Here is the data, call it table T
A B
-- --
1 14
2 15
3 16
4 1
4 3
4 6
4 9
4 12
4 15
I would like to get the value of A that has only one value and a B value of 15.
There are two rows where B=15 but there are 6 rows where A=4 and only one row where A=2.
So the correct SQL should return me the 2.
I have tried this but it returns both rows.
select A from T group by A,B having Count(A) = 1 and B = 15
This similarly fails:
select A from T where B = 15 group by A having count( A ) = 1
Try this:
select A
from T
group by A
having Count(A) = 1 and Max(B) = 15;
Your problem seems to be that you are grouping by both columns. You only want to group by A.
Admittedly, your query has group by A, T, but I think that is a typo, based on the described behavior.
You can check the count of B after grouping by A.
select A
from T
group by A
having Count(B) = 1 and max(B) = 15
Related
I have a table X in Postgres with the following entries
A B C
2 3 1
3 3 1
0 4 1
1 4 1
2 4 1
3 4 1
0 5 1
1 5 1
2 5 1
3 5 1
0 2 2
1 2 3
I would like to find out the entries having maximum of Column C for every kind of A and B i.e (group by B) with the most efficient query possible and return corresponding A and B.
Expected Output:
A B C
1 2 3
2 3 1
0 4 1
0 5 1
Please help me with this problem . Thank you
demo: db<>fiddle
Using DISTINCT ON:
SELECT DISTINCT ON (B)
A, B, C
FROM
my_table
ORDER BY B, C DESC, A
DISTINCT ON gives you exactly the first row for an ordered group. In this case B is grouped.
After ordering B (which is necessary): We first order the maximum C (with DESC) to the top of each group. Then (if there are tied MAX(C) values) we order the A to get the minimum A to the top.
Seems like it is a greatest n per group problem:
WITH cte AS (
SELECT *, RANK() OVER (PARTITION BY B ORDER BY C DESC, A ASC) AS rnk
FROM t
)
SELECT *
FROM cte
WHERE rnk = 1
You're not clear which A needs to be considered, the above returns the row with smallest A.
itseems to me you need max()
select A,B, max(c) from table_name
group by A,B
this will work:
select * from (SELECT t.*,
rank() OVER (PARTITION BY A,B order by C) rank
FROM tablename t)
where rank=1 ;
I have columns name, timestamp, doing. I've already sorted by name, then by timestamp, and I expect that moving down the doing column within a group with the same name looks like A, A, A, B, B, A, A, ... - alternating series of A and B. I need to get only the rows which comprise the first B row after a transition from A to B within a group with the same name.
name timestamp doing
1 1 A
1 2 A
1 3 B
1 4 B
1 5 A
2 2 B
2 4 A
2 6 B
2 8 A
I would like to return
name timestamp doing
1 3 B
2 6 B
But not
2 2 B
because it is not a transition from A to B within name = 2
I think you just want lag():
select t.*
from (select t.*,
lag(doing) over (partition by name order by timestamp) as prev_doing
from t
) t
where prev_doing = 'A' and doing = 'B';
I have a table with 2 columns A, and B that represent a connection graph between the two.
A B
1 3
2 5
4 2
3 5
2 3
I need to find how many instances of column A occur in column B (including 0)
So for the example above I would need the result set
A OccurencesInB
1 0
2 1
3 2
4 0
The best I have so far is
SELECT B, COUNT(*) AS TABLE_COUNT
FROM TABLE
GROUP BY B
ORDER BY TABLE_COUNT DESC
This does not find the instances of A that do not occur in B, which is crucial.
Any assistance will be greatly appreciated!
Use a correlated sub-query:
SELECT A,
TABLE_COUNT = (SELECT COUNT(*)
FROM TableName t2
WHERE t2.B = t1.A)
FROM TableName t1
GROUP BY A
ORDER BY TABLE_COUNT DESC, A
Result:
A TABLE_COUNT
3 2
2 1
1 0
4 0
By using SQlite, I'd like to get all rows that show in a specific column only one single distinct value. Like from following table:
A B
1 2
2 1
3 2
4 3
5 1
6 1
7 2
8 4
9 2
Here I'd like to get only row Nr. 4 an 8 as there values (3 and 4) occur only once in the entire column.
You could use a query like this:
SELECT *
FROM mytable
WHERE B IN (SELECT B FROM mytable GROUP BY B HAVING COUNT(DISTINCT A)=1)
Please see fiddle here.
Subquery will return all B values that are present only once (you could also use HAVING COUNT(*)=1 in this case), the outer query will return all rows where B is returned by the subquery.
ID? A B C
--- -- -- --
1 J 1 B
2 J 1 S
3 M 1 B
4 M 1 S
5 M 2 B
6 M 2 S
7 T 1 B
8 T 2 S
9 C 1 B
10 C 1 S
11 C 2 B
12 N 1 S
13 N 2 S
14 N 3 S
15 Q 1 S
16 Q 1 S
17 Z 1 B
I need to find unique values with multiple column with some added condition. The unique value are combination of Col A,B and C.
If Col A has only two rows (like record 1 and 2) and the Column B is same on both data and there is a different value as in Column C then i dont need those records.
If Col A has only multiple rows (like record 3 to 6 ) with different Col B and C combination we want to see those values.
If Col A has multiple rows (like record 7 to 8 ) with different Col B and C combination we want to see those values.
If Col A has only multiple rows (like record 9 to 11 ) with similar/different Col B and C combination we want to see those values.
If Col A has only multiple rows (like record 12onwards ) with similar Col C and similar or different Column B we dont need those values...
If single value like Row 17 there is no need to display either
Tried a lot but not getting exact answer any help is greatly appreciated..
Trying to go through all the logic, I think you want all rows where the values of both columns A and B differ. An easy way to see whether records differ is by looking at the min and max values. And, you can do this using analytic functions:
select A, B, C
from (select t.*,
count(*) over (partition by A) as Acnt,
min(B) over (partition by A) as Bmin,
max(B) over (partition by A) as Bmax,
min(C) over (partition by A) as Cmin,
max(C) over (partition by A) as Cmax
from t
) t
where (Bmin <> Bmax or Cmin <> Cmax)
Your example data does not have any actual duplicates, so I don't think a count(distinct) is necessary. Your rules say nothing about what to do when A only appears once. This version will filter those rows out.