I have this select:
"Select * from table" that return:
Id
Value
1
1
1
1
2
10
2
10
My goal is create a sum from each Value group by id like this:
Id
Value
Sum
1
1
2
1
1
2
2
10
20
2
10
20
I Have tried ways like:
SELECT Id,Value, (SELECT SUM(Value) FROM Table V2 WHERE V2.Id= V.Id GROUP BY IDRNC ) FROM Table v;
But the is not grouping by id.
Id
Value
Sum
1
1
1
1
1
1
2
10
10
2
10
10
Aggregation aggregates rows, reducing the number of records in the output. In this case you want to apply the result of a computation to each of your records, task carried out by the corresponding window function.
SELECT table.*, SUM(Value) OVER(PARTITION BY Id) AS sum_
FROM table
Check the demo here.
Your attempt looks correct.
Can you try the below query :
It works for me :
SELECT Id, Value,
(SELECT SUM(Value) FROM Table V2 WHERE V2.Id= V.Id GROUP BY ID) as sum
FROM Table v;
You can do it using inner join to join with selection grouped by id :
select t.*, sum
from _table t
inner join (
select id, sum(Value) as sum
from _table
group by id
) as s on s.id = t.id
You can check it here
Your select is ok if you adjust it just a little:
SELECT Id,Value, (SELECT SUM(Value) FROM Table V2 WHERE V2.Id= V.Id GROUP BY IDRNC ) FROM Table v;
GROUP BY IDRNC is a mistake and should be GROUP BY ID
you should give an alias to a sum column ...
subquery selecting the sum does not have to have self table alias to be compared with outer query that has one (this is not a mistake - works either way)
Test:
WITH
a_table (ID, VALUE) AS
(
Select 1, 1 From Dual Union All
Select 1, 1 From Dual Union All
Select 2, 10 From Dual Union All
Select 2, 10 From Dual
)
SELECT ID, VALUE, (SELECT SUM(VALUE) FROM a_table WHERE ID = v.ID GROUP BY ID) "ID_SUM" FROM a_table v;
ID VALUE ID_SUM
---------- ---------- ----------
1 1 2
1 1 2
2 10 20
2 10 20
Related
I have the following data:
ID Site
2 NULL
2 32
3 6
4 7
8 12
8 13
9 14
9 14
Result should be:
ID Site
2 NULL
2 32
8 12
8 13
Note that the result find unique combinations of ID and Site that repeat more than once for a given ID.
I did the following query but does not return the result:
select distinct id, site
from Table1
group by id, site
having count(*) > 1
order by id
SELECT
ID,
site
FROM table1
WHERE ID IN (
SELECT ID
FROM (
SELECT ID ,site
FROM table1
GROUP BY ID ,site
) x
GROUP BY ID
HAVING count(*)>1
)
See: DBFIDDLE
The SELECT ID, site FROM table1 GROUP BY ID, site will select the distinct values.
Then, using HAVING count(*) > 1, only the IDs that appear more than once are filtered.
P.S. You should try to avoid using DISTINCT and GROUP BY in one query. It makes life so much more complicated when you do that ... 😉
One way to do it is to do the select distinct in a CTE, then use the count window function to get the desired result:
with u as (
select distinct *
from Table1
), v as (
select *
, count(*) over(partition by ID) as cnt
from u
)
select ID, Site
from v
where cnt > 1;
Fiddle
My table
id name num
1 a 3
2 b 4
I need to return every row num number of times. I do it this way.
select DB.BAN_KEY as BAN_KEY, DB.CUST_FULLNAME as CUST_FULLNAME
from TST_DIM_BAN_SELECTED DB
inner join (select rownum rn from dual connect by level < 10) a
on a.rn <= DB.N
There resulting table looks like this.
id name
1 a
1 a
1 a
2 b
2 b
2 b
2 b
But I also need every row in the group to be numbered like this.
id name row_num
1 a 1
1 a 2
1 a 3
2 b 1
2 b 2
2 b 3
2 b 4
How can I do it?
You don't need an inner join to a dummy table or an analytic function to generate the row numbers; you could just use connect by (and its corresponding level function) on the table itself, like so:
WITH tst_dim_ban_selected AS (SELECT 1 ban_key, 'a' cust_fullname, 3 n FROM dual UNION ALL
SELECT 2 ban_key, 'b' cust_fullname, 4 n FROM dual)
-- end of mimicking your table with data in it. See SQL below
SELECT db.ban_key,
db.cust_fullname,
LEVEL row_num
FROM tst_dim_ban_selected db
CONNECT BY LEVEL <= db.n
AND PRIOR db.ban_key = db.ban_key -- assuming this is the primary key
AND PRIOR sys_guid() IS NOT NULL;
BAN_KEY CUST_FULLNAME ROW_NUM
---------- ------------- ----------
1 a 1
1 a 2
1 a 3
2 b 1
2 b 2
2 b 3
2 b 4
If you have other columns than ban_key in the table's primary key, you need to make sure they are included in the connect by clause's list of prior <column> = <column>s. This is so the connect by can identify each row uniquely, meaning that it's looping just over that row and no others. The PRIOR sys_guid() IS NOT NULL is required to prevent connect by loops from occurring.
You can use analytic function for this:
Select id, name,
row_number() over (partition by id, name order by id, name)
From(/* your query */) t;
This can be done without subquery:
Select id, name,
row_number() over (partition by id, name order by id, name)
From /* joins */
You could use this:
SELECT db.ban_key AS ban_key, db.cust_fullname AS cust_fullname,
ROW_NUMBER() OVER (PARTITION BY db.n ORDER BY db.ban_key) AS row_num
FROM tst_dim_ban_selected db
INNER JOIN (SELECT rownum rn FROM dual CONNECT BY level < 10) a
ON a.rn <= db.n;
Use a recursive sub-query factoring clause:
WITH split ( id, name, rn, n ) AS (
SELECT BAN_KEY, CUST_FULLNAME, 1, N
FROM TST_DIM_BAN_SELECTED
UNION ALL
SELECT id, name, rn + 1, n
FROM split
WHERE rn < n
)
SELECT id, name, rn
FROM split;
I want to add an extra column, where the max values of each group (ID) will appear.
Here how the table looks like:
select ID, VALUE from mytable
ID VALUE
1 4
1 1
1 7
2 2
2 5
3 7
3 3
Here is the result I want to get:
ID VALUE max_values
1 4 7
1 1 7
1 7 7
2 2 5
2 5 5
3 7 7
3 3 7
Thank you for your help in advance!
Your previous questions indicate that you are using SQL Server, in which case you can use window functions:
SELECT ID,
Value,
MaxValue = MAX(Value) OVER(PARTITION BY ID)
FROM mytable;
Based on your comment on another answer about first summing value, you may need to use a subquery to actually get this:
SELECT ID,
Date,
Value,
MaxValue = MAX(Value) OVER(PARTITION BY ID)
FROM ( SELECT ID, Date, Value = SUM(Value)
FROM mytable
GROUP BY ID, Date
) AS t;
There is no need to use GROUP BY in subselect.
select ID, VALUE,
(select MAX(VALUE) from mytable where ID = t.ID) as MaxValue
from mytable t
Use this query.
SELECT ID
,value
,(
SELECT MAX(VALUE)
FROM GetMaxValue gmv
WHERE gmv.ID = gmv1.ID
GROUP BY ID
) as max_value
FROM GetMaxValue gmv1
ORDER BY ID
Try it with a sub select and group by, then grab the MAX of this group:
select
ID,
VALUE,
(select MAX(VALUE)
from mytable
group by ID
having ID = t.ID
) as max_values
from mytable t
Edit:
I built a SQL fiddle, which shows that my solution works, but also VDohnal is correct and doesn't need the group by, so I'll upvote his answer.
Given a table with multiple rows of an int field and the same identifier, is it possible to return the 2nd maximum and 2nd minimum value from the table.
A table consists of
ID | number
------------------------
1 | 10
1 | 11
1 | 13
1 | 14
1 | 15
1 | 16
Final Result would be
ID | nMin | nMax
--------------------------------
1 | 11 | 15
You can use row_number to assign a ranking per ID. Then you can group by id and pick the rows with the ranking you're after. The following example picks the second lowest and third highest :
select id
, max(case when rnAsc = 2 then number end) as SecondLowest
, max(case when rnDesc = 3 then number end) as ThirdHighest
from (
select ID
, row_number() over (partition by ID order by number) as rnAsc
, row_number() over (partition by ID order by number desc) as rnDesc
) as SubQueryAlias
group by
id
The max is just to pick out the one non-null value; you can replace it with min or even avg and it would not affect the outcome.
This will work, but see caveats:
SELECT Id, number
INTO #T
FROM (
SELECT 1 ID, 10 number
UNION
SELECT 1 ID, 10 number
UNION
SELECT 1 ID, 11 number
UNION
SELECT 1 ID, 13 number
UNION
SELECT 1 ID, 14 number
UNION
SELECT 1 ID, 15 number
UNION
SELECT 1 ID, 16 number
) U;
WITH EX AS (
SELECT Id, MIN(number) MinNumber, MAX(number) MaxNumber
FROM #T
GROUP BY Id
)
SELECT #T.Id, MIN(number) nMin, MAX(number) nMax
FROM #T INNER JOIN
EX ON #T.Id = EX.Id
WHERE #T.number <> MinNumber AND #T.number <> MaxNumber
GROUP BY #T.Id
DROP TABLE #T;
If you have two MAX values that are the same value, this will not pick them up. So depending on how your data is presented you could be losing the proper result.
You could select the next minimum value by using the following method:
SELECT MAX(Number)
FROM
(
SELECT top 2 (Number)
FROM table1 t1
WHERE ID = {MyNumber}
order by Number
)a
It only works if you can restrict the inner query with a where clause
This would be a better way. I quickly put this together, but if you can combine the two queries, you will get exactly what you were looking for.
select *
from
(
select
myID,
myNumber,
row_number() over (order by myID) as myRowNumber
from MyTable
) x
where x.myRowNumber = 2
select *
from
(
select
myID,
myNumber,
row_number() over (order by myID desc) as myRowNumber
from MyTable
) y
where x.myRowNumber = 2
let the table name be tblName.
select max(number) from tblName where number not in (select max(number) from tblName);
same for min, just replace max with min.
As I myself learned just today the solution is to use LIMIT. You order the results so that the highest values are on top and limit the result to 2. Then you select that subselect and order it the other way round and only take the first one.
SELECT somefield FROM (
SELECT somefield from table
ORDER BY somefield DESC LIMIT 2)
ORDER BY somefield ASC LIMIT 1
I have a simple query:
select id, count(*) n
from mytable
group by id
Is it possible to include also the sum(n) in the same query? So the result would look something like this:
id n
---- -----------
1 12
2 1
3 14
4 1
5 2
6 6
Sum=36
You can use a common table expression to do this:
--
; WITH cte as (SELECT id
,count(*) n
FROM mytable
GROUP BY id)
SELECT id, n FROM cte
UNION ALL
SELECT 'Sum', SUM(n) from cte
You can also use ROLLUP: (this may not be exactly correct syntax)
SELECT id
,count(*) n
FROM mytable
GROUP BY id
WITH ROLLUP