I need to select content statistics group By Date.
Here example of records :
id cid viewCount created_at
1 1 50 31-12-2018 18:00:00
2 1 50 01-01-2019 18:00:00
3 2 50 01-01-2019 18:00:00
4 2 100 01-01-2019 19:00:00
5 2 150 01-01-2019 20:00:00
6 3 1000 01-01-2019 15:00:00
Need to return :
id cid viewCount date
1 1 50 31-12-2018
2 1 50 01-01-2019
5 2 150 01-01-2019
6 3 1000 01-01-2019
I tried the following code
$qb = $this->createQueryBuilder('c');
$qb->select('a.id as id')
->addSelect('COALESCE(SUM(a.viewCount),0) as viewCount')
->addSelect('DATE_FORMAT(a.createdAt, \'%d-%m-%Y\') as date');
->innerJoin('c.analytics', 'a')
->groupBy('c.cid')
->addGroupBy('date')
->orderBy('a.createdAt', 'ASC');
return:
id cid viewCount date
1 1 50 31-12-2018
2 1 50 01-01-2019
3 2 50 01-01-2019
4 2 100 01-01-2019
5 2 150 01-01-2019
6 3 1000 01-01-2019
I have tried to create a subquery :
$qbLastHour = $this->createQueryBuilder('cc');
$qbLastHour->select('MAX(DATE_FORMAT(aa.createdAt, \'%H\'))')
->innerJoin('cc.analytics', 'aa')
->where('cc.id=c.id')
->groupBy('cc.cid')
->addGroupBy('s');
$qb->addSelect(sprintf("(%s) AS r", $qbLastHour->getDQl()));
But something go wrong because i dont groupBy date at the subquery.
If someone can help me. Thank you
Update
Here is an attempt, in sql again, to select only one row per date and cid based on the max time per day
SELECT id, c.cid, viewCount, max_date
FROM content a
JOIN content_analytic c ON a.id = c.content_id
RIGHT JOIN (SELECT c.cid, DATE_FORMAT(created_at, '%d-%m-%Y') dt, MAX(created_at) max_date
FROM content a
JOIN content_analytic c ON a.id = c.content_id
GROUP BY dt, c.cid) x ON x.max_date = a.created_at and x.cid = c.cid
This is how I believe the query should be in pure sql
SELECT c.cid, COALESCE(SUM(a.viewCount), 0), DATEFORMAT(a.created_at, ‘%d-%m-%Y’) as date
FROM content a
INNER JOIN content_analytic c ON a.id = c.content_id
GROUP BY c.cid, date
ORDER BY date
Related
I have a table 'customer' which contains 4 columns
name day product price
A 2021-04-01 p1 100
B 2021-04-01 p1 100
C 2021-04-01 p2 120
A 2021-04-01 p2 120
A 2021-04-02 p1 100
B 2021-04-02 p3 80
C 2021-04-03 p2 120
D 2021-04-03 p2 120
C 2021-04-04 p1 100
With a command
SELECT COUNT(name)
FROM (SELECT name
FROM customer
WHERE day > '2021-03-28'
AND day < '2021-04-09'
GROUP BY name
HAVING COUNT(name) > 2)
I could count number of customer that bought something more than twice in a period of time.
I would like to know in each day (GROUP BY over day) how many customers bought something with this condition that in a period they bought something more than twice.
Suggested Edit:
For above example A and C are valid agents by the condition.
The desired output will be:
day how_many
2021-04-01 2
2021-04-02 1
2021-04-03 1
2021-04-04 1
I interpret your question as wanting to know how many customers made more than one purchase on each day. If so, one method uses two levels of aggregation:
select day,
sum(case when day_count >= 2 then 1 else 0 end)
from (select c.name, c.day, count(*) as day_count
from customer c
group by c.name, c.day
) nc
group by day
order by day;
I'm trying to add new columns of first values of the day for location and weight.
For instance, the original data format is:
id dttm location weight
--------------------------------------------
1 1/1/20 11:10:00 A 40
1 1/1/20 19:07:00 B 41.1
2 1/1/20 08:01:00 B 73.2
2 1/1/20 21:00:00 B 73.2
2 1/2/20 10:03:00 C 74
I want each id to have only one day record, such as:
id dttm location weight
--------------------------------------------
1 1/1/20 11:10:00 A 40
2 1/1/20 08:01:00 B 73.2
2 1/2/20 10:03:00 C 74
I have other columns in my data set that I'm using location and weight to create, so I don't think I can just filter for 'first' records of the day.. Is it possible to write query to recognize first record of the day for those two columns and create new column with those values?
You can use row_number():
select t.*
from (select t.*,
row_number() over (partition by id, ddtm::date order by dttm) as seqnum
from t
) t
where seqnum = 1;
I have a table called Inventory with the below columns
item warehouse date sequence number value
111 100 2019-09-25 12:29:41.000 1 10
111 100 2019-09-26 12:29:41.000 1 20
222 200 2019-09-21 16:07:10.000 1 5
222 200 2019-09-21 16:07:10.000 2 10
333 300 2020-01-19 12:05:23.000 1 4
333 300 2020-01-20 12:05:23.000 1 5
Expected Output:
item warehouse date sequence number value
111 100 2019-09-26 12:29:41.000 1 20
222 200 2019-09-21 16:07:10.000 2 10
333 300 2020-01-20 12:05:23.000 1 5
Based on item and warehouse, i need to pick latest date and latest sequence number of value.
I tried with below code
select item,warehouse,sequencenumber,sum(value),max(date) as date1
from Inventory t1
where
t1.date IN (select max(date) from Inventory t2
where t1.warehouse=t2.warehouse
and t1.item = t2.item
group by t2.item,t2.warehouse)
group by t1.item,t1.warehouse,t1.sequencenumber
Its working for latest date but not for latest sequence number.
Can you please suggest how to write a query to get my expected output.
You can use row_number() for this:
select *
from (
select
t.*,
row_number() over(
partition by item, warehouse
order by date desc, sequence_number desc, value desc
) rn
from mytable t
) t
where rn = 1
There are two tables.
In the first I have columns:
id - a person
time - the time of receiving the bonus (timestamp)
money - size of bonus
And the second:
id
time - time of getting a rank (timestamp)
range - military rank (int)
The task is to withdraw the amount and number of bonuses received by people in the rank of captain (range = 7) with aggregation by day.
I have no ideas how to do a table with this data. I can summarize data by all days such as
SELECT DISTINCTROW Payment.user_id AS user_id, Sum(IIf(IsNull(Payment.money),0,Payment.money)) AS [Sum - money], Count(Payment.money) AS [Count - Payment], Format(Payment.time, "Short Date") as day
FROM Payment
GROUP BY Payment.user_id, Format (Payment.time, "Short Date")
Having ((Count(Payment.money) > 0));
Can you help me with second part and summarize them? thanks
For example: first table (Payment):
user_id time money
a 01.01.10 00:00:00 15,00
a 01.01.10 10:00:00 2,00
a 03.01.10 00:00:00 3,00
c 04.01.10 00:00:00 4,00
c 04.01.10 00:05:00 5,00
d 06.01.10 00:00:00 6,00
e 07.01.10 00:00:00 7,00
e 08.01.10 00:00:00 8,00
The second one:
user_id time range
a 01.01.10 00:00:00 6
a 01.01.10 09:00:00 7
a 04.01.10 00:00:00 8
b 04.01.10 00:00:00 4
c 04.01.10 00:05:00 7
d 06.01.10 00:00:00 5
e 07.01.10 00:00:00 6
f 08.01.10 00:00:00 6
g 08.01.10 00:00:00 7
I expected:
user_id time sum
a 01.01.10 2
a 03.01.10 3
c 04.01.10 5
Here is one possible method using joins:
select t1.user_id, datevalue(p.time) as [time], sum(p.money) as [sum]
from
(
(select t.user_id, t.time from rank t where t.range = 7) t1
inner join payment p on t1.user_id = p.user_id
)
left join
(select t.user_id, t.time from rank t where t.range > 7) t2 on p.user_id = t2.user_id
where
p.time >= t1.time and (t2.user_id is null or p.time < t2.time)
group by
t1.user_id, datevalue(p.time)
I have assumed that your second table is called rank (this was not stated in your question).
Here, the subquery t1 obtains the set of users with range = 7 (captain), and the subquery t2 obtains the set of users with range > 7. I then select all records with a payment date greater than or equal to the date of promotion to captain, but less than any subsequent promotion (if it exists).
This yields the following result:
+---------+------------+------+
| user_id | time | sum |
+---------+------------+------+
| a | 01/01/2010 | 2.00 |
| a | 03/01/2010 | 3.00 |
| c | 04/01/2010 | 5.00 |
+---------+------------+------+
Unless I have misunderstood, I would argue that your expected result is incorrect as the payment below occurs before user_id = c achieved the rank of captain:
c 04.01.10 00:00:00 4,00
c 04.01.10 00:05:00 7
This question already has answers here:
Calculate a Running Total in SQL Server
(15 answers)
Closed 3 years ago.
I have the following table:
________________________
date | amount
________________________
01-01-2019 | 10
01-01-2019 | 10
01-01-2019 | 10
01-01-2019 | 10
02-01-2019 | 5
02-01-2019 | 5
02-01-2019 | 5
02-01-2019 | 5
03-01-2019 | 20
03-01-2019 | 20
These are mutation values by date. I would like my query to return the summed amount by date. So for 02-01-2019 I need 40 ( 4 times 10) + 20 ( 4 times 5). For 03-01-2019 I would need ( 4 times 10) + 20 ( 4 times 5) + 40 ( 2 times 20) and so on. Is this possible in one query? How do I achieve this?
My current query to get the individual mutations:
Select s.date,
Sum(s.amount) As Sum_amount
From dbo.Financieel As s
Group By s.date
You can try below -
DEMO
select dateval,
SUM(amt) OVER(ORDER BY dateval ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) as amt
from
(
SELECT
dateval,
SUM(amount) amt
FROM t2 group by dateval
)A
OUTPUT:
dateval amt
01/01/2019 00:00:00 40
01/02/2019 00:00:00 60
01/03/2019 00:00:00 100
Try this below script to get your desired output-
SELECT A.date,
(SELECT SUM(amount) FROM <your_table> WHERE Date <= A.Date) C_Total
FROM <your_table> A
GROUP BY date
ORDER BY date
Output is-
date C_Total
01-01-2019 40
02-01-2019 60
03-01-2019 100
I suggest to use a window function, like this:
select date, sum(amount) over( order by date)
from table