Consider following table:
Number | Value
1 a
1 b
1 a
2 a
2 a
3 c
4 a
5 d
5 a
I want to choose every row, where the value for one number is the same, so my result should be:
Number | Value
2 a
3 c
4 a
I manage to get the right numbers by using nested
SQL-Statements like below. I am wondering if there is a simpler solution for my problem.
SELECT
a.n,
COUNT(n)
FROM
(
SELECT number n , value k
FROM testtable
GROUP BY number, value
) a
GROUP BY n
HAVING COUNT(n) = 1
You can try this
SELECT NUMBER,MAX(VALUE) AS VALUE FROM TESTTABLE
GROUP BY NUMBER
HAVING MAX(VALUE)=MIN(VALUE)
You can try also this:
SELECT DISTINCT t.number, t.value
FROM testtable t
LEFT JOIN testtable t_other
ON t.number = t_other.number AND t.value <> t_other.value
WHERE t_other.number IS NULL
Another alternative using exists.
select distinct num, val from testtable a
where not exists (
select 1 from testtable b
where a.num = b.num
and a.val <> b.val
)
http://sqlfiddle.com/#!9/dd080dd/5
Related
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've the following table with 3 columns: Id, FeatureName and Value:
Id FeatureName Value
-- ----------- -----
1 AAA 10
1 ABB 12
1 BBB 12
2 AAA 15
2 ABB 12
2 ACD 7
3 AAA 10
3 ABB 12
3 CCC 12
.............
Each Id has different features and each Feature has a value for that Id.
I need to write a query which gives me the Ids that have exactly the same features and values than a given one, but only taking into account those whose name starts with 'A'. For example, in the top table, I can use that query to search for all the Ids that have the same features. For example, features with values where Id=1 would result Id=3 with same features starting with 'A' and same values for these features.
I found a couple of different ways to do this, but all of them go very slow when the table has lots of rows (more than hundred of thousands)
The way I obtain the best performance is using the next query:
select a2.Id
from (select a.FeatureName, a.Value
from Table1 a
where a.Id = 1) a1,
(select a.Id, a.FeatureName, a.Value
from Table1 a
where a.FeatureName like 'A%') a2
where a1.FeatureName = a2.FeatureName
and a1.value = a2.value
group by a2.Id
having count(*) = 2
intersect
select a.Id
from Table1 a
where a.FeatureName like 'A%'
group by a.Id
having count(*)= 2
where #nFeatures is the number of features starting by 'A' in Id=1. I counted them before calling this query. I make the intersection to avoid results that have the same parameters than Id=1 but also some others whose name starts with 'A'.
I think that the slowest part is the second subquery:
select a.Id, a.FeaureName, a.Value
from MyTable a
where a.FeatureName = 'A%'
but I don't know how to make it faster. Maybe I will have to play with the indexes.
Any idea of how could I write a fast query for this purpose?
So you want all rows where the combination of FeatureName and Value is not unique? You can use EXISTS:
SELECT t.*
FROM dbo.Table1 t
WHERE t.FeatureName LIKE 'A%'
AND EXISTS(SELECT 1 FROM dbo.Table1 t2
WHERE t.Id <> t2.ID
AND t.FeatureName = t2.FeatureName
AND t.Value = t2.Value)
Demo
how could I write a fast query for this purpose?
If it's not fast enough create an index on FeatureName + Value.
I tried to eliminate the join with MyTable again to select the data for the ID's that have matching FeatureName and Value values. Here's the query:
with joined_set as
(
SELECT
mt1.*, mt2.id as mt2_id, mt2.featurename as mt2_FeatureName, mt2.value as mt2_value
from
(
select *
from mytable
where featurename like 'A%'
) mt1
left join
(
select *
from mytable
where featurename like 'A%'
) mt2
on mt2.id <> mt1.id and mt2.FeatureName = mt1.featurename and mt2.value = mt1.value
)
select distinct id
from joined_set
where id not in
(select id
from joined_set
group by id
having SUM(
CASE
WHEN mt2_id is null THEN 1
ELSE 0
END
) <> 0
);
Here is the SQL Fiddle demo. It has an extra condition in the inline view mt2, to perform this search only for id = 1.
I'm a little dense this morning, I'm not sure if you wanted just the ID's or...
Here's my take on it...
You could probably move the where FeatureName like 'A%' into the inner query to filter the data on the initial table scan.
with dupFeatures (FeatureName, Value, dupCount)
as
(
select FeatureName, Value, count(*) as dupCount from MyTable
group by FeatureName, Value
having count(*) > 1
)
select MyTable.Id, dupFeatures.FeatureName,dupFeatures.Value
from dupFeatures
join MyTable on (MyTable.FeatureName = dupFeatures.FeatureName and
MyTable.Value = dupFeatures.Value )
where dupFeatures.FeatureName like 'A%'
order by FeatureName, Value, Id
A general solution is
With Rows As (
select id
, FeatureName
, Value
, rows = Count(id) OVER (PARTITION BY id)
FROM test
WHERE FeatureName LIKE 'A%')
SELECT a.id aID, b.id bID
FROM Rows a
INNER JOIN Rows b ON a.id < b.id and a.FeatureName = b.FeatureName
and a.rows = b.rows
GROUP BY a.id, b.id
ORDER BY a.id, b.id
to limit the solution to a group just add a WHERE condition on the main query for a.ID. The CTE is needed to get the correct number of rows for each id
SQLFiddle demo, in the demo I changed little the test data to have a another couple of ID with only one of the FeatureName of 1 and 3
I am searching for an implementation of the following pseodo-code:
SELECT A, B, C
FROM X
HAVING COUNT(A,B) > 1
Here is an example of what the code should do:
Assume table X looks as follows:
A B C D
--------------
1 1 0 2
1 1 1 1
2 1 1 0
The first and the second row have the same entries in columns A and B, the third column is identical in column B but different in column A. The desired output is columns A,B, and C of rows 1 and 2:
1 1 0
1 1 1
How could this be implemented? The problem with my pseodo-code is, that COUNT accepts either a single column or all columns (*), but it can't take two out of 4 columns. GROUP BY has the same property.
You can do this with an exists clause. This should work in all databases:
select a, b, c
from x
where exists (select 1
from x x2
where x.a = x2.a and x.b = x2.b and x.c <> x2.c
);
This assumes that the rows have difference c values.
This will perform best with an index on x(a, b).
For RDMS that supports analytic functions, you can do
SELECT a,b,c
FROM
(
SELECT a, b, c, count(1) OVER(PARTITION BY a,b) cnt
FROM X
)t1
WHERE t1.cnt >1
If analytic/windows function are not available , join should do the job
SELECT t1.a, t1.b, t1.c
FROM X t1
INNER JOIN
(
SELECT a,b
FROM X
GROUP BY a,b
HAVING COUNT(1) >1
)t2 ON (t2.a=t1.a AND t2.b=t1.b)
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 )
)
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