I'll try to explain the type of the query that I want:
Assume I have a table like this:
| ID | someID | Number |
|----|--------|--------|
| 1 | 1 | 10 |
| 2 | 1 | 11 |
| 3 | 1 | 14 |
| 4 | 2 | 10 |
| 5 | 2 | 13 |
Now, I want to find the someID that have a specific numbers (For example query for numbers 10, 11, 14 will return someID 1 and query for numbers 10, 13 will return 2). But, if someID contains all the query numbers but also more numbers, it will not return by the query. (For example query for 10, 11 will return nothing).
Is it possible?
SELECT t1.someId
FROM yourTable t1
WHERE t1.number IN (10,14,11)
GROUP BY t1.someID
HAVING COUNT(DISTINCT t1.ID) = (SELECT COUNT(DISTINCT t2.ID) FROM yourTable t2 WHERE t1.someID=t2.someID)
Example Fiddle
select someID
from yourtable
where number in (10,11,14)
and not exists (select * from yourtable t2 where number not in(10,11,14)
and t2.someid=yourtable.someid)
group by someID
having count(distinct ID) = 3
Where 3 is the number of items you are querying for
Yes, once you get the query numbers into a table variable (say it's called #QNums, with one column named QNum)) try
Select distinct someId
From table t
Where exists (Select * from #QNums
where QNum = t.Number)
And not Exists (Select * From table t2
Where someId = t.someId
And not exists(Select * From #QNums
where QNum = t3.Number))
Related
I have a table that looks like this:
ID | PARENT_ID
--------------
0 | NULL
1 | 0
2 | NULL
3 | 1
4 | 2
5 | 4
6 | 3
Being an SQL noob, I'm not sure if I can accomplish what I would like in a single command.
What I would like is to start at row 6, and recursively follow the "history", using the PARENT_ID column to reference the ID column.
The result (in my mind) should look something like:
6|3
3|1
1|0
0|NULL
I already tried something like this:
SELECT T1.ID
FROM Table T1, Table T2
WHERE T1.ID = 6
OR T1.PARENT_ID = T2.PARENT_ID;
but that just gave me a strange result.
With a recursive cte.
If you want to start from the maximum id:
with recursive cte (id, parent_id) as (
select t.*
from (
select *
from tablename
order by id desc
limit 1
) t
union all
select t.*
from tablename t inner join cte c
on t.id = c.parent_id
)
select * from cte
See the demo.
If you want to start specifically from id = 6:
with recursive cte (id, parent_id) as (
select *
from tablename
where id = 6
union all
select t.*
from tablename t inner join cte c
on t.id = c.parent_id
)
select * from cte;
See the demo.
Results:
| id | parent_id |
| --- | --------- |
| 6 | 3 |
| 3 | 1 |
| 1 | 0 |
| 0 | |
Let's say we have 2 tables:
Table1: Table2:
id | t2id id | col
---------- ----------
1 | 1 1 | a
2 | 2 2 | b
3 | 2 3 | c
4 | 1 4 | d
5 | 3 5 | e
6 | 3 6 | f
7 | 4 7 | g
8 | 5 8 | h
9 | 1 9 | i
10 | 4 10 | j
My question is:
Is there any short way to put limit for distinct results of Table1.t2id column?
For example: if limit = 2 then all rows with t2id from 1 to 2 (or any other values) are selected.
Expected result (with limit = 2):
Res:
id | t2id
----------
1 | 1
2 | 2
3 | 2
4 | 1
9 | 1
Note:
Any information or suggestion are accepted
You could use just the where clause
Select id,t2id
from table1
where t2id<=2
Or you can use where .. between
Select id,t2id
from table1
where t2id between 1 and 2
I believe you want to:
Create a subquery with all the columns you need + this one: DENSE_RANK() OVER (ORDER BY Table1.t2id) AS MyRank
outside of the sub-query, add a where on MyRank
Complete solution:
SELECT id, tb2id
FROM (
SELECT id, tb2id, DENSE_RANK() OVER (ORDER BY Table1.t2id) AS MyRank
FROM table1
) MySubQuery
WHERE MyRank <= 2
This will adapt to JOINs with table2 (with potential multiplicity increase) and non-consecutive values in tb2id.
You can also use in:
select t1.*
from table1 t1
where t1.t2_id in (select t2.id from table2 t2 limit 2);
The advantage of this approach is that it is easy to make it random:
select t1.*
from table1 t1
where t1.t2_id in (select t2.id from table2 t2 order by random() limit 2);
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
I have a long table like the following. The table adds two similar rows after the id changes. E.g in the following table when ID changes from 1 to 2 a duplicate record is added. All I need is a SELECT query to skip this and all other duplicate records only if the ID changes.
# | name| id
--+-----+---
1 | abc | 1
2 | abc | 1
3 | abc | 1
4 | abc | 1
5 | abc | 1
5 | abc | 2
6 | abc | 2
7 | abc | 2
8 | abc | 2
9 | abc | 2
and so on
You could use NOT EXISTS to eliminate the duplicates:
SELECT *
FROM yourtable AS T
WHERE NOT EXISTS
( SELECT 1
FROM yourtable AS T2
WHERE T.[#] = T2.[#]
AND T2.ID > T.ID
);
This will return:
# name ID
------------------
. ... .
4 abc 1
5 abc 2
6 abc 2
. ... .
... (Some irrelevant rows have been removed from the start and the end)
If you wanted the first record to be retained, rather than the last, then just change the condition T2.ID > T.ID to T2.ID < T.ID.
You can use the following CTEs to simulate LAG window function not available in SQL Server 2008:
;WITH CTE_RN AS (
SELECT *, ROW_NUMBER() OVER (ORDER BY [#], id) AS rn
FROM #mytable
), CTE_LAG AS (
SELECT t1.[#], t1.name,
t1.id AS curId, t2.id AS prevId,
t1.[#] AS cur#, t2.[#] AS lag#
FROM CTE_RN t1
LEFT JOIN CTE_RN t2 ON t1.rn = t2.rn + 1 )
You can now filter out the 'duplicate' records using the above CTE_LAG and the following predicate in your WHERE clause:
;WITH (
... cte definitions here
) SELECT *
FROM CTE_LAG
WHERE (NOT ((prevId <> curId) AND (cur# = lag#))) OR (prevId IS NULL)
If prevId <> curId and cur# = lag#, then there is a change in the value of the id column and the following record has the same [#] value as the previous one, i.e. it is a duplicate.
Hence, using NOT on (prevId <> curId) AND (cur# = lag#), filters out all 'duplicate' records. This means record (5, abc, 2) will be eliminated.
SQL Fiddle Demo here
P.S. You can also add column name in the logical expression of the WHERE clause, depending on what defines a 'duplicate'.
So I achieved it by using the following query in SQL server.
select #, name, id
from table
group by #, name, id
having count(*) > 0
Following / copying computhomas's question, but adding some twists...
I have the following table in MSSQL2008
id | business_key | result | date
1 | 1 | 0 | 9
2 | 1 | 1 | 8
3 | 2 | 1 | 7
4 | 3 | n | 6
5 | 4 | 1 | 5
6 | 4 | 0 | 4
And now i want to group based on the business_key returning the complete entry with the newest date.
So my expected result is:
id | business_key | result | date
1 | 1 | 0 | 9
3 | 2 | 1 | 7
4 | 3 | n | 6
5 | 4 | 1 | 5
I also bet that there is a way to achieve that, i just can't find / see / think of it at the moment.
edit: sorry about this, I actually meant something else from original question I did. I felt like editing this might be better than accepting a solution and making another question. my original problem was that I am not filtering by id.
SELECT t.*
FROM
(
SELECT *, ROW_NUMBER() OVER
(
PARTITION BY [business_key]
ORDER BY [date] DESC
) AS [RowNum]
FROM yourTable
) AS t
WHERE t.[RowNum] = 1
SELECT
*
FROM
mytable
WHERE
ID IN (SELECT MAX(ID) FROM mytable GROUP BY business_key)
SELECT
MAX(T1.id) AS [id],
T1.business_key,
T1.result
FROM
dbo.My_Table T1
LEFT OUTER JOIN dbo.My_Table T2 ON
T2.business_key = T1.business_key AND
T2.id > T1.id
WHERE
T2.id IS NULL
GROUP BY T1.business_key,
T1.result
ORDER BY MAX(T1.id)
Edited based on clarifications
SELECT M1.*
FROM My_Table M1
INNER JOIN
(
SELECT [business_key], MAX([date]) as MaxDate
FROM My_Table
GROUP BY [business_key]
) M2 ON M1.business_key = M2.business_key AND M1.[date] = M2.MaxDate
ORDER BY M1.[id]
Assuming the combination of business_key & date is unique then....
Working example (3rd time is a charm):
declare #src as table(id int, business_key int,result int,[date] int)
insert into #src
SELECT 1,1,0,9
UNION SELECT 2,1,1,8
UNION SELECT 3,2,1,7
UNION SELECT 4,3,1,6
UNION SELECT 5,4,1,5
UNION SELECT 6,4,0,4
;with bkdate(business_key,[date])
AS
(
select business_key,MAX([date])
from #src
group by business_key
)
select src.* from #src src
inner join bkdate
ON src.[date] = bkdate.date
and src.business_key = bkdate.business_key
order by id
How about (edited after question change):
with latestdate as (
select business_key, maxdate=max(date)
from the_table
group by business_key
), latest as (
select ID = max(id)
from the_table
inner join latestdate
on the_table.business_key=latestdate.business_key
and the_table.date=latestdate.maxdate
group by the_table.business_key
)
select the_table.*
from the_table
inner join latest
on latest.id=the_table.id