I have this query that shows the number of flights each month, but the month appears in number format and I want to convert the month number as text.
Here is the query:
select to_char(f.departuretime, 'yyyy-mm') month, count(*) numberofflights
from flight f
group by to_char(f.departuretime, 'yyyy-mm')
order by numberofflights desc;
output:
MONTH NUMBEROFFLIGHTS
2022-05 7
2022-11 5
2022-08 3
... ...
I want to display the months like "2022-MAY" or just "MAY" and so on.
You can use the month format instead of MM to get the month's name instead of its number:
select to_char(f.departuretime, 'yyyy-MONTH') month, count(*) numberofflights
from flight f
group by to_char(f.departuretime, 'yyyy-MONTH')
order by numberofflights desc;
Related
I have a dataset that's just a list of orders made by customers each day.
order_date
month
week
customer
2022-10-06
10
40
Paul
2022-10-06
10
40
Edward
2022-10-01
10
39
Erick
2022-09-26
9
39
Divine
2022-09-23
9
38
Alice
2022-09-21
9
38
Evelyn
My goal is to calculate the total number of unique customers within a two-week period. I can count the number of customers within a month or week period but not two weeks. Also, the two weeks are in a rolling order such that weeks 40 and 39 (as in the sample above) is one window period while weeks 39 and 38 is the next frame.
So far, this is how I am getting the monthly and weekly numbers. Assume that the customer names are distinct per day.
select order_date,
month,
week,
COUNT(DISTINCT customer) over (partition by month) month_active_outlets,
COUNT(DISTINCT customer) OVER (partition by week) week active outlets,
from table
Again, I am unable to calculate the unique customer names within a two-week period.
I think the easiest would be to create your own grouper in a subquery and then use that to get to your count. Currently, COUNT UNIQUE and ORDER BY in the window is not supported, therefore that approach wouldn't work.
A possible query could be:
WITH
week_before AS (
SELECT
EXTRACT(WEEK from order_date) as week, --to be sure this is the same week format
month,
CONCAT(week,'-', EXTRACT(WEEK FROM DATE_SUB(order_date, INTERVAL 7 DAY))) AS two_weeks,
customer
FROM
`test`.`Basic`)
SELECT
two_weeks,
COUNT(DISTINCT customer) AS unique_customer
FROM
week_before
GROUP BY
two_weeks
The window function is the right tool. To obtain the 2 week date, we first extract the week number of the year:
mod(extract(week from order_date),2)
If the week number is odd (modulo 2) we add a week. Then we trunc to the start of (the even) week.
date_trunc(date_add(order_date,interval mod(extract(week from order_date),2) week),week )
with tbl as
(Select date("2022-10-06") as order_date, "Paul" as customer
union all select date("2022-10-06"),"Edward"
union all select date("2022-10-01"),"Erick"
union all select date("2022-09-26"),"Divine"
union all select date("2022-09-23"),"Alice"
union all select date("2022-09-21"),"Evelyn"
)
select *,
date_trunc(order_date,month) as month,
date_trunc(order_date,week) as week,
COUNT(DISTINCT customer) OVER week2 as customer_2weeks,
string_agg(cast(order_date as string)) over week2 as list_2weeks,
from tbl
window week2 as (partition by date_trunc(date_add(order_date,interval mod(extract(week from order_date),2) week),week ))
The first days of a year are counted to the last week of the previous year:
select order_date,
extract(isoweek from order_date),
date_trunc(date_add(order_date,interval mod(extract(week from order_date),2) week),week)
from
unnest(generate_date_array(date("2021-12-01"),date("2023-01-14"))) order_date
order by 1
i am trying to work with this query to produce a list of all 11 years and 12 months within the years with the sales data for each month. Any suggestions? this is my query so far.
SELECT
distinct(extract(year from date)) as year
, sum(sale_dollars) as year_sales
from `project-1-349215.Dataset.sales`
group by date
it just creates a long list of over 2000 results when i am expecting 132 max one for each month in the years.
You should change your group by statement if you have more results than you expected.
You can try:
group by YEAR(date), MONTH(date)
or
group by EXTRACT(YEAR_MONTH FROM date)
A Grouping function is for takes a subsection of the date in your case year and moth and collect all rows that fit, and sum it up,
So a s´GROUp BY date makes no sense, what so ever as you don't want the sum of every day
So make this
SELECT
extract(year from date) as year
,extract(MONTH from date) as month
, sum(sale_dollars) as year_sales
from `project-1-349215.Dataset.sales`
group by 1,2
Or you can combine both year and month
SELECT
extract(YEAR_MONTH from date) as year
, sum(sale_dollars) as year_sales
from `project-1-349215.Dataset.sales`
group by 1
I have a table in Presto with this schema:
created_at Record
timestamp String
created_at has records from 2020 to 2022.
What's the best way to get the total number of record by month, like this output:
Date N_Records
2020-01 1000
2020-02 1500
----
2022-03 3000
What I did so far:
select date_format(created_at, '%b') month, count(*) count
from table
group by date_format(created_at, '%b')
order by 1 asc
Problems with my code:
I don't have the respective year and the results are not sorted by asc month.
Can someone help with to improve my query?
You can use format including 4-digit year and 2-digit month:
select date_format(created_at, '%Y-%m') Date, count(*) N_Records
from table
group by 1
order by 1 asc
This is what my table looks like:
NOTE: Don't worry about the BMI field being empty in some rows. We assume that each row is a reading. I have omitted some columns for privacy reasons.
I want to get a count of the number of active customers per month. A customer is active if they have at least 18 readings in total (1 reading per day for 18 days in a given month). How do I write this SQL query? Assume the table name is 'cust'. I'm using SQL Server. Any help is appreciated.
Presumably a patient is a customer in your world. If so, you can use two levels of aggregation:
select yyyy, mm, count(*)
from (select year(createdat) as yyyy, month(createdat) as mm,
patient_id,
count(distinct convert(date, createdat)) as num_days
from t
group by year(createdat), month(createdat), patient_id
) ymp
where num_days >= 18
group by yyyy, mm;
You need to group by patient and the month, then group again by just the month
SELECT
mth,
COUNT(*) NumPatients
FROM (
SELECT
EOMONTH(c.createdat) mth
FROM cust c
GROUP BY EOMONTH(c.createdat), c.patient_id
HAVING COUNT(*) >= 18
-- for distinct days you could change it to:
-- HAVING COUNT(DISTINCT CAST(c.createdat AS date)) >= 18
) c
GROUP BY mth;
I'm trying to find the number of employees joined over a calender year, broken down on a monthly basis. So if 15 employees had joined in January, 30 in February and so on, the output I'd like would be
Month | Employees
------|-----------
Jan | 15
Feb | 30
I've come up with a query to fetch it for a particular month
SELECT * FROM (
SELECT COUNT(EMP_NO), EMP_JN_DT
FROM EMP_REG WHERE
EMP_JN_DT between '01-NOV-09' AND '30-NOV-09'
GROUP BY EMP_JN_DT )
ORDER BY 2
How do I extend this for the full calender year?
SELECT Trunc(EMP_JN_DT,'MM') Emp_Jn_Mth,
Count(*)
FROM EMP_REG
WHERE EMP_JN_DT between date '2009-01-01' AND date '2009-12-31'
GROUP BY Trunc(EMP_JN_DT,'MM')
ORDER BY 1;
If you do not have anyone join in a particular month then you'd get no row returned. To over come this you'd have to outerjoin the above to a list of months in the required year.
SELECT to_date(EMP_JN_DT,'MON') "Month", EMP_NO "Employees"
FROM EMP_REG
WHERE EMP_JN_DT between date '2009-01-01' AND date '2009-12-31'
GROUP by "Month"
ORDER BY 1;
http://www.techonthenet.com/oracle/functions/extract.php
There is a function that returns month. What you need to do is just put it in group by
The number of employees in January can be selected in the following way:
SELECT EXTRACT(MONTH FROM HIREDATE) AS MONTH1, COUNT(*)
FROM employee
WHERE EXTRACT(MONTH FROM HIREDATE)=1
GROUP BY EXTRACT(MONTH FROM HIREDATE)