I have a table like this:
id employee_id contract_id month year d1 d2 d3
1 25 1 11 2011 1 01 01
2 16 5 11 2011 1 11 0
3 29 3 11 2011 1 001 100
1 25 4 11 2011 0 11 011
Suppose I need data for month='11' AND year='2011', then for all rows having the same 'employee_id', the data should merge like this:
id employee_id contract_id month year d1 d2 d3
1 25 1,4 11 2011 1,0 01,11 01,011
2 16 5 11 2011 1 11 0
3 29 3 11 2011 1 001 100
I was trying GROUP_CONCAT but couldn't figure out the query. Please help.
SELECT
id,
employee_id,
GROUP_CONCAT(contract_id SEPARATOR ',') AS contract_ids,
`month`,
`year`,
GROUP_CONCAT(d1 SEPARATOR ',') AS d1s,
GROUP_CONCAT(d2 SEPARATOR ',') AS d2s,
GROUP_CONCAT(d3 SEPARATOR ',') AS d3s
FROM
`table`
WHERE
`month` = 11 AND `year` = 2011
GROUP BY
employee_id
SELECT *,
GROUP_CONCAT(`d2`) as `d2`,
GROUP_CONCAT(`d3`) as `d3`,
GROUP_CONCAT(`contract_id`) as contract_id
FROM table WHERE month='11' AND year='2011' GROUP BY employee_id;
Related
I have 1 table named ItemDelivery. I wanted to get the count of items that has DeliveryDate and the items that has been receivedDate per month. Some items deliveryDate month have different receiveDate month such as items scheduled for delivery on the later part of the month would be received on early days of succeeding month. Some may take months to be delivered for overseas.
This is the data:
id iditem deliveryDate receiveDate
1 2 2021-01-03 2021-01-05
2 2 2021-01-03
3 3 2021-02-05 2021-02-06
4 5 2021-02-05
5 4 2021-02-20 2021-03-01
6 3 2021-03-15 2021-04-08
I would like to have
Mo Delivery Recieve
Jan 2 1
Feb 3 1
Mar 1 1
Apr 0 1
This query gives 1 columns only
select date_format(deliveryDate,'%b') as mo ,
count(id) as delivery
from ItemDelivery
where year(deliveryDate)=2021
group by month(deliveryDate)
union all
select date_format(receiveDate,'%b') as mo ,
count(id) as received
from ItemDelivery
where year(receiveDate)=2021
group by month(receiveDate)
Output:
Mo Delivery
Jan 2
Feb 3
Mar 1
Jan 1
Feb 1
Mar 1
Apr 1
This query also have different output
SELECT d1.mo, d1.delivery, d2.received
FROM
(SELECT month(deliveryDate) as mo, count(id) AS delivery
FROM ItemDelivery
WHERE year(deliveryDate)=2021 group by month(deliveryDate)) as d1,
(SELECT month(receiveDate) as mo, count(id) AS received
FROM ItemDelivery
WHERE year(receiveDate)=2021 group by month(receiveDate)) as d2
Output:
mo delivery received
1 2 1
2 3 1
3 1 1
1 2 1
2 3 1
3 1 1
1 2 1
2 3 1
3 1 1
1 2 1
2 3 1
3 1 1
This has also the same output except if I use condition d1.mo=d2.mo:
select d1.mo, d1.delivery, d2.received
from
(SELECT month(deliveryDate) as mo, count(id) as delivery
FROM ItemDelivery
WHERE year(deliveryDate)=2021 group by month(deliveryDate)) d1
inner join
(SELECT month(receiveDate) as mo, count(id) as received
FROM ItemDelivery
WHERE year(receiveDate)=2021 group by month(receiveDate)) d2
Any suggestions ?
SELECT
date_format(eventDate,'%b') AS mo,
SUM(delivery) AS delivery,
SUM(receive) AS receive
FROM
(
SELECT deliveryDate AS eventDate, 1 AS delivery, 0 AS receive FROM ItemDelivery
UNION ALL
SELECT receiveDate AS eventDate, 0 AS delivery, 1 AS receive FROM ItemDelivery
)
AS rotated
WHERE
eventDate >= '2021-01-01'
AND eventDate < '2022-01-01'
GROUP BY
month(eventDate)
I have a set of data that looks like this
Cust Id order date Ordered Product
1 Jan 2 1
1 Jan 5 2
1 March 14 1
1 September 9 1
1 December 12 2
2 Jan 5 1
2 Feb 13 2
3 March 12 2
3 April 5 3
3 June 10 2
and my output should look like this
Cust Id order Date Order product
1 Jan 31 3
1 feb 29 0
1 Mar31 1
1 Apr 30 0
1 May 31 0
1 June 30 0
1 July 31 0
1 Aug 31 0
1 Sept 30 1
1 oct 31 0
1 Nov 30 0
1 Dec 31 2
and I have got this far
1 January 31 3
1 March 31 1
1 September 30 1
1 December 31 2
and my code is
select customer_id,
date_format(last_day(order_date), '%M %d') as new_months,
sum(products_ordered) as total
from amazon_test
where customer_id =1
group by new_months, customer_id;
I currently stuck at the part where I need to have all the months and '0' as the output since no orders were made.
If you are running MySQL 8.0, one option is to use a recursive query to generate the months, and then bring the table with a left join:
with recursive months as (
select customer_id, date_format(min(order_date), '%Y-%m-01') order_date, max(order_date) max_order_date
from amazon_test
group by customer_id
union all
select customer_id, order_date + interval 1 month, max_order_date
from months
where order_date + interval 1 month < max_order_date
)
select
m.customer_id,
date_format(last_day(m.order_date), '%M %d') new_months,
coalesce(sum(t.products_ordered), 0) ordered_products
from months m
left join amazon_test t
on t.customer_id = m.customer_id
and t.order_date >= m.order_date
and t.order_date < m.order_date + interval 1 month
where m.customer_id = 1
group by m.customer_id, m.order_date
order by m.customer_id, m.order_date
I phrased the query so it actually operates on all customer_ids - if you remove the where clause in the outer query, you do get the results for all customers. If you really want the results for only one customer, you can optimize the query by pushing the where filter to the anchor of the recusive query.
Demo on DB Fiddle:
customer_id | new_months | ordered_products
----------: | :----------- | ---------------:
1 | January 31 | 3
1 | February 29 | 0
1 | March 31 | 1
1 | April 30 | 0
1 | May 31 | 0
1 | June 30 | 0
1 | July 31 | 0
1 | August 31 | 0
1 | September 30 | 1
1 | October 31 | 0
1 | November 30 | 0
1 | December 31 | 2
I'm trying to display zero on missing month, but didn't succeed.
Table:
clicks | impressions | ctr | position | month | year
111 2709 4 20 3 2015
101 2695 3 20 6 2015
76 2714 2 21 7 2015
.
.
.
64 1212 4 25 11 2015
81 1905 4 24 12 2015
Required output:
clicks | impressions | ctr | position | month | year
0 0 0 0 1 2015
0 0 0 0 2 2015
111 2709 4 20 3 2015
0 0 0 0 4 2015
0 0 0 0 5 2015
101 2695 3 20 6 2015
.
.
.
64 1212 4 25 11 2015
81 1905 4 24 12 2015
Like #jarlh suggested ,you can do it like this: creating a 'table' that contains all month availabe(you will have to populate it your self, add which month and years you want) and then left join to the original table and when value not exists, put 0.
select coalese(s.clicks,0) as clicks,
coalese(s.impressions ,0) as impressions,
coalese(s.ctr ,0) as ctr ,
coalese(s.position ,0) as position ,
t.month,
t.year
from(
SELECT 1 as month_num,2015 as year_num
union SELECT 2,2015
union select 3,2015
union select 4,2015 ....) t
LEFT OUTER JOIN YourTable s
ON(t.month_num = s.month and t.year_num = s.year)
I have a table having structure as follows
id cust_id target month year fiscal_ID
1 234 50 4 2013 1
2 234 50 5 2013 1
3 234 50 6 2013 1
4 234 150 7 2013 1
5 234 150 8 2013 1
6 234 150 9 2013 1
I need to get the result as follows
cust_id target quarter year fiscal_ID
234 150 Q1 2013 1
234 450 Q2 2013 1
months 4,5,6 in Q1, 7,8,9 in Q2 etc
Since you are storing the month and year in separate columns, one way you can get the result is to use a derived table that references the month and quarter and you join to that data:
select t.cust_id,
sum(target) target,
d.qtr,
t.year,
t.fiscal_id
from yourtable t
inner join
(
select 4 mth, 'Q1' qtr union all
select 5 mth, 'Q1' qtr union all
select 6 mth, 'Q1' qtr union all
select 7 mth, 'Q2' qtr union all
select 8 mth, 'Q2' qtr union all
select 9 mth, 'Q2'
) d
on t.month = d.mth
group by t.cust_id, d.qtr, t.year, t.fiscal_id;
See SQL Fiddle with Demo.
I have a MYSQL table like this:
id | userid | score | datestamp |
-----------------------------------------------------
1 | 1 | 5 | 2012-12-06 03:55:16
2 | 2 | 0,5 | 2012-12-06 04:25:21
3 | 1 | 7 | 2012-12-06 04:35:33
4 | 3 | 12 | 2012-12-06 04:55:45
5 | 2 | 22 | 2012-12-06 05:25:11
6 | 1 | 16,5 | 2012-12-06 05:55:21
7 | 1 | 19 | 2012-12-06 13:55:16
8 | 2 | 8,5 | 2012-12-07 06:27:16
9 | 2 | 7,5 | 2012-12-07 08:33:16
10 | 1 | 10 | 2012-12-07 09:25:19
11 | 1 | 6,5 | 2012-12-07 13:33:16
12 | 3 | 6 | 2012-12-07 15:45:44
13 | 2 | 4 | 2012-12-07 16:05:16
14 | 2 | 34 | 2012-12-07 18:33:55
15 | 2 | 22 | 2012-12-07 18:42:11
I would like to display user scores like this:
if a user on a certain day has more than 3 scores it would get only highest 3, repeat that for every day for this user and then add all days together. I want to display this sum for every user.
EDIT:
So in the example above for user 1 on 06.12. I would add top 3 scores together and ignore 4th score, then add to that number top 3 from the next day and so on. I need that number for every user.
EDIT 2:
Expected output is:
userid | score
--------------------
1 | 59 //19 + 16.5 + 7 (06.12.) + 10 + 6.5 (07.12.)
2 | 87 //22 + 0.5 (06.12.) + 34 + 22 + 8.5 (07.12.)
3 | 18 //12 (06.12.) + 6 (07.12.)
I hope this is more clear :)
I would really appreciate the help because I am stuck.
Please take a look at the following code, if your answer to my comment is yes :) Since your data all in 2012, and month of november, I took day.
SQLFIDDLE sample
Query:
select y.id, y.userid, y.score, y.datestamp
from (select id, userid, score, datestamp
from scores
group by day(datestamp)) as y
where (select count(*)
from (select id, userid, score, datestamp
from scores group by day(datestamp)) as x
where y.score >= x.score
and y.userid = x.userid
) =1 -- Top 3rd, 2nd, 1st
order by y.score desc
;
Results:
ID USERID SCORE DATESTAMP
8 2 8.5 December, 07 2012 00:00:00+0000
20 3 6 December, 08 2012 00:00:00+0000
1 1 5 December, 06 2012 00:00:00+0000
Based on your latter updates to question.
If you need some per user by year/month/day and then find highest, you may simply add aggregation function like sum to the above query. I am reapeating myself, since your sample data is for just one year, there's no point group by year or month. That's why I took day.
select y.id, y.userid, y.score, y.datestamp
from (select id, userid, sum(score) as score,
datestamp
from scores
group by userid, day(datestamp)) as y
where (select count(*)
from (select id, userid, sum(score) as score
, datestamp
from scores
group by userid, day(datestamp)) as x
where y.score >= x.score
and y.userid = x.userid
) =1 -- Top 3rd, 2nd, 1st
order by y.score desc
;
Results based on sum:
ID USERID SCORE DATESTAMP
1 1 47.5 December, 06 2012 00:00:00+0000
8 2 16 December, 07 2012 00:00:00+0000
20 3 6 December, 08 2012 00:00:00+0000
UPDATED WITH NEW SOURCE DATA SAMPLE
Simon, please take a look at my own sample. As your data was changing, I used mine.
Here is the reference. I have used pure ansi style without any over partition or dense_rank.
Also note the data I used are getting top 2 not top 3 scores. You can change is accordingly.
Guess what, the answer is 10 times simpler than the first impression your first data gave....
SQLFIDDLE
Query to 1:
-- for top 2 sum by user by each day
SELECT userid, sum(Score), datestamp
FROM scores t1
where 2 >=
(SELECT count(*)
from scores t2
where t1.score <= t2.score
and t1.userid = t2.userid
and day(t1.datestamp) = day(t2.datestamp)
order by t2.score desc)
group by userid, datestamp
;
Results for query 1:
USERID SUM(SCORE) DATESTAMP
1 70 December, 06 2012 00:00:00+0000
1 30 December, 07 2012 00:00:00+0000
2 22 December, 06 2012 00:00:00+0000
2 25 December, 07 2012 00:00:00+0000
3 30 December, 06 2012 00:00:00+0000
3 30 December, 07 2012 00:00:00+0000
Final Query:
-- for all two days top 2 sum by user
SELECT userid, sum(Score)
FROM scores t1
where 2 >=
(SELECT count(*)
from scores t2
where t1.score <= t2.score
and t1.userid = t2.userid
and day(t1.datestamp) = day(t2.datestamp)
order by t2.score desc)
group by userid
;
Final Results:
USERID SUM(SCORE)
1 100
2 47
3 60
Here goes a snapshot of direct calculations of data I used.
SELECT
*
FROM
table1
LEFT JOIN
(SELECT * FROM table1 ORDER BY score LIMIT 3) as lr on DATE(lr.datestamp) = DATE(table1.datastamp)
GROUP BY
datestamp