I have a table with 4 columns huge number of records. It has the following structure:
DATE_ENTERED EMP_NAME DATA ORIGINATED
01-JAN-20 A 545454 APPLE
I want to calculate no of records for every first day of every month in a year
is there any way can we fetch the data for every first day of month.
In oracle you can use TRUNC function on the date as follows:
SELECT TRUNC(DATE_ENTERED), COUNT(1) AS CNT
FROM YOUR_TABLE
WHERE TRUNC(DATE_ENTERED) = TRUNC(DATE_ENTERED, 'MON')
GROUP BY TRUNC(DATE_ENTERED, 'MON')
Please note that the TRUNC(DATE_ENTERED, 'MON') returns the first day of the month for DATE_ENTERED.
Cheers!!
SELECT Year, Month, COUNT(*)
FROM
(
SELECT
YEAR(DATE_ENTERED) Year
MONTH(DATE_ENTERED) Month
DAY(DATE_ENTERED) Day
FROM your_table
WHERE DAY(DATE_ENTERED) = 1
) A
GROUP BY Year, Month
Generally WHERE DAY(DATE_ENTERED) = 1 will get you the records only for dates at the start of each month. Thus using Year and Month function you can group them by in order to get a count for each year and each month
You mean something like
SELECT COUNT(*)
FROM Table
WHERE DAY(DATE_ENTERED) = 1 AND
YEAR(DATE_ENTERED) = Some_Year
GROUP BY DATE_ENTERED
You can also use DATE_ENTERED BETWEEN 'YYYY0101' and 'YYYY1231' (replace the YYYY with the year you want to retrieve data for) instead of YEAR(DATE_ENTERED) = Some_Year, if performance is an issue.
You can use something like this:
select * from your_table
where DAY(DATE_ENTERED) = 1
and DATE_ENTERED between '2020-01-01' and '2020-12-31'
for number of count use this:
select count(*) from your_table
where DAY(DATE_ENTERED)= 1
and DATE_ENTERED between '2020-01-01' and '2020-12-31'
UPDATE
select * from your_table where Extract(day FROM DATE_ENTERED) = 1 and DATE_ENTERED between '01-JAN-20 ' and '01-DEC-20 ';
this is how the data looks like:
For the list of records
select count(*) from your_table where Extract(day FROM DATE_ENTERED) = 1 and DATE_ENTERED between '01-JAN-20 ' and '01-DEC-20 ';
UPDATE-2
select EXTRACT(month from DATE_ENTERED) as Count,
to_char(to_date(DATE_ENTERED, 'DD-MM-YYYY'), 'Month') from your_table
where Extract(day FROM DATE_ENTERED) = 1 and DATE_ENTERED between '01-JAN-20
'and '01-DEC-20 ' group by EXTRACT(month from DATE_ENTERED),
to_char(to_date(DATE_ENTERED, 'DD-MM-YYYY'), 'Month');
Here is the output:
Related
I am trying to get the frequency of unique ID values for each month of the last year. However, I don't get the outcome.. including the error message "SELECT list expression references column user_id which is neither grouped nor aggregated".
How can I get the count of unique IDs in each month and them group them by month?
What I tried:
SELECT
user_id,
EXTRACT(MONTH FROM date) as month
FROM
TABLE
WHERE
date >= '2020-09-01'
GROUP BY
month
I want something like this:
month
count of unique user_id
1
300
2
200
...
...
12
250
You would use GROUP BY and COUNT(DISTINCT):
SELECT EXTRACT(MONTH FROM date) as month, COUNT(DISTINCT user_id)
FROM TABLE
WHERE date >= '2020-09-01'
GROUP BY 1;
I would advise you to include the year in the query. In BigQuery, this is simplest using DATE_TRUNC():
SELECT DATE_TRUNC(date, MONTH) as month, COUNT(DISTINCT user_id)
FROM TABLE
WHERE date >= '2020-09-01'
GROUP BY 1;
I have a table like this, I hope to count the number of ids by month. I used the following code but it does not work.
id date_time
1390880502018723840,2021-05-08
1390881127930372100,2021-05-08
1390881498270736386,2021-05-08
SELECT twitter.tweets.id
WHERE Month(twitter.tweets.date_time)=01 AND Year(twitter.tweets.date_time)=2021 ;
you have to use count() function and to_char to get year month part of date in one column:
SELECT count(witter.tweets.id)
WHERE to_char(twitter.tweets.date_time,'YYYY-MM')= '2021-01';
you can generalize it for all the month/year by using group by :
SELECT to_char(twitter.tweets.date_time,'YYYY-MM') , count(witter.tweets.id)
group by to_char(twitter.tweets.date_time,'YYYY-MM');
To get counts for all months since Jan 2021:
SELECT date_trunc('month', date_time), count(*)
FROM twitter.tweets
WHERE date_time >= '2021-01-01'
GROUP BY 1;
If id can be NULL (which should be disallowed for an id column), use the slightly more expensive count(id) instead.
Count of distinct IDs:
SELECT date_trunc('month', date_time), count(DISTINCT id)
FROM twitter.tweets
WHERE date_time >= '2021-01-01'
GROUP BY 1;
For only Jan 2021:
SELECT count(DISTINCT id)
FROM twitter.tweets
WHERE date_time >= '2021-01-01'
WHERE date_time < '2021-02-01';
SELECT EVENT_DT - ((EVENT_DT -DATE'1900-01-07') MOD 7) AS dates,
CLSFD_USER_ID AS user_id,
COUNT(DISTINCT CLSFD_USER_ID) AS number_of_user_ids,
COUNT(DISTINCT CLSFD_CAS_AD_ID) AS number_of_ads,
SUM(IMPRSN_CNT) AS number_of_impressions
FROM clsfd_access_views.CLSFD_CAS_AD_HST
WHERE CLSFD_SITE_ID = 3001
AND datum >= '2017-01-01'
GROUP BY 1,2
I want to have the total number of unique users during each month of the year 2017. I tried:
GROUP BY EXTRACT(MONTH FROM datum), 2
But this returns an error. What would be the most efficient code to retrieve the total number of user ids, ads, and impressions, per month.
It doesn't make sense to me to be aggregating by users, since they are what you are trying to count. Try grouping by the month and year alone:
SELECT
EXTRACT(YEAR FROM EVENT_DT) || '-' || EXTRACT(MONTH FROM EVENT_DT) AS month,
COUNT(DISTINCT CLSFD_USER_ID) AS number_of_user_ids,
COUNT(DISTINCT CLSFD_CAS_AD_ID) AS number_of_ads,
SUM(IMPRSN_CNT) AS number_of_impressions
FROM clsfd_access_views.CLSFD_CAS_AD_HST
WHERE
CLSFD_SITE_ID = 3001 AND
datum >= '2017-01-01' AND datum < '2018-01-01'
GROUP BY
EXTRACT(YEAR FROM EVENT_DT) || '-' || EXTRACT(MONTH FROM EVENT_DT);
Note that I changed your restriction on datum to also exclude any year greater than 2017.
If you want this values to be included in current query, then you should use analytical functions. For example "total number of unique users during each month" would be something like:
select count(distinct user_id) over(partition by EXTRACT(MONTH FROM datum))
Be aware that those values will be repeated for each user.
tbl_totalMonth has id,time, date and kwh column.
I want to get the last recorded data of the months and group it per month so the result would be the name of the month and kwh.
the result should be something like this:
month | kwh
------------
January | 150
February | 400
the query I tried: (but it returns the max kwh not the last kwh recorded)
SELECT DATENAME(MONTH, a.date) as monthly, max(a.kwh) as kwh
from tbl_totalMonth a
WHERE date > = DATEADD(yy,DATEDIFF(yy,0, GETDATE() -1 ),0)
group by DATENAME(MONTH, a.date)
I suspect you need something quite different:
select *
from (
select *
, row_number() over(partition by month(a.date), year(a.date) order by a.date DESC) as rn
from tbl_totalMonth a
WHERE date > = DATEADD(yy,DATEDIFF(yy,0, GETDATE() -1 ),0)
) d
where rn = 1
To get "the last kwh recorded (per month)" you need to use row_number() which - per month - will order the rows (descending) and give each one a row number. When that number is 1 you have "the most recent" row for that month, and you won't need group by at all.
You could use group by and month
select datename(month, date), sum(kwh)
from tbl_totalMonth
where date = (select max(date) from tbl_totalMonth )
group by datename(month, date)
if you need only the last row for each month then youn should use
select datename(month, date), khw
from tbl_totalMonth a
inner join (
select max(date) as max_date
from tbl_totalMonth
group by month(date)) t on t.max_date = a.date
I have a table, lets say "Records" with structure:
id date
-- ----
1 2012-08-30
2 2012-08-29
3 2012-07-25
I need to write an SQL query in PostgreSQL to get record_id for MIN date in each month.
month record_id
----- ---------
8 2
7 3
as we see 2012-08-29 < 2012-08-30 and it is 8 month, so we should show record_id = 2
I tried something like this,
SELECT
EXTRACT(MONTH FROM date) as month,
record_id,
MIN(date)
FROM Records
GROUP BY 1,2
but it shows 3 records.
Can anybody help?
SELECT DISTINCT ON (EXTRACT(MONTH FROM date))
id,
date
FROM Records1
ORDER BY EXTRACT(MONTH FROM date),date
SQLFiddle http://sqlfiddle.com/#!12/76ca2/3
UPD: This query:
1) Orders the records by month and date
2) For every month picks the first record (the first record has MIN(date) because of ordering)
Details here http://www.postgresql.org/docs/current/static/sql-select.html#SQL-DISTINCT
This will return multiples if you have duplicate minimum dates:
Select
minbymonth.Month,
r.record_id
From (
Select
Extract(Month From date) As Month,
Min(date) As Date
From
records
Group By
Extract(Month From date)
) minbymonth
Inner Join
records r
On minbymonth.date = r.date
Order By
1;
Or if you have CTEs
With MinByMonth As (
Select
Extract(Month From date) As Month,
Min(date) As Date
From
records
Group By
Extract(Month From date)
)
Select
m.Month,
r.record_id
From
MinByMonth m
Inner Join
Records r
On m.date = r.date
Order By
1;
http://sqlfiddle.com/#!1/2a054/3
select extract(month from date)
, record_id
, date
from
(
select
record_id
, date
, rank() over (partition by extract(month from date) order by date asc) r
from records
) x
where r=1
order by date
SQL Fiddle
select distinct on (date_trunc('month', date))
date_trunc('month', date) as month,
id,
date
from records
order by 1, 3 desc
I think you need use sub-query, something like this:
SELECT
EXTRACT(MONTH FROM r.date) as month,
r.record_id
FROM Records as r
INNER JOIN (
SELECT
EXTRACT(MONTH FROM date) as month,
MIN(date) as mindate
FROM Records
GROUP BY EXTRACT(MONTH FROM date)
) as sub on EXTRACT(MONTH FROM r.date) = sub.month and r.date = sub.mindate