I have a table below. There are 3 types(A,B,C) available. I want to create general a Sql query. If row has type B or type C, the row with type B or C should be listed. If row has just type A, the row with type A should be listed.
Table;
Number
Type
1
A
1
B
2
A
3
A
3
C
4
A
5
A
6
A
6
B
6
C
Expected result when the query run;
Number
Type
1
B
2
A
3
C
4
A
5
A
6
B
6
C
How can I create the query? Thank you in advance for your help.
I would assign each type a precedence, and only return the types of the highest precedence.
Where two types can be the same precedence (because you want to return both), RANK() (rather than ROW_NUMBER()) will ensure both are assigned the same value.
WITH
precedence AS
(
SELECT
*,
RANK()
OVER (
PARTITION BY Number
ORDER BY CASE Type WHEN 'C' THEN 2
WHEN 'B' THEN 2
WHEN 'A' THEN 1
ELSE 0
END
DESC
)
AS row_precedence
FROM
your_table
)
SELECT
*
FROM
precedence
WHERE
row_precedence = 1
One option to make it a little neater could be to use APPLY (or a join on a lookup table) to derive the integers outside of the window function's code...
WITH
precedence AS
(
SELECT
*,
RANK()
OVER (
PARTITION BY Number
ORDER BY type_precedence.value DESC
)
AS row_precedence
FROM
your_table
CROSS APPLY
(
SELECT
CASE Type WHEN 'C' THEN 2
WHEN 'B' THEN 2
WHEN 'A' THEN 1
ELSE 0
END
AS value
)
AS type_precedence
)
SELECT
*
FROM
precedence
WHERE
row_precedence = 1
Demo: https://dbfiddle.uk/2Abwpa8p
Related
Consider the following table
id attribute
1 a
1 a
1 b
2 a
2 a
3 c
4 a
I want to select the ids that have attribute of 'a' only, ie 2 and 4.
Cant select 1 because 1 has 'a' and 'b', cant select 3 because it has 'c' only. We select 2 and 4 because it has 'a' value only.
You can use
SELECT id
FROM YourTable
GROUP BY id
HAVING MAX(attribute) = 'a' AND MIN(attribute) = 'a'
AND COUNT(*) = COUNT(attribute)
the
COUNT(*) = COUNT(attribute)
is to discard any id that have NULL attribute as well as a. Remove this if that is not the semantics you want or the column is not nullable anyway.
Please test this:
SELECT id
FROM attribute
GROUP BY id
HAVING
COUNT(DISTINCT attribute) = 1 AND MIN(attribute)= 'a';
For example:
This is the original result
Alpha Beta
A 1
B 2
B 3
C 4
After Order by the number of Alpha, this is the result I want
Alpha Beta
B 2
B 3
A 1
C 4
I tried to use GroupBy and OrderBy, but ACCESS always ask me to include all columns.
Why is 'B' placed before 'A' ? I don't understand this order..
Any way, doesn't seem like you need a group by, not from your data sample, but for your desired result you can use CASE EXPRESSION :
SELECT t.alpha,t.beta FROM YourTable t
ORDER BY CASE WHEN t.alpha = 'B' THEN 1 ELSE 0 END DESC,
t.aplha,
t.beta
EDIT: Use this query:
SELECT t.alpha,t.beta FROM YourTable t
INNER JOIN(SELECT s.alpha,count(*) as cnt
FROM YourTable s
GROUP BY s.alpha) t2
ON(t.aplha = t2.alpha)
ORDER BY t2.cnt,t.alpha,t.beta
The query counts number of rows for every distinct Alpha and sorts. General Sql, tweak for ACCESS if needed.
SELECT t1.alpha,t1.beta
FROM t t1
JOIN (
SELECT t2.alpha, count(t2.*) AS n FROM t t2 GROUP BY t2.alpha
) t3 ON t3.alpha = t1.alpha
ORDER BY t3.n, t1.alpha, t1.beta
I would like to write an Oracle query which returns a specific set of information. Using the table below, if given an id, it will return the id and value of B. Also, if B=T, it will return the next row as well. If that next row has a B=T, it will return that, and so on until a F is encountered.
So, given 3 it would just return one row: (3,F). Given 4 it would return 3 rows: ((4,T),(5,T),(6,F))
id B
1 F
2 F
3 F
4 T
5 T
6 F
7 T
8 F
Thank you in advance!
Use a sub-query to find out at what point you should stop, then return all row from your starting point to the calculated stop point.
SELECT
*
FROM
yourTable
WHERE
id >= 4
AND id <= (SELECT MIN(id) FROM yourTable WHERE b = 'F' AND id >= 4)
Note, this assumes that the last record is always an 'F'. You can deal with the last record being a 'T' using a COALESCE.
SELECT
*
FROM
yourTable
WHERE
id >= 4
AND id <= COALESCE(
(SELECT MIN(id) FROM yourTable WHERE b = 'F' AND id >= 4),
(SELECT MAX(id) FROM yourTable )
)
This is my example data (there are no indexes and I do not want to create any):
CREATE TABLE tblTest ( a INT , b INT );
INSERT INTO tblTest ( a, b ) VALUES
( 1 , 2 ),
( 5 , 1 ),
( 1 , 4 ),
( 3 , 2 )
I want the minimum value in of both column a and column b which is greater then a given value. E.g. if the given value is 3 then I want 4 to be returned.
This is my current solution:
SELECT MIN (subMin) FROM
(
SELECT MIN (a) as subMin FROM tblTest
WHERE a > 3 -- Returns 5
UNION
SELECT MIN (b) as subMin FROM tblTest
WHERE b > 3 -- Returns 4
)
This searches the table twice - once to get min(a) once to get min(b).
I believe it should be faster to do this with just one pass. Is this possible?
You want to use conditional aggregatino for this:
select min(case when a > 3 then a end) as minA,
min(case when b > 3 then b end) as minB
from tblTest;
To get the minimum of both values, you can use a SQLite extension, which handles multiple values for min():
select min(min(case when a > 3 then a end),
min(case when b > 3 then b end)
)
from tblTest
The only issue is that the min will return NULL if either argument is NULL. You can fix this by doing:
select coalesce(min(min(case when a > 3 then a end),
min(case when b > 3 then b end)
),
min(case when a > 3 then a end),
min(case when b > 3 then b end)
)
from tblTest
This version will return the minimum value, subject to your conditions. If one of the conditions has no rows, it will still return the minimum of the other value.
From the top of my head, you could modify the table and add a min value column to store the minimum value of the two columns. then query that column.
Or you can do this:
select min(val)
from
(
select min(col1, col2) as val
from table1
)
where
val > 3
The outer SELECT, queries the memory, not the table itself.
Check SQL Fiddle
I will keep this simple- I would like to know if there is a good way to select all the values in a column when it never has a null in another column. For example.
A B
----- -----
1 7
2 7
NULL 7
4 9
1 9
2 9
From the above set I would just want 9 from B and not 7 because 7 has a NULL in A. Obviously I could wrap this as a subquery and USE the IN clause etc. but this is already part of a pretty unique set and am looking to keep this efficient.
I should note that for my purposes this would only be a one-way comparison... I would only be returning values in B and examining A.
I imagine there is an easy way to do this that I am missing, but being in the thick of things I don't see it right now.
You can do something like this:
select *
from t
where t.b not in (select b from t where a is null);
If you want only distinct b values, then you can do:
select b
from t
group by b
having sum(case when a is null then 1 else 0 end) = 0;
And, finally, you could use window functions:
select a, b
from (select t.*,
sum(case when a is null then 1 else 0 end) over (partition by b) as NullCnt
from t
) t
where NullCnt = 0;
The query below will only output one column in the final result. The records are grouped by column B and test if the record is null or not. When the record is null, the value for the group will increment each time by 1. The HAVING clause filters only the group which has a value of 0.
SELECT B
FROM TableName
GROUP BY B
HAVING SUM(CASE WHEN A IS NULL THEN 1 ELSE 0 END) = 0
If you want to get all the rows from the records, you can use join.
SELECT a.*
FROM TableName a
INNER JOIN
(
SELECT B
FROM TableName
GROUP BY B
HAVING SUM(CASE WHEN A IS NULL THEN 1 ELSE 0 END) = 0
) b ON a.b = b.b