE())
Related
I have a date field in one of my tables and the column name is from_dt. Now I have to compare a month and year combination against this from_dt field and check whether the month has already passed. The current database function uses separate conditions for the month and the year, but this is wrong as it will compare month and year separately. The current code is like this
SELECT bill_rate, currency FROM table_name WHERE
emp_id = employee_id_param
AND EXTRACT(MONTH FROM from_dt) <= month_param
AND EXTRACT(YEAR FROM from_dt) <= year_param
Now the fromt_dt field has value 2021-10-11. If I give month_param as 01 and year_param as 2022, this condition will not work as the month 10 is greater than 1, which I have given. Basically, I need to check whether 01-2022 (Jan 2022) is greater than r equal to 2021-10-01(October 1st, 2021). It would be very much helpful if someone can shed some light here.
If you just want to check whether one date is >= then another:
# select '2022-01-01'::date >= '2021-10-11'::date;
?column?
----------
t
If you want to restrict to year/month then:
select date_trunc('month','2022-01-01'::date) >= date_trunc('month', '2021-10-11'::date);
?column?
----------
t
Where the date_trunc components are:
select date_trunc('month','2022-01-01'::date) ;
date_trunc
------------------------
2022-01-01 00:00:00-08
select date_trunc('month','2021-10-11'::date) ;
date_trunc
------------------------
2021-10-01 00:00:00-07
See Postgres date_trunc for more information.
Assuming the given year_param and month_param are integers you can use the make_date function to create the first of the year_month and date_trunc to get the first on the month from the table. Just compare those values. (See date functions) So:
select bill_rate, currency
from table_name
where emp_id = employee_id_param
and date_trunc('month',from_dt) =
make_date( year_param, month_param, 01);
I have a table in SQL Server 2012 with a month column stored as nvarchar(255):
"January", "February", "March"
And another column in this table with year stored separately as float
"2012","2013,"2014".
I do not have a day column so I want to create a combined month date column with the day starting as 1.
So for month and year fields January 2012. I want to show '2012-01-01'
How can I do such and add that into my current table?
I want to find the maximum row for a record in my table for each employee.
so for an [employee #], [month],[year]. what is latest record so for example below:
1. 102, Jan, 2019
2. 102, feb, 2019
I want to only see the second record which is the latest.
SQL Server has pretty flexible conversion to date. So, just convert the columns to a date:
select convert(date, month + ' ' + year)
You can get the maximum as:
select empid, max(convert(date, month + ' ' + year))
from t
group by empid;
If you really like, you can change the format for output purposes. I would advise you to stick with a date, though.
Note: This assumes that your internationalization settings are set to English -- which seems reasonable if you are storing month names in English.
Fix your design! The way you store data makes it really inefficient to interpret it. Here, I think the simplest option is datefromparts() and a 12-branches case expression.
Assuming that the (float) year is stored in column col_year and the (string) month is in col_month:
select t.*,
datefromparts(
cast(col_year as int),
case col_month
when 'January' then 1
when 'February' then 2
...
when 'December' then 12
end,
1
) as date_col
from mytable t
I am new to SQL, here is my problem.
I have a table with daily dates:
Date:
20190101
20190102
20190103
.
**20190131**
20190201
20190202
20190203
.
**20190228**
20190301
20190302
20190303
.
**20190331**
I want to select only the month-end dates, what would be the code to do that?
thanks
I am using MS SQL Studio.
One method in standard SQL would be:
select t.*
from t
where extract(month from date + interval '1' day) <> extract(month from date);
Date/time functions vary significantly by database, so the exact functions might not match your database. However, the idea is simple: add one day and see if the month changes.
In standard SQL, you could do:
select date
from mytable
where date = date_trunc('month', date) + interval '1' month - interval '1' day
Edit
In SQL Server, you can just use eomonth(). Given a date, this functions returns the corresponding end of month, which you can compare against the date. So:
select date
from mytable
where date = eomonth(date)
I am facing some problem with the hive code.
My FROM TABLE is partitioned based on month, year and day. I came up with the following code to get the data I need. The logic is something like if the current mth is 01 then change the month to 12 and the year to yr - 1
else change month to mth - 1 and keep the year as is.
set hivevar:yr=2019;
set hivevar:mth=03;
set hivevar:dy=29;
SELECT * from
FROM table
WHERE
month = case when cast('${mth}' as int) = 01 then 12 else cast((cast('${mth}' as int) - 1) as string) end
AND year = case when cast('${mth}' as int) = 01 then cast((cast('${yr}' as int) - 1) as string) else '${yr}' end;
It is not working, my select * is coming empty. Please help.
desc table
From what i understand, you are trying to get data from the previous month given a date. If so, you can use inbuilt date functions to do it.
select *
from table
where concat_ws('-',year,month,day) >= add_months(date_add(concat_ws('-','${yr}','${mth}','${dy}'),1-'${dy}'), -1)
and concat_ws('-',year,month,day) < date_add(concat_ws('-','${yr}','${mth}','${dy}'),1-'${dy}')
The solution assumes year, month and day are of the format yyyy, MM and dd. If not, adjust them as needed
Also, you should consider storing date as a column even though you have it partitioned by year,month and day.
I have a date column in the format YY-MON-DD, e.g. 25-JUN-05. Is it possible to isolate this into 3 separate columns for year, month and day? Where month is converted from text to numerical, e.g. Year: 25, Month: 06, Day: 05?
MS SQL SERVER
As Nebi suggested, you can use DATEPART and extract each part and store it into different columns.
SELECT DATEPART(DAY,'2008-10-22'); -- Returns DAY part i.e 22
SELECT DATEPART(MONTH,'2008-10-22'); -- Returns MONTH part i.e 10
SELECT DATEPART(YEAR,'2008-10-22'); -- Returns YEAR part i.e 2008
Try with the below script,if you are using SQL Server.
SELECT 'Year: '+CAST(LEFT(YourdateColumn,2) as VARCHAR(2))+', Month: ' +CAST(MONTH('01-'+SUBSTRING(YourdateColumn,4,3)+'-15')as VARCHAR(2))+', Day:'+CAST(RIGHT(YourdateColumn,2)as VARCHAR(2))
FROM Yourtable
sample output :
You didn't specify your DBMS.
The following is standard SQL assuming that column really is a DATE column
select extract(year from the_column) as the_year,
extract(month from the_column) as the_month,
extract(day from the_column) as the_day
from the_table;