This question already has an answer here:
increment row number when value of field changes in Oracle
(1 answer)
Closed 2 years ago.
Suppose I have Oracle or Postgresql database.
ID IdExample OrderByColumn What I want
---------- ---------- ---------- ----------
1 1 1300 1
2 1 2450 1
3 2 5000 2
4 2 4800 2
5 1 5100 3
6 1 6000 3
7 4 7000 4
8 1 8000 5
How do count the changes that are in idExample, data is sorted by OrderByColumn
I need output new column that is represented by "what I want"
pay attention to "1" in IdExample. It repeats but I wants to iterate.
The query should execute quickly with the table having tens of thousands of records.
THANKS
You need to use lag and sum analytical function as follows:
Select t.*,
sum(case when lg is null or lg <> idexample then 1 else 0 end)
over (order by id) as result
from
(Select t.*,
lag(idexample) over (order by id) as lg
From your_table t) t
Related
This question already has answers here:
Removing duplicate rows from table in Oracle
(24 answers)
Closed 6 months ago.
I'm new to sql and I can't work out how to delete duplicate rows, I have a table like this called 'till_total':
till_id
total
1
80
1
80
1
60
2
30
2
30
2
50
I want to only delete full duplicate rows so the table ends up like this
till_id
total
1
80
1
60
2
30
2
50
I wrote this code to try and do it
SELECT till_id, total, COUNT(*) AS CNT
FROM till_total
GROUP BY till_id, total
HAVING COUNT(*) > 1
ORDER BY till_id;
But that seems to delete all rows where the till_id is repeated. Could anyone help me with this?
Good, old ROWID approach:
Before:
SQL> select * from till_total;
TILL_ID TOTAL
---------- ----------
1 80
1 80
1 60
2 30
2 30
2 50
6 rows selected.
Delete duplicates:
SQL> delete from till_total a
2 where a.rowid > (select min(b.rowid)
3 from till_total b
4 where b.till_id = a.till_id
5 and b.total = a.total
6 );
2 rows deleted.
After:
SQL> select * from till_total;
TILL_ID TOTAL
---------- ----------
1 80
1 60
2 30
2 50
SQL>
WITH till_total AS (
SELECT till_id
row_number() OVER(PARTITION BY till_id ORDER BY desc) AS row
FROM TABLE
)
DELETE till_total WHERE row > 1
This might work for you, deleting rows that are more than 1 duplicate, not less than 1.
I have data that looks like this:
ID num_of_days
1 0
2 0
2 8
2 9
2 10
2 15
3 10
3 20
I want to add another column that increments in value only if the num_of_days column is divisible by 5 or the ID number increases so my end result would look like this:
ID num_of_days row_num
1 0 1
2 0 2
2 8 2
2 9 2
2 10 3
2 15 4
3 10 5
3 20 6
Any suggestions?
Edit #1:
num_of_days represents the number of days since the customer last saw a doctor between 1 visit and the next.
A customer can see a doctor 1 time or they can see a doctor multiple times.
If it's the first time visiting, the num_of_days = 0.
SQL tables represent unordered sets. Based on your question, I'll assume that the combination of id/num_of_days provides the ordering.
You can use a cumulative sum . . . with lag():
select t.*,
sum(case when prev_id = id and num_of_days % 5 <> 0
then 0 else 1
end) over (order by id, num_of_days)
from (select t.*,
lag(id) over (order by id, num_of_days) as prev_id
from t
) t;
Here is a db<>fiddle.
If you have a different ordering column, then just use that in the order by clauses.
The first six balls mean first over, next six balls mean second over & so on than how to get average runs for each over.
input as
Ball no Runs
1 4
2 6
3 3
4 2
5 6
6 1
1 2
2 4
3 6
4 3
5 1
6 1
1 2
output should be:
Over no avg runs
1 3.66
2 2.83
As Gordon Linoff suggested, SQL table represents unordered sets, So you have to use an ordered column in your table. If you can use such a column you may use below query -
SELECT Over_no AVG(Runs) avg_runs
FROM (SELECT Ball_no, Runs, CEIL(ROW_NUMBER() OVER(ORDER BY ORDER_COLUMN, Ball_no) RN / 6) Over_no
FROM YOUR_TABLE)
GROUP BY Over_no;
I have managed to solve my problem with the following query:
SELECT ROWNUM OVER_NO, AVG_RUNS
FROM(
SELECT ROWNUM RN,
ROUND(AVG(RUNS)OVER(ORDER BY ROWNUM RANGE BETWEEN CURRENT ROW AND 5 FOLLOWING),2) AVG_RUNS
FROM TABLE_NAME
)
WHERE RN=1 OR RN=7;
This question already has answers here:
Select first row in each GROUP BY group?
(20 answers)
Closed 7 years ago.
So... i got a table like this:
id group number year
1 1 1 2000
2 1 2 2000
3 1 1 2001
4 2 1 2000
5 2 2 2000
6 2 1 2001
7 2 2 2001
8 2 3 2001
And i need to select the bigger number of the bigger year for each group. So i expect the result of the exemple to be:
3 1 1 2001
8 2 3 2001
any ideias?
OBS: using Postgres
SELECT *
FROM (
SELECT *,
row_number() over (partition by "group" order by "year" desc, "number" desc ) x
FROM table1
) x
WHERE x = 1;
demo: http://sqlfiddle.com/#!15/cd78e/2
If it's just certain rows you want to get you can use DISTINCT. If you want different maximums on the same rows you could use GROUP BY
SELECT DISTINCT ON ("group") * FROM tbl
ORDER BY "group", year DESC, id DESC;
I currently have a table in SQL that looks like this
PRODUCT_ID_1 PRODUCT_ID_2 SCORE
1 2 10
1 3 100
1 10 3000
2 10 10
3 35 100
3 2 1001
That is, PRODUCT_ID_1,PRODUCT_ID_2 is a primary key for this table.
What I would like to do is use this table to add in a row to tell whether or not the current row is the one that maximizes SCORE for a value of PRODUCT_ID_1.
In other words, what I would like to get is the following table:
PRODUCT_ID_1 PRODUCT_ID_2 SCORE IS_MAX_SCORE_FOR_ID_1
1 2 10 0
1 3 100 0
1 10 3000 1
2 10 10 1
3 35 100 0
3 2 1001 1
I am wondering how I can compute the IS_MAX_SCORE_FOR_ID_1 column and insert it into the table without having to create a new table.
You can try like this...
Select PRODUCT_ID_1, PRODUCT_ID_2 ,SCORE,
(Case when b.Score=
(Select Max(a.Score) from TableName a where a.PRODUCT_ID_1=b. PRODUCT_ID_1)
then 1 else 0 End) as IS_MAX_SCORE_FOR_ID_1
from TableName b
You can use a window function for this:
select product_id_1,
product_id_2,
score,
case
when score = max(score) over (partition by product_id_1) then 1
else 0
end as is_max_score_for_id_1
from the_table
order by product_id_1;
(The above is ANSI SQL and should run on any modern DBMS)