I hope you can suppor me with a piece of code I'm writing. I'm working with the following query:
SELECT case_id, case_date, people_id FROM table_1;
and I've to search in the DB how many times the same people_id is repeted in the DB, (different case_id) considering the case_date -90days timeframe. Any advise on how to address that?
Data sample
Additional info: as results I'm expecting to have the list of people_id with how many cases received in the 90 days from the last case_date.
expected result sample:
The way I understood the question, it would be something like this:
select people_id,
case_id,
count(*)
from table_1
where case_date >= trunc(sysdate) - 90
group by people_id,
case_id
You want to filter WHERE the case_date is greater than or equal to 90 days before the start of today and then GROUP BY the people_id and COUNT the number of DISTINCT (different) case_id:
SELECT people_id,
COUNT( DISTINCT case_id ) AS number_of_cases
FROM table_1
WHERE case_date >= TRUNC( SYSDATE ) - INTERVAL '90' DAY
GROUP BY
people_id;
If you only want to count repeated case_id per person_id then:
SELECT person_id,
COUNT(*) AS number_of_repeated_cases
FROM (
SELECT case_id,
person_id,
FROM table_1
WHERE case_date >= TRUNC( SYSDATE ) - INTERVAL '90' DAY
GROUP BY
people_id,
case_id
HAVING COUNT(*) >= 2
)
GROUP BY
people_id;
I think you want window functions:
select t.*,
count(*) over (partition by people_idorder by case_date
range between interval '90' day preceding and current row
) as person_count_90_day
from t;
Related
I have a table in the snowflake with a time range from for example 2019.01 to 2020.01. An ID can appear multiple times (match with) on any of the dates.
For example:
my_table: two columns dddate and id
dddate
id
2019-02-03
607
2019-01-07
356
2019-08-06
491
2019-01-01
607
2019-12-17
529
2019-04-15
356
......
Is there a way I can find the total number of IDs that appeared at least one time in the current month that also appeared at least one time in the previous three months, and group by month to show each month's number count starting from 2019-04 (The first month that has previous three months data available in the table) until 2020-01.
I am thinking of some code like this:
WITH PREV_THREE AS (
SELECT
DATE_TRUNC('MONTH', dddate) AS MONTH,
ID AS CURR_ID
FROM my_table mt
INNER JOIN
(
(
SELECT
MONTH(DATEADD(DATE_TRUNC('MONTH', dddate), -1, GETDATE())) AS PREV_MONTH,
ID AS PREV_3_MON_ID
FROM my_table
)
UNION ALL
(
SELECT
MONTH(DATEADD(DATE_TRUNC('MONTH', dddate), -2, GETDATE())) AS PREV_MONTH,
ID AS PREV_3_MON_ID
FROM my_table
)
UNION ALL
(
SELECT
MONTH(DATEADD(DATE_TRUNC('MONTH', dddate), -3, GETDATE())) AS PREV_MONTH,
ID AS PREV_3_MON_ID
FROM my_table
)
) AS PREV_3_MON
ON mt.CURR_ID = PREV_3_MON.PREV_3_MON_ID
)
SELECT MONTH, COUNT(DISTINCT ID) AS COUNTER
FROM PREV_THREE
GROUP BY 1
ORDER BY 1
However, it somehow returns an error and doesn't seem working. Could anyone please help me with this? Thank you in advance!
You can use lag():
select distinct id
from (select t.*,
lag(dddate) over (partition by id order by dddate) as prev_dddate
from my_table t
) t
where dddate >= date_trunc('MONTH', current_date) and
prev_dddate < date_trunc('MONTH', current_date) and
prev_dddate >= date_trunc('MONTH', current_date) - interval '3 month';
You can do this for multiple months as:
select date_trunc('MONTH', dddate), count(distinct id)
from (select t.*,
lag(dddate) over (partition by id order by dddate) as prev_dddate
from my_table t
) t
where prev_dddate < date_trunc('MONTH', date_trunc('MONTH', dddate)) and
prev_dddate >= date_trunc('MONTH', date_trunc('MONTH', dddate)) - interval '3 month'
group by date_trunc('MONTH', dddate);
Even if an id appears multiple times in one month, one of those will be first and the lag() will identify the most recent previous month.
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.)
The best way to explain what I need is showing, so, here it is:
Currently I have this query
select
date_
,count(*) as count_
from table
group by date_
which returns me the following database
Now I need to get a new column, that shows me the count off all the previous 7 days, considering the row date_.
So, if the row is from day 29/06, I have to count all ocurrencies of that day ( my query is already doing it) and get all ocurrencies from day 22/06 to 29/06
The result should be something like this:
If you have values for all dates, without gaps, then you can use window functions with a rows frame:
select
date,
count(*) cnt
sum(count(*)) over(order by date rows between 7 preceding and current row) cnt_d7
from mytable
group by date
order by date
you can try something like this:
select
date_,
count(*) as count_,
(select count(*)
from table as b
where b.date_ <= a.date_ and b.date_ > a.date - interval '7 days'
) as count7days_
from table as a
group by date_
If you have gaps, you can do a more complicated solution where you add and subtract the values:
with t as (
select date_, count(*) as count_
from table
group by date_
union all
select date_ + interval '8 day', -count(*) as count_
from table
group by date_
)
select date_,
sum(sum(count_)) over (order by date_ rows between unbounded preceding and current row) - sum(count_)
from t;
The - sum(count_) is because you do not seem to want the current day in the cumulated amount.
You can also use the nasty self-join approach . . . which should be okay for 7 days:
with t as (
select date_, count(*) as count_
from table
group by date_
)
select t.date_, t.count_, sum(tprev.count_)
from t left join
t tprev
on tprev.date_ >= t.date_ - interval '7 day' and
tprev.date_ < t.date_
group by t.date_, t.count_;
The performance will get worse and worse as "7" gets bigger.
Try with subquery for the new column:
select
table.date_ as groupdate,
count(table.date_) as date_count,
(select count(table.date_)
from table
where table.date_ <= groupdate and table.date_ >= groupdate - interval '7 day'
) as total7
from table
group by groupdate
order by groupdate
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 want to find customers where for example, system by error registered duplicates of an order.
It's pretty easy, if reg_date is EXACTLY the same but I have no idea how to implement it in query to count as duplicate if for example there was up to 1 second difference between transactions.
select * from
(select customer_id, reg_date, count(*) as cnt
from orders
group by 1,2
) x where cnt > 1
Here is example dataset:
https://www.db-fiddle.com/f/m6PhgReSQbVWVZhqe8n4mi/0
CUrrently only customer's 104 orders are counted as duplicates because its reg_date is identical, I want to count also orders 1,2 and 4,5 as there's just 1 second difference
demo:db<>fiddle
SELECT
customer_id,
reg_date
FROM (
SELECT
*,
reg_date - lag(reg_date) OVER (PARTITION BY customer_id ORDER BY reg_date) <= interval '1 second' as is_duplicate
FROM
orders
) s
WHERE is_duplicate
Use the lag() window function. It allows to have a look hat the previous record. With this value you can do a diff and filter the records where the diff time is more than one second.
Try this following script. This will return you day/customer wise duplicates.
SELECT
TO_CHAR(reg_date :: DATE, 'dd/mm/yyyy') reg_date,
customer_id,
count(*) as cnt
FROM orders
GROUP BY
TO_CHAR(reg_date :: DATE, 'dd/mm/yyyy'),
customer_id
HAVING count(*) >1