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';
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 the following situation:
ID DATE_TIME AMOUNT
23 14-MAY-2021 10:47:01 5
23 14-MAY-2021 11:49:52 3
23 14-MAY-2021 12:03:18 4
How can get the sum of the amount and take the DATE by day not hourly?
Example:
ID DATE_TIME TOTAL
23 20210514 12
I tried this way but i got error:
SELECT DISTINCT ID, TO_CHAR(DATE_TIME, 'YYYYMMDD'), SUM(AMOUNT) AS TOTAL FROM MY_TABLE
WHERE ID ='23' AND DATE_TIME > SYSDATE-1
GROUP BY TOTAL, DATE_TIME
You don't need DISTINCT if you use GROUP BY - anything that is grouped must be distinct unless it joined to something else later on that caused it to repeat again
You were almost there too
SELECT ID, TO_CHAR(DATE_TIME, 'YYYYMMDD') AS DATE_TIME, SUM(AMOUNT) AS TOTAL
FROM MY_TABLE
WHERE ID ='23' AND DATE_TIME > SYSDATE-1
GROUP BY ID, TO_CHAR(DATE_TIME, 'YYYYMMDD')
You need to group by the output of the function, not the input. Not every database can GROUP BY aliases used in the select (technically the SELECT hasn't been done by the time the GROUP is done so the aliases don't exist yet, and you wouldnt group by the total because that's an aggregate (the result of summing up every various value in the group)
If you need to do further work with that date, don't convert it to a string.. Cut the time off using TRUNC:
SELECT ID, TRUNC(DATE_TIME) as DATE_TIME, SUM(AMOUNT) AS TOTAL
FROM MY_TABLE
WHERE ID ='23' AND DATE_TIME > SYSDATE-1
GROUP BY ID, TRUNC(DATE_TIME)
TRUNC can cut a date down to other parts, for example TRUNC(DATE_TIME, 'HH24') will remove the minutes and seconds but leave the hours
Convert the DATE column to a string with the required accuracy and then group on that:
SELECT ID,
TO_CHAR("DATE", 'YYYY-MM-DD'),
SUM(AMOUNT) AS TOTAL FROM MY_TABLE
WHERE ID ='23'
AND "DATE" > SYSDATE-1
GROUP BY ID, TO_CHAR("DATE", 'YYYY-MM-DD')
or truncate the value so that the time component is set to midnight for each date:
SELECT ID,
TRUNC("DATE"),
SUM(AMOUNT) AS TOTAL FROM MY_TABLE
WHERE ID ='23'
AND "DATE" > SYSDATE-1
GROUP BY ID, TRUNC("DATE")
(Note: DATE is a keyword and cannot be used as an identifier unless you use a quoted-identifier; and you would need to use the quotes, and the exact case, everytime you refer to the column. You would be better to rename the column to something else that is not a keyword.)
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:
I have the following:
id date_start date_end
----- ---------- --------
1a3 2001-12-12 2002-12-12
23b 2005-01-24 2008-11-02
11ad 2012-01-15 2014-13-09
19d 2015-01-23 2016-02-04
And I want to get the count of each person where the date range includes the first day of each year.
for example I can do:
select count(distinct id) from table
where '2001-01-01' between date_start and date_end
but I want to produce the count for all years from 2000-2015. I want to avoid manually doing:
select count(distinct id) from table
where '2001-01-01' between date_start and date_end
select count(distinct id) from table
where '2002-01-01' between date_start and date_end
select count(distinct id) from table
where '2003-01-01' between date_start and date_end
I am just having trouble visualizing the group by clause for this.If I had just year I could do:
select count(distinct id), year from table
group by year
however I cannot fit the where '2001-01-01' between date_start and date_end into this group clause.
can anyone help?
Thanks!
You can use left join. Here is a method:
select y.yyyy, count(distinct t.id)
from ((select '2001-01-01' as yyyy from sys.sysdummy) union all
(select '2002-01-01' as yyyy from sys.sysdummy) union all
(select '2003-01-01' as yyyy from sys.sysdummy)
) y left join
table t
on y.yyyy between date_start and date_end
group by y.yyyy
order by y.yyyy;
Well you only have to check and count if the the year of date_start is different from the one of date_end.
select count(*)
from table
where year(date_start) < year(date_end)
I hope you don't mind me answering my own question. Using Michael's logic I was able to combine this with a dummy table - an idea from Gordon's answer. here is what I did:
with dummy(yr) as (
select 2000 from SYSIBM.SYSDUMMY1
union all
select yr + 1 from dummy where yr < 2015
)
select d.yr, count(distinct t.id)
from table t, dummy d
where d.yr between year(t.date_start) + 1 and year(t.date_end)
group by d.yr
order by d.yr
as I say I hope the community does not mind me taking the parts of both posters answers to get the solution I was looking for. I did not want hardcoding and I think the answer I post satisfies this while using a simple piece of date arithmetic logic.
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