I have the following query:
set #cumulativeSum := 0;
select
(#cumulativeSum:= #cumulativeSum + (count(distinct `ce`.URL, `ce`.`IP`))) as `uniqueClicks`,
cast(`ce`.`dt` as date) as `createdAt`
from (SELECT DISTINCT min((date(CODE_EVENTS.CREATED_AT))) dt, CODE_EVENTS.IP, CODE_EVENTS.URL
FROM CODE_EVENTS
GROUP BY CODE_EVENTS.IP, CODE_EVENTS.URL) as ce
join ATTACHMENT on `ce`.URL = ATTACHMENT.`ID`
where ATTACHMENT.`USER_ID` = 6
group by cast(`ce`.`dt` as date)
ORDER BY ce.URL;
It works almost ok, I would like to have as result set a date and amount of cumulative sum as uniqueClicks, the problem is that in my result set it is not added up together.
uniqueClicks createdAt
1 2018-02-01
3 2018-02-03
1 2018-02-04
and I'd like to have
uniqueClicks createdAt
1 2018-02-01
4 2018-02-03
5 2018-02-04
I believe you can obtain a rolling sum of the unique clicks without needing to resort to dynamic SQL:
SELECT
t1.CREATED_AT,
(SELECT SUM(t2.uniqueClicks) FROM
(
SELECT CREATED_AT, COUNT(DISTINCT IP, URL) uniqueClicks
FROM CODE_EVENTS
GROUP BY CREATED_AT
) t2
WHERE t2.CREATED_AT <= t1.CREATED_AT) uniqueClicksRolling
FROM
(
SELECT DISTINCT CREATED_AT
FROM CODE_EVENTS
) t1
ORDER BY t1.CREATED_AT;
The subquery aliased as t2 computes the number of unique clicks on each given day which appears in your table. The distinct count of IP and URL is what determines the number of clicks. We can then subquery this intermediate table and sum clicks for all days up and including the current date. This is essentially cursor style action, and can replace your use of session variables.
Related
Hi i have a table like so it includes a user id, date, amount and active flag
id
date
amount
active
1001
2017-07-12
10
1
1001
2017-07-12
5
0
1001
2017-07-12
12
0
1001
2017-05-05
5
0
1001
2017-06-01
11
0
my requirement is to get the total amount for this particular user for the whole day that he was active, so since the user was active on the date of '2017-07-12' i should be able to get all the amount for that particular date so my amount for this particular user would be 27.
What would be a right query to perform this action in mysql by looking at the active flag and how would i go about to do it?
We can use an aggregation approach here:
SELECT id, SUM(amount) AS total_amount
FROM
(
SELECT *
FROM yourTable t1
WHERE active = 1 OR
EXISTS (SELECT 1
FROM yourTable t2
WHERE t2.date = t1.date AND
t2.active = 1)
) t
GROUP BY id
ORDER BY id;
Demo
Either
SELECT id, `date`, SUM(amount) amount
FROM table
GROUP BY 1, 2
HAVING SUM(active)
or
SELECT id, `date`, SUM(amount) * (SUM(active) > 0) amount
FROM table
GROUP BY 1, 2
depends on desired output (does "non-active" dates must be skipped at all, or they'd be returned with zero amount).
Another solution:
select tbl.id,
tbl.`date`,
sum(amount) as tot_date_amount
from tbl
inner join (select `date`
from tbl
where active = 1
) as t2 on tbl.`date`=t2.`date`
group by id,`date`;
https://dbfiddle.uk/?rdbms=mysql_8.0&fiddle=c173dcc9a72490146ff7c094a10b08b6
The subquery will select only the dates where active = 1 .Using inner join will return only the sum for the active = 1 dates
price | date | product_id
100 | 2020-09-21 | 1
400 | 2020-09-20 | 2
300 | 2020-09-20 | 3
200 | 2020-09-19 | 1
400 | 2020-09-18 | 2
I add an entry into this table every day with a product's price that day.
Now I want to get most price drops for the last week (all dates up to 2020-09-14), in this example it would only return the product_id = 1, because that's the only thing that changed.
I think I have to join the table to itself, but I'm not getting it to work.
Here's something that I wanted to return the most price changes over the last day, however it's not working.
select pt.price, pt.date, pt.product_id, (pt.price - py.price) as change
from prices as pt
inner join (
select *
from prices
where date > '2020-09-20 19:33:43'
) as py
on pt.product_id = py.product_id
where pt.price - py.price > 0
order by change
I understand that you want to count how many times the price of each product changed over the last 7 days.
A naive approach would use aggregation and count(distinct price) - but it fails when a product's price changes back and forth.
A safer approach is window functions: you can use lag() to retrieve the previous price, and compare it against the current price; it is then easy to aggregate and count the price changes:
select product_id, sum(price <> lag_price) cnt_price_changes
from (
select t.*, lag(price) over(partition by product_id order by date) lag_price
from mytable t
where date >= current_date - interval 7 day
) t
group by product_id
order by price_changes desc
Try using MAX() and MIN() instead....
select MAX(pt.price), MIN(pt.price), MAX(pt.price) - MIN(pt.price) as change
from prices as pt
inner join (
select *
from prices
where date > '2020-09-20 19:33:43'
) as py
on pt.product_id = py.product_id
order by change
Instead of subtracting every row by every other row to get the result, you can find the max and min's easily by means of MAX() and MIN(), and, ultimately, **MAX() - MIN()**. Relevant lines from the linked MySQL documentation...
MAX(): Returns the maximum value of expr.
MIN(): Returns the minimum value of expr.
You won't be able to pull the other fields (id's, dates) since this is a GROUP BY() implied by the MAX() and MIN(), but you should then be able to get that info by query SELECT * FROM ... WHERE price = MAX_VALUE_JUST_ACQUIRED.
This examples will get you results per WeekOfYear and WeekOfMonth regarding the lowering of the price per product.
SELECT
COUNT(m1.product_id) as total,
m1.product_id,
WEEK(m1.ddate) AS weekofyear
FROM mytest m1
WHERE m1.price < (SELECT m2.price FROM mytest m2 WHERE m2.ddate<m1.ddate AND m1.product_id=m2.product_id LIMIT 0,1)
GROUP BY weekofyear,m1.product_id
ORDER BY weekofyear;
SELECT
COUNT(m1.product_id) as total,
m1.product_id,
FLOOR((DAYOFMONTH(ddate) - 1) / 7) + 1 AS weekofmonth
FROM mytest m1
WHERE m1.price < (SELECT m2.price FROM mytest m2 WHERE m2.ddate<m1.ddate AND m1.product_id=m2.product_id LIMIT 0,1)
GROUP BY weekofmonth,m1.product_id
ORDER BY weekofmonth;
Try this out in SQLFiddle.
I have done an unsuccessful attempt to merge values from two different tables.The first one displays just as I would like but the 2nd one displays only the first row every time.
Select * From (Select date as Date, Sum(qty) as qtySum, Sum(weight)
as weightSum From stock_list Group by date) as A,
(Select Sum(weight) as weightSum,Count(barcode)
as barcodeCount From selected_items Group by date) as B Group by date;
This is the output that I get.
These is my selected_items table.
This is my stock_list.
Both my queries work individually and I get the correct output only when I try to run them together it gives a problem for the 2nd query.Can anyone point out my mistake or show me a better way to do it.
This is what my final objective is
The first problem is that you are grouping by date in your subquery B, but you don't select it, so your result set might be something like:
weightSum barcodeCount
---------------------------
26 8
9 14
4 7
This is the result for 3 dates, but you have no idea which date which row refers to.
Your next problem is that you are using a cross join, because there is no link between your two queries, this means if your first query returns:
Date qtySum weightSum
----------------------------------------
2016-01-20 1 1
2016-01-21 2 2
After you have done this cross join you end up:
Date qtySum a.weightSum b.weightSum barcodeCount
--------------------------------------------------------------------------
2016-01-20 1 1 26 8
2016-01-20 1 1 9 14
2016-01-20 1 1 4 7
2016-01-21 2 2 26 8
2016-01-21 2 2 9 14
2016-01-21 2 2 4 7
So every row from A is matched with every row from B giving you 6 total rows.
Your third problem is that you then group by date, but don't perform any aggregates, without delving too much into the fine print of the SQL Standard, the group by clause, and functional dependency, lets simplify it to MySQL allows this, but you shouldn't do it unless you understand the limitations (This is covered in more detail on this in this answer). Anything in the select that is not in a group by clause should probably be within an aggregate.
So, due to MySQL's GROUP BY Extension by selecting everything and grouping only by date, what you are effectively saying is take 1 row per date, but you have no control over which row, it might be the first row from each group as displayed above, so the result you would get is:
Date qtySum a.weightSum b.weightSum barcodeCount
--------------------------------------------------------------------------
2016-01-20 1 1 26 8
2016-01-21 2 2 26 8
Which I think is why you are ending up with all the same values from the subquery B repeated.
So that covers what is wrong, now on to a solution, assuming there will be dates in stock_list that don't exist in selected_items, and vice versa you would need a full join, but since this is not supported in MySQL you would have to use UNION, the simplest way would be:
SELECT t.Date,
SUM(t.StockQuantity) AS StockQuantity,
SUM(t.StockWeight) AS StockWeight,
SUM(t.SelectedWeight) AS SelectedWeight,
SUM(t.BarcodeCount) AS BarcodeCount
FROM ( SELECT date,
SUM(qty) AS StockQuantity,
SUM(weight) AS StockWeight,
0 AS SelectedWeight,
0 AS BarcodeCount
FROM stock_list
GROUP BY Date
UNION ALL
SELECT date,
0 AS StockQuantity,
0 AS StockWeight,
SUM(weight) AS SelectedWeight,
COUNT(BarCode) AS BarcodeCount
FROM selected_items
GROUP BY Date
) AS t
GROUP BY t.Date;
EDIT
I can't test this, nor am I sure of your exact logic, but you can use variables to calculate a running total in MySQL. This should give an idea of how to do it:
SELECT Date,
StockQuantity,
StockWeight,
SelectedWeight,
BarcodeCount,
(#w := #w + StockWeight - SelectedWeight) AS TotalWeight,
(#q := #q + StockQuantity - BarcodeCount) AS TotalQuantity
FROM ( SELECT t.Date,
SUM(t.StockQuantity) AS StockQuantity,
SUM(t.StockWeight) AS StockWeight,
SUM(t.SelectedWeight) AS SelectedWeight,
SUM(t.BarcodeCount) AS BarcodeCount
FROM ( SELECT date,
SUM(qty) AS StockQuantity,
SUM(weight) AS StockWeight,
0 AS SelectedWeight,
0 AS BarcodeCount
FROM stock_list
GROUP BY Date
UNION ALL
SELECT date,
0 AS StockQuantity,
0 AS StockWeight,
SUM(weight) AS SelectedWeight,
COUNT(BarCode) AS BarcodeCount
FROM selected_items
GROUP BY Date
) AS t
GROUP BY t.Date
) AS t
CROSS JOIN (SELECT #w := 0, #q := 0) AS v
GROUP BY t.Date;
You could use a join. However, if the set of dates is not the same in both tables, then you would want a full outer join. But that is not available in MySQL. Instead:
select date, sum(qtySum), sum(weightsum1), sum(weightsum2), sum(barcodeCount)
from ((Select date as Date, Sum(qty) as qtySum, Sum(weight) as weightSum1,
NULL as weightsum2, NULL as barcodeCount
From stock_list
Group by date
) union all
(Select date, null, null, Sum(weight), Count(barcode) as barcodeCount
From selected_items
Group by date
)
) t
Group by date;
I'm not sure how your desired output corresponds to the query you have provided. But this should aggregate and combine the data from the two tables by date, so you can finalize the query.
You can use FULL JOIN operator to solve your task if some dates may not exists in both tables.
Select ISNULL(A.date, B.date) AS date,
A.qtySum,
A.weightSum,
B.weightSum,
B.barcodeCount
From
(Select date as Date,
Sum(qty) as qtySum,
Sum(weight) as weightSum
From stock_list
Group by date) as A
FULL JOIN
(Select date,
Sum(weight) as weightSum,
Count(barcode) as barcodeCount
From selected_items
Group by date) as B ON A.date = B.date
If I have a table(Oracle or MySQL), which stores the date user logins.
So how can I write a SQL(or something else) to find the users who have continuously login for n days.
For example:
userID | logindate
1000 2014-01-10
1000 2014-01-11
1000 2014-02-01
1000 2014-02-02
1001 2014-02-01
1001 2014-02-02
1001 2014-02-03
1001 2014-02-04
1001 2014-02-05
1002 2014-02-01
1002 2014-02-03
1002 2014-02-05
.....
We can see that user 1000 has continually logined for two days in 2014, and user 1001 has continually logined for 5 days. and user 1002 never continuously logins.
The SQL should be extensible , which means I can pick every number of n, and modify a little or pass a new parameter, and the results is as expected.
Thank you!
As we don't know what dbms you are using (you named both MySQL and Oracle), here are are two solutions, both doing the same: Order the rows and subtract rownumber days from the login date (so if the 6th record is 2014-02-12 and the 7th is 2014-02-13 they both result in 2014-02-06). So we group by user and that groupday and count the days. Then we group by user to find the longest series.
Here is a solution for a dbms with analytic window functions (e.g. Oracle):
select userid, max(days)
from
(
select userid, groupday, count(*) as days
from
(
select
userid, logindate - row_number() over (partition by userid order by logindate) as groupday
from mytable
)
group by userid, groupday
)
group by userid
--having max(days) >= 3
And here is a MySQL query (untested, because I don't have MySQL available):
select
userid, max(days)
from
(
select
userid, date_add(logindate, interval -row_number day) as groupday, count(*) as days
from
(
select
userid, logindate,
#row_num := #row_num + 1 as row_number
from mytable
cross join (select #row_num := 0) r
order by userid, logindate
)
group by userid, groupday
)
group by userid
-- having max(days) >= 3
I think the following query will give you a very extensible parametrization:
select z.userid, count(*) continuous_login_days
from
(
with max_dates as
( -- Get max date for every user ID
select t.userid, max(t.logindate) max_date
from test t
group by t.userid
),
ranks as
( -- Get ranks for login dates per user
select t.*,
row_number() over
(partition by t.userid order by t.logindate desc) rnk
from test t
)
-- So here, we select continuous days by checking if rank inside group
-- (per user ID) matches login date compared to max date
select r.userid, r.logindate, r.rnk, m.max_date
from ranks r, max_dates m
where m.userid = r.userid
and r.logindate + r.rnk - 1 = m.max_date -- here is the key
) z
-- Then we only group by user ID to get the number of continuous days
group by z.userid
;
Here is the result:
USERID CONTINUOUS_LOGIN_DAYS
1 1000 2
2 1001 5
3 1002 1
So you can just choose by querying field CONTINUOUS_LOGIN_DAYS.
EDIT : If you want to choose from all ranges (not only the last one), my query structure no longer works because it relied on the last range. But here is a workaround:
with w as
( -- Parameter
select 2 nb_cont_days from dual
)
select *
from
(
select t.*,
-- Get number of days around
(select count(*) from test t2
where t2.userid = t.userid
and t2.logindate between t.logindate - nb_cont_days + 1
and t.logindate) m1,
-- Get also number of days more in the past, and in the future
(select count(*) from test t2
where t2.userid = t.userid
and t2.logindate between t.logindate - nb_cont_days
and t.logindate + 1) m2,
w.nb_cont_days
from w, test t
) x
-- If these 2 fields match, then we have what we want
where x.m1 = x.nb_cont_days
and x.m2 = x.nb_cont_days
order by 1, 2
You just have to change the parameter in the WITH clause, so you can even create a function from this query to call it with this parameter.
SELECT userID,count(userID) as numOfDays FROM LOGINTABLE WHERE logindate between '2014-01-01' AND '2014-02-28'
GROUP BY userID
In this case you can check the login days per user, in a specific period
I want to make a MySQL to get daily differential values from a table who looks like this:
Date | VALUE
--------------------------------
"2011-01-14 19:30" | 5
"2011-01-15 13:30" | 6
"2011-01-15 23:50" | 9
"2011-01-16 9:30" | 10
"2011-01-16 18:30" | 15
I have made two subqueries. The first one is to get the last daily value, because I want to compute the difference values from this data:
SELECT r.Date, r.VALUE
FROM table AS r
JOIN (
SELECT DISTINCT max(t.Date) AS Date
FROM table AS t
WHERE t.Date < CURDATE()
GROUP BY DATE(t.Date)
) AS x USING (Date)
The second one is made to get the differential values from the result of the first one (I show it with "table" name):
SELECT Date, VALUE - IFNULL(
(SELECT MAX( VALUE )
FROM table
WHERE Date < t1.table) , 0) AS diff
FROM table AS t1
ORDER BY Date
At first, I tried to save the result of first query in a temporary table but it's not possible to use temporary tables with the second query. If I use the first query inside the FROM of second one between () with an alias, the server complaints about table alias doesn't exist. How can get a something like this:
Date | VALUE
---------------------------
"2011-01-15 00:00" | 4
"2011-01-16 00:00" | 6
Try this query -
SELECT
t1.dt AS date,
t1.value - t2.value AS value
FROM
(SELECT DATE(date) dt, MAX(value) value FROM table GROUP BY dt) t1
JOIN
(SELECT DATE(date) dt, MAX(value) value FROM table GROUP BY dt) t2
ON t1.dt = t2.dt + INTERVAL 1 DAY