I have a test table
ID V_ID
1 1
1 2
I want max(V_ID) and resulr should be V_ID 2
select Id,max(V_ID) from test
group by Id,value
I am trying simple query but it's still pulling two records. Is there any other simple query 1) we can try rank 2)?
You should be grouping only by the ID column:
SELECT ID, MAX(V_ID)
FROM test
GROUP BY IdD;
A more general pattern for this type of problem uses ROW_NUMBER to find the entire record for each Id having the max value of V_ID:
SELECT ID, V_ID
FROM
(
SELECT *, ROW_NUMBER() OVER (PARTITION BY ID ORDER BY V_ID DESC) rn
FROM test
) t
WHERE rn = 1;
Related
I tried to do it with this query, but it's not working...
DELETE FROM employee
WHERE ( SELECT * FROM
(SELECT row_number() OVER (partition by id) rn FROM employee) alias
) > 1;
Please click on this link to view the table
The above query is not working and giving this error message:
Error Code: 1242. Subquery returns more than 1 row
try like below by using subquery
delete from
(
select *.row_number() over (partition by id order by id) rn
from employee
) alias where rn > 1;
You are matching an integer (1) with set of rows returned from the subquery, which SQL will not allow
You can match an integer (1) with a single value returned from the subquery.
Use below query (using CTE) to remove duplicates.
;WITH TempEmp (id,duplicateRecCount)
AS
(
SELECT id,ROW_NUMBER() OVER(PARTITION by id ORDER BY id)
AS duplicateRecCount
FROM employee
)
DELETE FROM TempEmp
WHERE duplicateRecCount > 1
I've got a table with a lot of columns in it and I want to run a query to find the most common value in each column.
Ordinarily for a single column, I'd run something like:
SELECT country
FROM users
GROUP BY country
ORDER BY count(*) DESC
LIMIT 1
Does PostgreSQL have a built in function for doing this or can anyone suggest a query I could run to achieve this?
Using the same query, for more than one column you should do:
SELECT *
FROM
(
SELECT country
FROM users
GROUP BY 1
ORDER BY count(*) DESC
LIMIT 1
) country
,(
SELECT city
FROM users
GROUP BY 1
ORDER BY count(*) DESC
LIMIT 1
) city
This works for any type and will return all the values in the same row, with the columns having its original name.
For more columns just had more subquerys as:
,(
SELECT someOtherColumn
FROM users
GROUP BY 1
ORDER BY count(*) DESC
LIMIT 1
) someOtherColumn
Edit:
You could reach it with window functions also. However it will not be better in performance nor in readability.
Starting from PG 9.4 there is aggregate function for this:
mode() WITHIN GROUP (ORDER BY sort_expression)
returns the most frequent input value (arbitrarily choosing the first one if there are multiple equally-frequent results)
And for earlier versions, you could create one...
CREATE OR REPLACE FUNCTION mode_array(anyarray)
RETURNS anyelement AS
$BODY$
SELECT a FROM unnest($1) a GROUP BY 1 ORDER BY COUNT(1) DESC, 1 LIMIT 1;
$BODY$
LANGUAGE SQL IMMUTABLE;
CREATE AGGREGATE mode(anyelement)(
SFUNC = array_append, --Function to call for each row. Just builds the array
STYPE = anyarray,
FINALFUNC = mode_array, --Function to call after everything has been added to array
INITCOND = '{}'--Initialize an empty array when starting
) ;
Usage: SELECT mode(column) FROM table;
If I were doing this, I'd write a query like this one:
SELECT 'country', country
FROM users
GROUP BY country
ORDER BY count(*) DESC
LIMIT 1
UNION ALL
SELECT 'city', city
FROM USERS
GROUP BY city
ORDER BY count(*) DESC
LIMIT 1
-- etc.
It should be noted this only works if all the columns are of compatible types. If they are not, you'll probably need a different solution.
This window function version will read the users table and the computed table once each. The correlated subquery version will read the users table once for each of the columns. If the columns are many as in the OPs case then my guess is that this is faster. SQL Fiddle
select distinct on (country_count, age_count) *
from (
select
country,
count(*) over(partition by country) as country_count,
age,
count(*) over(partition by age) as age_count
from users
) s
order by country_count desc, age_count desc
limit 1
I have a field called "Users", and I want to run SUM() on that field that returns the sum of all DISTINCT records. I thought that this would work:
SELECT SUM(DISTINCT table_name.users)
FROM table_name
But it's not selecting DISTINCT records, it's just running as if I had run SUM(table_name.users).
What would I have to do to add only the distinct records from this field?
Use count()
SELECT count(DISTINCT table_name.users)
FROM table_name
SQLFiddle demo
This code seems to indicate sum(distinct ) and sum() return different values.
with t as (
select 1 as a
union all
select '1'
union all
select '2'
union all
select '4'
)
select sum(distinct a) as DistinctSum, sum(a) as allSum, count(distinct a) as distinctCount, count(a) as allCount from t
Do you actually have non-distinct values?
select count(1), users
from table_name
group by users
having count(1) > 1
If not, the sums will be identical.
You can see for yourself that distinct works with the following example. Here I create a subquery with duplicate values, then I do a sum distinct on those values.
select DistinctSum=sum(distinct x), RegularSum=Sum(x)
from
(
select x=1
union All
select 1
union All
select 2
union All
select 2
) x
You can see that the distinct sum column returns 3 and the regular sum returns 6 in this example.
You can use a sub-query:
select sum(users)
from (select distinct users from table_name);
SUM(DISTINCTROW table_name.something)
It worked for me (innodb).
Description - "DISTINCTROW omits data based on entire duplicate records, not just duplicate fields." http://office.microsoft.com/en-001/access-help/all-distinct-distinctrow-top-predicates-HA001231351.aspx
;WITH cte
as
(
SELECT table_name.users , rn = ROW_NUMBER() OVER (PARTITION BY users ORDER BY users)
FROM table_name
)
SELECT SUM(users)
FROM cte
WHERE rn = 1
SQL Fiddle
Try here yourself
TEST
DECLARE #table_name Table (Users INT );
INSERT INTO #table_name Values (1),(1),(1),(3),(3),(5),(5);
;WITH cte
as
(
SELECT users , rn = ROW_NUMBER() OVER (PARTITION BY users ORDER BY users)
FROM #table_name
)
SELECT SUM(users) DisSum
FROM cte
WHERE rn = 1
Result
DisSum
9
If circumstances make it difficult to weave a "distinct" into the sum clause, it will usually be possible to add an extra "where" clause to the entire query - something like:
select sum(t.ColToSum)
from SomeTable t
where (select count(*) from SomeTable t1 where t1.ColToSum = t.ColToSum and t1.ID < t.ID) = 0
May be a duplicate to
Trying to sum distinct values SQL
As per Declan_K's answer:
Get the distinct list first...
SELECT SUM(SQ.COST)
FROM
(SELECT DISTINCT [Tracking #] as TRACK,[Ship Cost] as COST FROM YourTable) SQ
I made two queries that I thought should have the same result:
SELECT COUNT(*) FROM (
SELECT DISTINCT ON (id1) id1, value
FROM (
SELECT table1.id1, table2.value
FROM table1
JOIN table2 ON table1.id1=table2.id
WHERE table2.value = '1')
AS result1 ORDER BY id1)
AS result2;
SELECT COUNT(*) FROM (
SELECT DISTINCT ON (id1) id1, value
FROM (
SELECT table1.id1, table2.value
FROM table1
JOIN table2 ON table1.id1=table2.id
)
AS result1 ORDER BY id1)
AS result2
WHERE value = '1';
The only difference being that one had the WHERE clause inside SELECT DISTINCT ON, and the other outside that, but inside SELECT COUNT. But the results were not the same. I don't understand why the position of the WHERE clause should make a difference in this case. Can anyone explain? Or is there a better way to phrase this question?
here's a good way to look at this:
SELECT DISTINCT ON (id) id, value
FROM (select 1 as id, 1 as value
union
select 1 as id, 2 as value) a;
SELECT DISTINCT ON (id) id, value
FROM (select 1 as id, 1 as value
union
select 1 as id, 2 as value) a
WHERE value = 2;
The problem has to do with the unique conditions and what is visible where. It is behavior by design.
Lets say I have table with ID int, VALUE string:
ID | VALUE
1 abc
2 abc
3 def
4 abc
5 abc
6 abc
If I do select value, count(*) group by value I should get
VALUE | COUNT
abc 5
def 1
Now the tricky part, if there is count == 1 I need to get that ID from first table. Should I be using CTE? creating resultset where I will add ID string == null and run update b.ID = a.ID where count == 1 ?
Or is there another easier way?
EDIT:
I want to have result table like this:
ID VALUE count
null abc 5
3 def 1
If your ID values are unique, you can simply check to see if the max(id) = min(id). If so, then use either one, otherwise you can return null. Like this:
Select Case When Min(id) = Max(id) Then Min(id) Else Null End As Id,
Value, Count(*) As [Count]
From YourTable
Group By Value
Since you are already performing an aggregate, including the MIN and Max function is not likely to take any extra (noticeable) time. I encourage you to give this a try.
The way I would do it would indeed be a CTE:
using #group AS (SELECT value, Count(*) as count from MyTable GROUP BY value HAVING count = 1)
SELECT MyTable.ID, #group.value, #group.count from MyTable
JOIN #group ON #group.value = MyTable.value
When using group by, after the group by statement you can use a having clause.
So
SELECT [ID]
FROM table
GROUP BY [VALUE]
HAVING COUNT(*) = 1
Edit: with regards to your edited question: this uses some fun joins and unions
CREATE TABLE #table
(ID int IDENTITY,
VALUE varchar(3))
INSERT INTO #table (VALUE)
VALUES('abc'),('abc'),('def'),('abc'),('abc'),('abc')
SELECT * FROM (
SELECT Null as ID,VALUE, COUNT(*) as [Count]
FROM #table
GROUP BY VALUE
HAVING COUNT(*) > 1
UNION ALL
SELECT t.ID,t.VALUE,p.Count FROM
#table t
JOIN
(SELECT VALUE, COUNT(*) as [Count]
FROM #table
GROUP BY VALUE
HAVING COUNT(*) = 1) p
ON t.VALUE=p.VALUE
) a
DROP TABLE #table
maybe not the most efficient but something like this works:
SELECT MAX(Id) as ID,Value FROM Table WHERE COUNT(*) = 1 GROUP BY Value