I have two tables, the 1st contains transaction details and the 2nd contains user's orders :
id | transaction_date
1 | 2019-01-01
2 | 2019-02-01
3 | 2019-01-01
id | transaction_id | amount | user_id
15 1 7 1
20 2 15 1
25 3 25 1
And I would like to have this result, that is to say for all users orders have also the previous amount he paid based on the transaction date.
user_id | amount | previous amount
1 7 NULL
1 15 7
1 25 15
I tried multiple things including using the LAG function, but it doesn't seems to be possible with it because I have to join on another table to get the transaction_date. I think I should do a subquery with a left join but I don't figure out how to get only the previous order
Thanks
This is a join and lag():
select t2.user_id, t2.amount,
lag(t2.amount) over (partition by t2.user_id order by t1.date) as prev_amount
from table1 t1 join
table2 t2
on t2.transaction_id = t1.id;
Related
I have a table contains item_wise quantity at different hour of date. trying to add data for each hour(24 enteries in a day) with previous hour available quantity. For example for hour(2-10), it will be 5.
I created a table with hours enteries (1-24) & full join with shared table.
How can i add previous available entry. Need suggestion
item_id| date | hour| quantity
101 | 2022-04-25 | 2 | 5
101 | 2022-04-25 | 10 | 13
101 | 2022-04-25 | 18 | 67
101 | 2022-04-25 | 23 | 27
You can try to use generate_series to generate hours number, let it be the OUTER JOIN base table,
Then use a correlated-subquery to get your expect quantity column
SELECT t1.*,
(SELECT quantity
FROM T tt
WHERE t1.item_id = tt.item_id
AND t1.date = tt.date
AND t1.hour >= tt.hour
ORDER BY tt.hour desc
LIMIT 1) quantity
FROM (
SELECT DISTINCT item_id,date,v.hour
FROM generate_series(1,24) v(hour)
CROSS JOIN T
) t1
ORDER BY t1.hour
Provided the table of int 1 .. 24 is all24(hour) you can use lead and join
select t.item_id, t.date, all24.hour, t.quantity
from all24
join (
select *,
lead(hour, 1, 25) over(partition by item_id, date order by hour) - 1 nxt_h
from tbl
) t on all24.hour between t.hour and t.nxt_h
I am working with data similar to below,
week | product | sale
1 | ABC | 2
1 | ABC | 1
2 | ABC | 1
3 | ABC | 5
4 | ABC | 1
2 | DEF | 5
Let us say that is my Orders table named tblOrders. Now, in each row, I want to aggregate the total sales from last week for that product - for instance, if I am on week 2 of product "ABC", I need to show the aggregated sales amount of week 1 for product ABC. so, the output should look something like below,
week | product | sale | ProductPreviousWeekSales
1 | ABC | 2 | 0
1 | ABC | 1 | 0
2 | ABC | 1 | 3
3 | ABC | 5 | 1
4 | ABC | 1 | 5
2 | DEF | 5 | 0
I was originally thinking I could solve this using Aggregates and Window Function, but doesn't look to be so. Another thought I was having is to use Conditional Aggregate - something like sum(case when x=currentRow.x then sale else 0 end), but that wouldn't work too.
Here is the SQLFiddle for above sample - http://sqlfiddle.com/#!18/890b7/2
Note: I need to calculate similar value for Last 4 weeks, so trying to avoid doing this as a sub-query or multiple joins (if possible), as the data set I am working with is very large, and don't want to add to much performance overhead trying to incorporate this change.
Here is one approach which first aggregates your table in a separate CTE and uses LAG to find the previous week's amount, for each week and product:
WITH cte AS (
SELECT week, product,
LAG(SUM(sale)) OVER (PARTITION BY product ORDER BY week) AS lag_total_sales
FROM yourTable
GROUP BY week, product
)
SELECT t1.week, t1.product, t1.sale,
COALESCE(t2.lag_total_sales, 0) AS ProductPreviousWeekSales
FROM yourTable t1
INNER JOIN cte t2
ON t2.week = t1.week AND
t2.product = t1.product
ORDER BY
t1.product,
t1.week;
Demo
DISCLAIMER
The query I am showing below doesn't work in SQL Server, unfortunately. Up to SQL Server version 2019 the DBMS lacks full support of the RANGE clause that is essential for the query to work. Running the query in SQL Server results in
Msg 4194 Level 16 State 1 Line 1 RANGE is only supported with UNBOUNDED and CURRENT ROW window frame delimiters.
I am not deleting this answer, because this is standard SQL and the approach may help future readers. It runs fine in a lot of DBMS, and maybe a future version of SQL Server will be able to deal with this, too. I've added demos to show that it runs in PostgreSQL, MySQL and Oracle, but fails in SQL Server 2019.
ORIGINAL ANSWER
Your query shown in the fiddle (select a.*, sum(sale) over(partition by product) ProductPreviousWeekSales from tblOrder a) is merely lacking the appropriate windowing clause. As you are dealing with ties here (more than one row per product and week) this needs to be a RANGE clause:
select a.*,
sum(sale) over(partition by product
order by week range between 1 preceding and 1 preceding
) as ProductPreviousWeekSales
from tblOrder a
order by product, week;
(Use COALESCE if you want to see a zero instead of NULL.)
Demos:
https://dbfiddle.uk/?rdbms=postgres_13&fiddle=149eddbff82500d539b2c615f4167cff
https://dbfiddle.uk/?rdbms=mysql_8.0&fiddle=a8453970efac08ad69275914910bb13e
https://dbfiddle.uk/?rdbms=oracle_18&fiddle=64ed21150142caa0acb7f8c7ca7d9022
https://dbfiddle.uk/?rdbms=sqlserver_2019&fiddle=149eddbff82500d539b2c615f4167cff
You can do from following
; WITH cteorder AS
(
SELECT DISTINCT product, week FROM dbo.tblOrder
)
SELECT
cte.*,
SUM(ISNULL(b.sale,0)) ProductPreviousWeekSales
from tblOrder a
INNER JOIN cteorder cte ON cte.product = a.product AND cte.week = a.week
LEFT JOIN dbo.tblOrder b ON b.product = cte.product AND b.week = (a.week-1)
GROUP BY cte.product,
cte.week
You can run from : Fiddle
You need to select from TblOrders twice. Once, grouping by week and product and summing the sales, and the second time, a row-by-row scan against TblOrders, left-joining it with the grouping query on same product and week offset by 1:
If the join fails , the sales value of the joined grouping query returns NULL. You can put in 0 instead of NULL using COALESCE(), but ISNULL() has all chances of being faster, as it has a fixed number of parameters, while COALESCE() has a variable argument list, which comes at a certain cost.
WITH
tblorders(wk,product,sales) AS (
SELECT 1,'ABC',2
UNION ALL SELECT 1,'ABC',1
UNION ALL SELECT 2,'ABC',1
UNION ALL SELECT 3,'ABC',5
UNION ALL SELECT 4,'ABC',1
UNION ALL SELECT 2,'DEF',5
)
,
grp AS (
SELECT
wk
, product
, SUM(sales) AS sales
FROM tblorders
GROUP BY
wk
, product
)
SELECT
o.wk
, o.product
, o.sales
, ISNULL(g.sales,0) AS productpreviousweeksales
FROM tblorders o
LEFT
JOIN grp g
ON o.wk - 1 = g.wk
AND o.product= g.product
ORDER BY 2,1
;
wk | product | sales | productpreviousweeksales
----+---------+-------+--------------------------
1 | ABC | 2 | 0
1 | ABC | 1 | 0
2 | ABC | 1 | 3
3 | ABC | 5 | 1
4 | ABC | 1 | 5
2 | DEF | 5 | 0
Hi，I have a column as below
+--------+--------+
| day | amount|
+--------+---------
| 2 | 2 |
| 1 | 3 |
| 1 | 4 |
| 2 | 2 |
| 3 | 3 |
| 4 | 3 |
+--------+--------+
now I want something like this sum day 1- day2 as row one , sum day1-3 as row 2, and so on.
+--------+--------+
| day | amount|
+--------+---------
| 1-2 | 11 |
| 1-3 | 14 |
| 1-4 | 17 |
+--------+--------+
Could you offer any one help ,thanks!
with data as(
select 2 day, 2 amount from dual union all
select 1 day, 3 amount from dual union all
select 1 day, 4 amount from dual union all
select 2 day, 2 amount from dual union all
select 3 day, 3 amount from dual union all
select 4 day, 3 amount from dual)
select distinct day, sum(amount) over (order by day range unbounded preceding) cume_amount
from data
order by 1;
DAY CUME_AMOUNT
---------- -----------
1 7
2 11
3 14
4 17
if you are using oracle you can do something like the above
Assuming the day range in left column always starts from "1-", What you need is a query doing cumulative sum on the grouped table(dayWiseSum below). Since it needs to be accessed twice I'd put it into a temporary table.
CREATE TEMPORARY TABLE dayWiseSum AS
(SELECT day,SUM(amount) AS amount FROM table1 GROUP BY day ORDER BY day);
SELECT CONCAT("1-",t1.day) as day, SUM(t2.amount) AS amount
FROM dayWiseSum t1 INNER JOIN dayWiseSum
t2 ON t1.day > t2.day
--change to >= if you want to include "1-1"
GROUP BY t1.day, t1.amount ORDER BY t1.day
DROP TABLE dayWiseSum;
Here's a fiddle to test with:
http://sqlfiddle.com/#!9/c1656/1/0
Note: Since sqlfiddle isn't allowing CREATE statements, I've replaced dayWiseSum with it's query there. Also, I've used "Text to DDL" option to paste the exact text of the table from your question to generate the create table query :)
I want to create a view in my database, based on these three tables:
I would like to select the rows in table3 that has the highest value in Weight, for rows that has the same value in Count.
Then I want them grouped by Category_ID and ordered by Date, so that if two rows in table3 are identical, I want the newest.
Let me give you an example:
Table1
ID | Date | UserId
1 | 2015-01-01 | 1
2 | 2015-01-02 | 1
Table2
ID | table1_ID | Category_ID
1 | 1 | 1
2 | 2 | 1
Table3
ID | table2_ID | Count | Weight
1 | 1 | 5 | 10
2 | 1 | 5 | 20 <-- count is 5 and weight is highest
3 | 1 | 3 | 40
4 | 2 | 5 | 10
5 | 2 | 3 | 40 <-- newest of the two equal rows
Then the result should be row 2 and 5 from table 3.
PS I'm doing this in mssql.
PPS I'm sory if the title is not appropriate, but I did not know how to formulate a good one.
SELECT
*
FROM
(
SELECT
t3.*
,RANK() OVER (PARTITION BY [Count] ORDER BY [Weight] DESC, Date DESC) highest
FROM TABLE3 t3
INNER JOIN TABLE2 t2 ON t2.Id = t3.Table2_Id
INNER JOIN TABLE1 t1 ON t1.Id = t2.Table1_Id
) t
WHERE t.Highest = 1
This will group by the Count (which must be the same). Then it will determine which has the highest weight. If two of more of them have the same 'heighest' weight, it takes the one with the most recent date first.
You can use RANK() analytic function here, and give those rows a rank and than choose the first rank for each ID
Something like
select *
from
(select
ID, table2_ID, Count, Weight,
RANK() OVER (PARTITION BY ID ORDER BY Count, Weight DESC) as Highest
from table3)
where Highest = 1;
This is the syntax for Oracle, if you not using it look in the internet for the your syntax which should be almost the same
[Reframing prior question, which had been posed as a question about Cursors.]
I am looking for a way to select counts under certain date conditions.
Say there is a table, T1, with 2 fields (ID, Date). The ID is not a unique key. The table records events by id, and some ids occur frequently, some infrequently.
For example:
ID | Date
1 | 2010-01-01
2 | 2010-02-01
3 | 2010-02-15
2 | 2010-02-15
4 | 2010-03-01
I would like to create a new table with the following fields: ID, Date, Count of times ID appears in 6 months previous to Date, Count of times ID appears in 6 months after Date.
In essence, for every row in the existing table, I want to add a column that looks back for times the same ID has appeared in previous six months, and look ahead for times the same ID has appeared in following six months.
So the output for the example would hopefully look something like:
ID | Date | Lookback | Lookahead
1 | 2010-01-01 | 0 | 0
2 | 2010-02-01 | 0 | 1
3 | 2010-02-15 | 0 | 0
2 | 2010-02-15 | 1 | 0
4 | 2010-03-01 | 0 | 0
Is there a best way to formulate the appropriate query?
You can do this with a self join (Assuming you have a primary key of KeyID):
SELECT T.ID,
T.Date,
Lookback = COUNT(CASE WHEN t2.Date < T.Date THEN t2.ID END),
Lookahead = COUNT(CASE WHEN t2.Date > T.Date THEN t2.ID END)
FROM T
INNER JOIN T t2
ON t2.ID = t.ID
AND t2.Date >= DATEADD(MONTH, -6, T.Date)
AND T2.Date < DATEADD(MONTH, 6, T.Date)
GROUP BY T.ID, T.Date, T.KeyID;
Example on SQL Fiddle
The key is that it just joins all rows for the previous 6 months and the next 6 months, and counts the result. The COUNT(CASE WHEN... ensures that for the before column you are only counting the records before, and the after only the records after.