Does anyone happen to know a way of basically taking the 'Distinct' command but only using it on a single column. For lack of example, something similar to this:
Select (Distinct ID), Name, Term from Table
So it would get rid of row with duplicate ID's but still use the other column information. I would use distinct on the full query but the rows are all different due to certain columns data set. And I would need to output only the top most term between the two duplicates:
ID Name Term
1 Suzy A
1 Suzy B
2 John A
2 John B
3 Pete A
4 Carl A
5 Sally B
Any suggestions would be helpful.
select t.Id, t.Name, t.Term
from (select distinct ID from Table order by id, term) t
You can use row number for this
Select ID, Name, Term from(
Select ID, Name, Term, ROW_NUMBER ( )
OVER ( PARTITION BY ID order by Name) as rn from Table
Where rn = 1)
as tbl
Order by determines the order from which the first row will be picked.
Related
I have following data in the table.
Id Name
1 Abc
2 Abc
3 Xyz
4 Xyz
5 def
6 def
I want following results from the query
Id Name
1 Abc
2 Xyz
3 def
I want to avoid duplicates in the name column.
Any help is greatly appreciated.
Select distinct
id, name
from table A
will not work as ids are having a different values.
Use a group by instead.
select
min(id), [name]
from
tableA
group by [name]
Note that in your example, the ids that corresponds with Xyz are 3 and 4, so getting a 2 next to Xyz is only possible if you break the integrity of the table. If you are just looking for an auto number next to the ids you can do this:
SELECT row_number() OVER (ORDER BY min(id)) id,
name
FROM tableA
group by name
You can get your specific result using:
select row_number() over (order by min(id)) as id, name
from table A
group by name;
Renumbering the rows seems strange, but row_number() will do that.
I used the following query and it returned what I wanted it to return, but I'm having a tough time wrapping my head around what the query is doing.
Query is nothing fancier than what's in the title: select distinct(count(*)) from table1
Distinct is not required in your SQL ,as you are going to get only result, count(*) without group by clause returns, count of all rows within that table.
Hence try this :
select count(*) from table1
Distinct is used for finding distinct values from a group of values:
say you have table1 , with column1 as :
Column1
----------
a
a
b
b
a
c
following sqls are run you will get output as :
1) select count(*) from table1
output :6
2) select distinct(count(*)) from table1
output :6
3) select count( distinct column1) from table1
output :3
Usually distinct is used inside count preferably with a particular column .
select count( distinct column_name_n ) from table1
The distinct is redundant... Select Count(*) with only one table can only generate one value, so distinct (which would eliminate duplicates) is irelelvant.
If you had multiple outputs, (if for example you were grouping on something) then it would cause the query to only display one output row for every distinct value of count(*) that would other wise be generated...
if, for example, you had
name
Bob
Bob
Bob
Bob
Mary
Mary
Mary
Mary
Dave
Dave
Al
George
then
select count(*)
From table
group By name
would result in
4
4
2
1
1
but
select distinct count(*)
From table
group By name
would result in
4
2
1
ID Sum Name
a 10 Joe
a 8 Mary
b 21 Kate
b 110 Casey
b 67 Pierce
What would you recommend as the best way to
obtain for each ID the name that corresponds to the largest sum (grouping by ID).
What I tried so far:
select ID, SUM(Sum) s, Name
from Table1
group by ID, Name
Order by SUM(Sum) DESC;
this will arrange the records into groups that have the highest sum first. Then I have to somehow flag those records and keep only those. Any tips or pointers? Thanks a lot
In the end I'd like to obtain:
a 10 Joe
b 110 Casey
You want the row_number() function:
select id, [sum], name
from (select t.*]
row_number() over (partition by id order by [sum] desc) as seqnum
from table1
) t
where seqnum = 1;
Your question is more confusing than it needs to be because you have a column called sum. You should avoid using SQL reserved words for identifiers.
The row_number() function assigns a sequential number to a group of rows, starting with 1. The group is defined by the partition by clause. In this case, all rows with the same id are in the same group. The ordering of the numbers is determined by the order by clause, so the one with the largest value of sum gets the value of 1.
If you might have duplicate maximum values and you want all of them, use the related function rank() or dense_rank().
select *
from
(
select *
,rn = row_number() over (partition by Id order by sum desc)
from table
)x
where x.rn=1
demo
I am working on SQL Server. I have a table, that contains around 75000 records. Among them there are several duplicate records. So i wrote a query to know which record repeated how many times like,
SELECT [RETAILERNAME],COUNT([RETAILERNAME]) as Repeated FROM [Stores] GROUP BY [RETAILERNAME]
It gives me result like,
---------------------------
RETAILERNAME | Repeated
---------------------------
X | 4
---------------------------
Y | 6
---------------------------
Z | 10
---------------------------
Among 4 record(s) of X record, i need take only first record of X.
so here i want to retrieve all fields from first row of duplicate records. i.e. Take all records whose RETAILERNAME='X' we will get some no. of duplicate records, we need to get only first row from them.
Please guide me.
You could try using ROW_NUMBER.
Something like
;WITH Vals AS (
SELECT [RETAILERNAME],
ROW_NUMBER() OVER(PARTITION BY [RETAILERNAME] ORDER BY [RETAILERNAME]) RowID
FROM [Stores ]
)
SELECT *
FROm Vals
WHERE RowID = 1
SQL Fiddle DEMO
You can then also remove the duplicates if need be (BUT BE CAREFUL THIS IS PERMANENT)
;WITH Vals AS (
SELECT [RETAILERNAME],
ROW_NUMBER() OVER(PARTITION BY [RETAILERNAME] ORDER BY [RETAILERNAME]) RowID
FROM Stores
)
DELETE
FROM Vals
WHERE RowID > 1;
You Can write query as under
SELECT TOP 1 * FROM [Stores] GROUP BY [RETAILERNAME]
HAVING your condition
WITH cte
AS (SELECT [retailername],
Row_number()
OVER(
partition BY [retailername]
ORDER BY [retailername])'RowRank'
FROM [retailername])
SELECT *
FROM cte
I have a SQL Server 2008 database with the following information in a table:
ID Name
-- ----
1 John
2 Jill
3 John
4 Phil
5 Matt
6 Jill
I want to display the unique names in a drop down list. Because of this, I need just one of the IDs associated with the unique name. I know it's dirty. I didn't create this mess. I just need the unique names with one of the ids. How do I write a query that will do that? I know that the following won't work because of the ID field.
SELECT DISTINCT
[ID], [Name]
FROM
MyTable
SELECT MIN(ID) AS ID, [Name]
FROM MyTable
GROUP BY [Name]
This will return the first (i.e. MINimum) ID for each distinct Name
You could also do it with rank over function
SELECT
Id,
Name
FROM
(
SELECT
Id,
[Name],
RANK() OVER (PARTITION BY [Name] Order By Id) As Idx
FROM Test
) A
WHERE Idx = 1
To get understanding about rank over function read this:
http://msdn.microsoft.com/en-us/library/ms176102.aspx