I have a table that has 3 columns: user_id, date, amount. I need to find out on which date the amount reached 1 Million for the first time. The amount can go up or down on any given day.
I tried using partition by user_id order by date desc but I can't figure out how to find the exact date on which it reached 1 Million for the first time. I am exploring lead, lag functions. Any pointers would be appreciated.
You may use conditional aggregation as the following:
select user_id,
min(case when amount >= 1000000 then date end) as expected_date
from table_name
group by user_id
And if you want to check where the amount reaches exactly 1M, use case when amount = 1000000 ...
If you meant that the amount is a cumulative amount over the increasing of date, then query will be:
select user_id,
min(case when cumulative_amount >= 1000000 then date end) as expected_date
from
(
select *,
sum(amount) over (partition by user_id order by date) cumulative_amount
from table_name
) T
group by user_id;
Try this:
select date,
sum(amount) as totalamount
from tablename
group by date
having totalamount>=1000000
order by date asc
limit 1
This would summarize the amount for each day and return 1 record where it reached 1M for the first time.
Sample result on SQL Fiddle.
And if you want it to be grouped for both date and user_id, add user_id in select and group by clauses.
select user_id, date,
sum(amount) as totalamount
from tablename
group by user_id,date
having totalamount>=1000000
order by date asc
limit 1
Example here.
Related
Using SQL Server Management Studio.
Let's say I have a table with transactions that contains User, Date, Transaction amount. I want a query that will return the date when a certain amount is reached - let's say 100.
For example the same user performs 10 transactions for 10 EUR. I want the query to select the date of the last transaction because that's when his volume reached 100. Of course, once 100 is reached, the query shouldn't change the date with the latest transaction anymore, but leave it at when 100 was reached.
Wrote this on pgadmin but I think syntax should be the same.
with cumulative as
(
select customer_id,
sum(amount) over (partition by customer_id order by payment_date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) cum_amt,
payment_date
from payment
)
select customer_id
, min(payment_date) as threshold_reached
from cumulative
where cum_amt>=100
group by customer_id
case when sum(amt) over (partition by user order by date) - amt < 100
and sum(amt) over (partition by user order by date) >= 100
then 1 else 0 end
I have written the sql query:
SELECT id
date_diff("day", create_date, date) as day
action_type
FROM "my_database"
It brings this:
id day action_type
1 0 upload
1 0 upload
1 0 upload
1 1 upload
1 1 upload
2 0 upload
2 0 upload
2 1 upload
How to change my query to get table with unique days in column day and average number "upload" action_type among all id's. So desired result must look like this:
day avg_num_action
0 2.5
1 1.5
It is 2.5, because (3+2)/2 (3 uploads of id:1 and 2 uploads for id:2). same for 1.5
Please try this. Consider your given query as a table. If any WHERE condition needed then please enable this other wise disable where clause.
SELECT t.day
, COUNT(*) / COUNT(DISTINCT t.id) avg_num_action
FROM (SELECT id,
date_diff("day", create_date, date) as day,
action_type
FROM "my_database") t
WHERE t.action_type = 'upload'
GROUP BY t.day
Create a table from your given result set and write query based on that.
SELECT t.tday
, COUNT(*) / COUNT(DISTINCT t.id) avg_num_action
FROM my_database t
GROUP BY t.tday
Please check from url https://dbfiddle.uk/?rdbms=mysql_8.0&fiddle=871935ea2b919c4e24eb83fcbce78973
Update: I think my two-steps approach is more complicated than needed. Rahul Biswas shows how this can be done in one step. I suggest you use and accept his answer.
Original answer:
Two steps:
Count entries per ID and day
Take the average count per day
The query:
with rows as (select id, date_diff('day', create_date, date) as day from mytable)
, per_id_and_day as (select id, day, count(*) as cnt from rows group by id, day)
select day, avg(cnt)
from per_id_and_day
group by day
order by day;
You don't need a subquery for this logic:
SELECT date_diff("day", create_date, date) as day,
COUNT(*) * 1.0 / COUNT(DISTINCT id)
FROM "my_database"
GROUP BY date_diff("day", create_date, date)
My SQL isn't the best - I can get this working in C# but it seems more efficient to get it in my data layer - I've got a table Prices:
ID
Price
DateTime
Each row is exactly 1 hour from the next, so I have a snapshot of a price every hour.
I'm trying to work out which hour in a day over the entire dataset has the lowest price (on average).
So ideally I'm after a list of each hour in the day ranked by how cheap on average that hour is over the entire dataset - so a maximum of 24 rows (one for each hour in the day).
Any help would be greatly appreciated!
Thanks :D
Which database are you on?
Different DBs have different ways to extract date from a timestamp column.
Postgres has date(timestamp), In Oracle, you can use trunc(timestamp). Or most DBs have to_char/to_date. So you can try that.
Once you have extracted the date, you can try something like this -
select ID,
Price,
DateTime,
trunc(DateTime) as day,
rank() over (partition by trunc(DateTime) order by Price asc) as least_for_day
from Prices
Now you can use the "least_for_day" ranked column and select by day.
Again, depending on the DB, you can either directly qualify on the ranked column in the same SQL or use the above as a sub-query and filter for the rank.
You can use a query like below
select
hour,
avg(daily_rank) avg_rank
from
(
select *, hour= format((datetime as datetime),'HH'), daily_rank= dense_rank() over (partition by cast(datetime as date) order by price asc)
) t
group by hour
Thank you very much to #Many Manjunath and #DhruvJoshi. Final solution below;
WITH prices AS
(
SELECT
[Price],
[DateTime],
CAST([DateTime] AS TIME) 'Time',
CAST([DateTime] as date) 'Date',
rank() over (partition by cast([DateTime] as date) order by [Price] asc) as least_for_day
FROM [dbo].[Prices]
)
SELECT [Time], count(*) 'Qty Cheapest' FROM prices
WHERE least_for_day = 1
GROUP BY [Time]
ORDER BY 2 DESC
That returns 24 rows:
I like to write a sql query that counts the number of days each user used the application and how many concurrent days. A user can enter the app several times a day but that should count as 1.
My table looks like this:
id | bigint
user_id | bigint
action_date | timestamp without time zone
To count the number of days per user:
SELECT user_id, count(DISTINCT action_date::date) AS days
FROM user_action_tbl
GROUP BY user_id;
One way to do it
SELECT user_id, COUNT(*) days_total, SUM(conseq) days_consecutive
FROM
(
SELECT user_id,
CASE WHEN LEAD(date, 1) OVER (PARTITION BY user_id ORDER BY date) - date = 1 THEN 1 ELSE 0 END consecutive
FROM
(
SELECT user_id, action_date::date date
FROM table1
GROUP BY user_id, action_date::date
) q
) p
GROUP BY user_id
Here is a SQLFiddle demo
I have a database of transactions, accounts, profit/loss, and date. I need to find the dates which the largest profit occurs by account. I have already found a way to find these actually max/min values but I can't seem to be able to pull the actual date from it. My code so far is like this:
Select accountnum, min(ammount)
from table
where date > '02-Jan-13'
group by accountnum
order by accountnum
Ideally I would like to see account num, the min or max, and then the date which this occurred on.
Try something like this to get the min and max amount for each customer and the date it happened.
WITH max_amount as (
SELECT accountnum, max(amount) amount, date
FROM TABLE
GROUP BY accountnum, date
),
min_amount as (
SELECT accountnum, min(amount) amount, date
FROM TABLE
GROUP BY accountnum, date
)
SELECT t.accountnum, ma.amount, ma.date, mi.amount, ma.date
FROM table t
JOIN max_amount ma
ON ma.accountnum = t.accountnum
JOIN min_amount mi
ON mi.accountnum = t.accountnum
If you want the data for just this year you could add a where clause to the end of the statement
WHERE t.date > '02-Jan-13'
The easiest way to do this is using window/analytic functions. These are ANSI standard and most databases support them (MySQL and Access being two notable exceptions).
Here is one way:
select t.accountnum, min_amount, max_amount,
min(case when amount = min_amount then date end) as min_amount_date,
min(case when amount = min_amount then date end) as max_amount_date,
from (Select t.*,
min(amount) over (partition by accountnum) as min_amount,
max(amount) over (partition by accountnum) as max_amount
from table t
where date > '02-Jan-13'
) t
group by accountnum, min_amount, max_amount;
order by accountnum
The subquery calculates the minimum and maximum amount for each account, using min() as a window function. The outer query selects these values. It then uses conditional aggregation to get the first date when each of those values occurred.
;with cte as
(
select accountnum, ammount, date,
row_number() over (partition by accountnum order by ammount desc) rn,
max(ammount) over (partition by accountnum) maxamount,
min(ammount) over (partition by accountnum) minamount
from table
where date > '20130102'
)
select accountnum,
ammount as amount,
date as date_of_max_amount,
minamount,
maxamount
from cte where rn = 1