I have a table in which one of the columns is m_date with type timestamp with timezone.
I want to write a query to filter the rows where the year is 2020.
One way is to use a range query:
select *
from the_table
where m_date >= date '2020-01-01'
and m_date < date '2021-01-01';
This can use an index in m_date.
Alternatively (but slower!)
select *
from the_table
where extract(year from m_date) = 2020
Related
I have data in a table in SQL with dates, but how do I select only those that happen in 2021. (The dates look like 31-oct-2020) in the table. The dates are the actual date variable, not just text.
You should avoid storing your dates as text, but rather should use a proper date column. That being said, you may check the right 4 characters of the date string:
SELECT *
FROM yourTable
WHERE RIGHT(date_col, 4) = '2021';
If the column be an actual date type then use:
SELECT *
FROM yourTable
WHERE date_col >= '2021-01-01' AND date_col < '2022-01-01';
I suspect that your DB is Oracle after checking out your previous post. Then you can use
SELECT *
FROM yourTable
WHERE EXTRACT(year FROM dt) = 2021
or
SELECT *
FROM yourTable
WHERE TRUNC(dt,'YYYY') = date'2021-01-01'
or
SELECT *
FROM yourTable
WHERE dt BETWEEN date'2021-01-01' AND date'2021-12-31'
You can benefit the index if there's one on the date column(namely dt) by using the last SELECT statement
If using MSSQL, you can leverage the YEAR(...) to extract the year from a date.
Replace and , with the table name and date column name respectively.
select * from <tablename> where year(<datecolumn>) = 2021
I have a table tk. It has a column effective_date with a data type of date. The table also has some other columns. What I want to do is query the table such that output contains all dates having same year but month should differ
I have tried with below query in SQL Server, but it's not returning the desired result:
select * tk_id
from tk
group by tk_id, YEAR(effective_date);
Are you looking for a where clause?
select *
from t
where effective_date >= '2019-01-01' and effective_date < '2020-01-01'
I have a Postgres table that I'm trying to analyze based on some date columns.
I'm basically trying to count the number of rows in my table that fulfill this requirement, and then group them by month and year. Instead of my query looking like this:
SELECT * FROM $TABLE WHERE date1::date <= '2012-05-31'
and date2::date > '2012-05-31';
it should be able to display this for the months available in my data so that I don't have to change the months manually every time I add new data, and so I can get everything with one query.
In the case above I'd like it to group the sum of rows which fit the criteria into the year 2012 and month 05. Similarly, if my WHERE clause looked like this:
date1::date <= '2012-06-31' and date2::date > '2012-06-31'
I'd like it to group this sum into the year 2012 and month 06.
This isn't entirely clear to me:
I'd like it to group the sum of rows
I'll interpret it this way: you want to list all rows "per month" matching the criteria:
WITH x AS (
SELECT date_trunc('month', min(date1)) AS start
,date_trunc('month', max(date2)) + interval '1 month' AS stop
FROM tbl
)
SELECT to_char(y.mon, 'YYYY-MM') AS mon, t.*
FROM (
SELECT generate_series(x.start, x.stop, '1 month') AS mon
FROM x
) y
LEFT JOIN tbl t ON t.date1::date <= y.mon
AND t.date2::date > y.mon -- why the explicit cast to date?
ORDER BY y.mon, t.date1, t.date2;
Assuming date2 >= date1.
Compute lower and upper border of time period and truncate to month (adding 1 to upper border to include the last row, too.
Use generate_series() to create the set of months in question
LEFT JOIN rows from your table with the declared criteria and sort by month.
You could also GROUP BY at this stage to calculate aggregates ..
Here is the reasoning. First, create a list of all possible dates. Then get the cumulative number of date1 up to a given date. Then get the cumulative number of date2 after the date and subtract the results. The following query does this using correlated subqueries (not my favorite construct, but handy in this case):
select thedate,
(select count(*) from t where date1::date <= d.thedate) -
(select count(*) from t where date2::date > d.thedate)
from (select distinct thedate
from ((select date1::date as thedate from t) union all
(select date2::date as thedate from t)
) d
) d
This is assuming that date2 occurs after date1. My model is start and stop dates of customers. If this isn't the case, the query might not work.
It sounds like you could benefit from the DATEPART T-SQL method. If I understand you correctly, you could do something like this:
SELECT DATEPART(year, date1) Year, DATEPART(month, date1) Month, SUM(value_col)
FROM $Table
-- WHERE CLAUSE ?
GROUP BY DATEPART(year, date1),
DATEPART(month, date1)
I have a table that contains records with a column indicates the Date. Given a record, I would like to select all records that are in the same week as the record. How can SQL do that?
I should say that I'm using SQLite.
You can use DATEPART with wk to get the current week. Then just check for equality.
In this case, I have also checked yy to make sure that you do not check the year of a previous week.
SELECT *
FROM TABLE
WHERE DATEPART(wk, TABLE.DATECOLUMN)
= DATEPART(wk, (SELECT DATECOLUMN FROM TABLE WHERE ID = GivenID))
AND DATEPART(yy, TABLE.DATECOLUMN) = DATEPART(yy, (SELECT DATECOLUMN FROM TABLE WHERE ID = GivenID))
UPDATE FOR SQLITE
To do this in SQLLite, Refer to this SO question and then this article that states %W is what you use to get week and %Y for year. Which gives you:
SELECT *
FROM TABLE
WHERE STRFTIME('%W', TABLE.DATECOLUMN)
= STRFTIME('%W', (SELECT DATECOLUMN FROM TABLE WHERE ID = GivenID))
AND STRFTIME('%Y', TABLE.DATECOLUMN)
= STRFTIME('%Y', (SELECT DATECOLUMN FROM TABLE WHERE ID = GivenID))
Use the datediff() function:
datediff(ww,start_date,end_date) < 1
You can use BETWEEN to specify the records you want.
SELECT * FROM records WHERE datecolumn between 'YYYY/MM/DD' AND 'YYYY/MM/DD'
How do I get a maximium daily value of a numerical field over a year in MS-SQL
This would query the daily maximum of value over 2008:
select
datepart(dayofyear,datecolumn)
, max(value)
from yourtable
where '2008-01-01' <= datecolumn and datecolumn < '2009-01-01'
group by datepart(dayofyear,datecolumn)
Or the daily maximum over each year:
select
datepart(year,datecolumn),
, datepart(dayofyear,datecolumn)
, max(value)
from yourtable
group by datepart(year,datecolumn), datepart(dayofyear,datecolumn)
Or the day(s) with the highest value in a year:
select
Year = datepart(year,datecolumn),
, DayOfYear = datepart(dayofyear,datecolumn)
, MaxValue = max(MaxValue)
from yourtable
inner join (
select
Year = datepart(year,datecolumn),
, MaxValue = max(value)
from yourtable
group by datepart(year,datecolumn)
) sub on
sub.Year = yourtable.datepart(year,datecolumn)
and sub.MaxValue = yourtable.value
group by
datepart(year,datecolumn),
datepart(dayofyear,datecolumn)
You didn't mention which RDBMS or SQL dialect you're using. The following will work with T-SQL (MS SQL Server). It may require some modifications for other dialects since date functions tend to change a lot between them.
SELECT
DATEPART(dy, my_date),
MAX(my_number)
FROM
My_Table
WHERE
my_date >= '2008-01-01' AND
my_date < '2009-01-01'
GROUP BY
DATEPART(dy, my_date)
The DAY function could be any function or combination of functions which gives you the days in the format that you're looking to get.
Also, if there are days with no rows at all then they will not be returned. If you need those days as well with a NULL or the highest value from the previous day then the query would need to be altered a bit.
Something like
SELECT dateadd(dd,0, datediff(dd,0,datetime)) as day, MAX(value)
FROM table GROUP BY dateadd(dd,0, datediff(dd,0,datetime)) WHERE
datetime < '2009-01-01' AND datetime > '2007-12-31'
Assuming datetime is your date column, dateadd(dd,0, datediff(dd,0,datetime)) will extract only the date part, and then you can group by that value to get a maximum daily value. There might be a prettier way to get only the date part though.
You can also use the between construct to avoid the less than and greater than.
Group on the date, use the max delegate to get the highest value for each date, sort on the value, and get the first record.
Example:
select top 1 theDate, max(theValue)
from TheTable
group by theDate
order by max(theValue) desc
(The date field needs to only contain a date for this grouping to work, i.e. the time component has to be zero.)
If you need to limit the query for a specific year, use a starting and ending date in a where claues:
select top 1 theDate, max(theValue)
from TheTable
where theDate between '2008-01-01' and '2008-12-13'
group by theDate
order by max(theValue) desc