I need to be able to select only the first row for each name that has the greatest value.
I have a table with the following:
id name value
0 JOHN 123
1 STEVE 125
2 JOHN 127
3 JOHN 126
So I am looking to return:
id name value
1 STEVE 125
2 JOHN 127
Any idea on the MSSQL Syntax on how to perform this operation?
While you specified SQL Server, you did not specify the version. If you are using SQL Server 2005 or later, you can do something like:
With RankedItems As
(
Select id, name, value
, Row_Number() Over ( Partition By name Order By value Desc, id Asc ) As ItemRank
From Table
)
Select id, name, value
From RankedItems
Where ItemRank = 1
try:
SELECT
MIN(id) as id,dt.name,dt.value
FROM (SELECT
name,MAX(value) as value
FROM YourTable
GROUP BY name
) dt
INNER JOIN YourTable t ON dt.name=t.name and dt.value=t.value
GROUP BY dt.name,dt.value
try it out:
DECLARE #YourTable table (id int, name varchar(10), value int)
INSERT #YourTable VALUES (0, 'JOHN', 123)
INSERT #YourTable VALUES (1, 'STEVE', 125)
INSERT #YourTable VALUES (2, 'JOHN', 127)
INSERT #YourTable VALUES (3, 'JOHN', 126)
--extra data not in the question, shows why you need the outer group by
INSERT #YourTable VALUES (4, 'JOHN', 127)
INSERT #YourTable VALUES (5, 'JOHN', 127)
INSERT #YourTable VALUES (6, 'JOHN', 127)
INSERT #YourTable VALUES (7, 'JOHN', 127)
SELECT
MIN(id) as id,dt.name,dt.value
FROM (SELECT
name,MAX(value) as value
FROM #YourTable
GROUP BY name
) dt
INNER JOIN #YourTable t ON dt.name=t.name and dt.value=t.value
GROUP BY dt.name,dt.value
ORDER BY id
output:
id name value
----------- ---------- -----------
1 STEVE 125
2 JOHN 127
(2 row(s) affected)
You could do something like
SELECT id, name, value
FROM (SELECT id, name, value
ROWNUMBER() OVER (PARTITION BY name ORDER BY value DESC) AS r
FROM table) AS x
WHERE x.r = 1 ;
This will not work in SQL Server 2000 and earlier, but it will be incredibly fast in SQL Server 2005 and 2008
How about:
SELECT a.id, a.name, b.maxvalue
FROM mytbl a
INNER JOIN (SELECT id, max(value) as maxvalue
FROM mytbl
GROUP BY id) b ON b.id = a.id
SELECT a.id, a.name, a.value
FROM mytbl a
INNER JOIN (SELECT name, max(value) as maxvalue
FROM mytbl
GROUP BY name) b ON b.name = a.name and b.maxvalue = a.value
Related
Users table details
userid values (abc,xyz,abc,sdf)
master table details
(mid,priority)values(101,1),(102,2),(101,1),(103,1)
i need to count of mid based on userid (userid is names of users) group by priority(priority is int ) grouping like case priority =1 then 'Open', priority =2 then 'closed' etc using CTE(common table expression)
Select * from users
userid
abc
xyz
abc
sdf
Select * from master
mid Priority
101 1
102 2
101 1
103 1
(Priority 1= Open 2=Closed)
OUTPUT expected:
Userid count(mid) Priority
abc 2 Open
xyz 1 Closed
sdf 1 Open
Try this:
use db_test;
go
drop table dbo.users;
create table dbo.users
(
userid varchar(max) not null
)
;
insert into dbo.users
values
('abc'),
('xyz'),
('sdf')
create table dbo.master
(
mid int not null,
Priority int not null
)
;
insert into dbo.master
values
(101, 1),
(102, 2),
(101, 1),
(103, 1)
;
with cte1 as (
select userid, row_number() over(order by userid asc) as rn
from dbo.users
), cte2 as (
select mid, priority, dense_rank() over(order by mid asc) as rn
from dbo.master
)
select a.userid, count(*) as [count(mid)], b.priority
from cte1 a join cte2 b on a.rn = b.rn
group by a.userid, b.priority
This is my original data (anonymised):
id usage verified date
1 4000 Y 2015-03-20
2 5000 N 2015-06-20
3 6000 N 2015-07-20
4 7000 Y 2016-09-20
Original query:
SELECT
me.usage,
mes.verified,
mes.date
FROM
Table1 me,
Table2 mes,
Table3 m,
Table4 mp
WHERE
me.theFk=mes.id
AND mes.theFk=m.id
AND m.theFk=mp.id
How would I go about selecting the most recent verified and non-verified?
So I would be left with:
id usage verified date
1 6000 N 2015-07-20
2 7000 Y 2016-09-20
I am using Microsoft SQL Server 2012.
First, do not use implicit joins. This was discontinued more than 10 years ago.
Second, embrace the power of the CTE, the in clause and row_number:
with CTE as
(
select
me.usage,
mes.verified,
mes.date,
row_number() over (partition by Verified order by Date desc) as CTEOrd
from Table1 me
inner join Table2 mes
on me.theFK = mes.id
where mes.theFK in
(
select m.id
from Table3 m
inner join Table4 mp
on mp.id = m.theFK
)
)
select CTE.*
from CTE
where CTEOrd = 1
You can select the TOP 1 ordered by date for verified=N, union'd with the TOP 1 ordered by date for verified=Y.
Or in pseudo SQL:
SELECT TOP 1 ...fields ...
FROM ...tables/joins...
WHERE Verified = 'N'
ORDER BY Date DESC
UNION
SELECT TOP 1 ...fields ...
FROM ...tables/joins...
WHERE Verified = 'Y'
ORDER BY Date DESC
drop table #stack2
CREATE TABLE #stack2
([id] int, [usage] int, [verified] varchar(1), [date] datetime)
;
INSERT INTO #stack2
([id], [usage], [verified], [date])
VALUES
(1, 4000, 'Y', '2015-03-20 00:00:00'),
(2, 5000, 'N', '2015-06-20 00:00:00'),
(3, 6000, 'N', '2015-07-20 00:00:00'),
(4, 7000, 'Y', '2016-09-20 00:00:00')
;
;with cte as (select verified,max(date) d from #stack2 group by verified)
select row_number() over( order by s2.[verified]),s2.[usage], s2.[verified], s2.[date] from #stack2 s2 join cte c on c.verified=s2.verified and c.d=s2.date
As per the data shown i had written the query.
for your scenario this will be use full
WITH cte1
AS (SELECT me.usage,
mes.verified,
mes.date
FROM Table1 me,
Table2 mes,
Table3 m,
Table4 mp
WHERE me.theFk = mes.id
AND mes.theFk = m.id
AND m.theFk = mp.id),
cte
AS (SELECT verified,
Max(date) d
FROM cte1
GROUP BY verified)
SELECT Row_number()
OVER(
ORDER BY s2.[verified]),
s2.[usage],
s2.[verified],
s2.[date]
FROM cte1 s2
JOIN cte c
ON c.verified = s2.verified
AND c.d = s2.date
You can as the below Without join.
-- Mock data
DECLARE #Tbl TABLE (id INT, usage INT, verified CHAR(1), date DATETIME)
INSERT INTO #Tbl
VALUES
(1, 4000 ,'Y', '2015-03-20'),
(2, 5000 ,'N', '2015-06-20'),
(3, 6000 ,'N', '2015-07-20'),
(4, 7000 ,'Y', '2016-09-20')
SELECT
A.id ,
A.usage ,
A.verified ,
A.MaxDate
FROM
(
SELECT
id ,
usage ,
verified ,
date,
MAX(date) OVER (PARTITION BY verified) MaxDate
FROM
#Tbl
) A
WHERE
A.date = A.MaxDate
Result:
id usage verified MaxDate
----------- ----------- -------- ----------
3 6000 N 2015-07-20
4 7000 Y 2016-09-20
CREATE TABLE #Table ( ID INT ,usage INT, verified VARCHAR(10), _date DATE)
INSERT INTO #Table ( ID , usage , verified , _date)
SELECT 1,4000 , 'Y','2015-03-20' UNION ALL
SELECT 2, 5000 , 'N' ,'2015-06-20' UNION ALL
SELECT 3, 6000 , 'N' ,'2015-07-20' UNION ALL
SELECT 4, 7000 , 'Y' ,'2016-09-20'
SELECT ROW_NUMBER() OVER(ORDER BY usage) ID,usage , A.verified , A._date
FROM #Table
JOIN
(
SELECT verified , MAX(_date) _date
FROM #Table
GROUP BY verified
) A ON #Table._date = A._date
I want to create a table with a subset of records from a master table.
for example, i have:
id name code
1 peter 73
2 carl 84
3 jack 73
I want to store peter and carl but not jack because has same peter's code.
I need hight performance because i have 20M records.
I try this:
SELECT id, name, DISTINCT(code) INTO new_tab
FROM old_tab
WHERE (conditions)
but don't work.
Assuming you want to pick the row with the maximum id per code, then this should do it:
insert into new_tab (id, name, code)
(SELECT id, name, code
FROM
(
SELECT id, name, code, rank() as rnk OVER (PARTITION BY code ORDER BY id DESC)
FROM old_tab WHERE rnk = 1
)
)
and for the minimum id per code, just change the sort order in the rank from DESC to ASC:
insert into new_tab (id, name, code)
(SELECT id, name, code
FROM
(
SELECT id, name, code, rank() as rnk OVER (PARTITION BY code ORDER BY id ASC)
FROM old_tab WHERE rnk = 1
)
)
Using a derived table, you can find the minID for each code, then join back to that in the outer to get the rest of the columns for that ID from oldTab.
select id,name,code
insert into newTabFROM
from old_tab t inner join
(SELECT min(id) as minId, code
from old_tab group by code) x
on t.id = x.minId
WHERE (conditions)
Try this:
CREATE TABLE #Temp
(
ID INT,
Name VARCHAR(50),
Code INT
)
INSERT #Temp VALUES (1, 'Peter', 73)
INSERT #Temp VALUES (2, 'Carl', 84)
INSERT #Temp VALUES (3, 'Jack', 73)
SELECT t2.ID, t2.Name, t2.Code
FROM #Temp t2
JOIN (
SELECT t.Code, MIN(t.ID) ID
FROM #temp t
JOIN (
SELECT DISTINCT Code
FROM #Temp
) d
ON t.Code = d.Code
GROUP BY t.Code
) b
ON t2.ID = b.ID
This question already has answers here:
Select top 10 records for each category
(14 answers)
Closed 5 years ago.
Assume we have a table which has two columns, one column contains the names of some people and the other column contains some values related to each person. One person can have more than one value. Each value has a numeric type. The question is we want to select the top 3 values for each person from the table. If one person has less than 3 values, we select all the values for that person.
The issue can be solved if there are no duplicates in the table by the query provided in this article Select top 3 values from each group in a table with SQL . But if there are duplicates, what is the solution?
For example, if for one name John, he has 5 values related to him. They are 20,7,7,7,4. I need to return the name/value pairs as below order by value descending for each name:
-----------+-------+
| name | value |
-----------+-------+
| John | 20 |
| John | 7 |
| John | 7 |
-----------+-------+
Only 3 rows should be returned for John even though there are three 7s for John.
In many modern DBMS (e.g. Postgres, Oracle, SQL-Server, DB2 and many others), the following will work just fine. It uses CTEs and ranking function ROW_NUMBER() which is part of the latest SQL standard:
WITH cte AS
( SELECT name, value,
ROW_NUMBER() OVER (PARTITION BY name
ORDER BY value DESC
)
AS rn
FROM t
)
SELECT name, value, rn
FROM cte
WHERE rn <= 3
ORDER BY name, rn ;
Without CTE, only ROW_NUMBER():
SELECT name, value, rn
FROM
( SELECT name, value,
ROW_NUMBER() OVER (PARTITION BY name
ORDER BY value DESC
)
AS rn
FROM t
) tmp
WHERE rn <= 3
ORDER BY name, rn ;
Tested in:
Postgres
Oracle
SQL-Server
In MySQL and other DBMS that do not have ranking functions, one has to use either derived tables, correlated subqueries or self-joins with GROUP BY.
The (tid) is assumed to be the primary key of the table:
SELECT t.tid, t.name, t.value, -- self join and GROUP BY
COUNT(*) AS rn
FROM t
JOIN t AS t2
ON t2.name = t.name
AND ( t2.value > t.value
OR t2.value = t.value
AND t2.tid <= t.tid
)
GROUP BY t.tid, t.name, t.value
HAVING COUNT(*) <= 3
ORDER BY name, rn ;
SELECT t.tid, t.name, t.value, rn
FROM
( SELECT t.tid, t.name, t.value,
( SELECT COUNT(*) -- inline, correlated subquery
FROM t AS t2
WHERE t2.name = t.name
AND ( t2.value > t.value
OR t2.value = t.value
AND t2.tid <= t.tid
)
) AS rn
FROM t
) AS t
WHERE rn <= 3
ORDER BY name, rn ;
Tested in MySQL
I was going to downvote the question. However, I realized that it might really be asking for a cross-database solution.
Assuming you are looking for a database independent way to do this, the only way I can think of uses correlated subqueries (or non-equijoins). Here is an example:
select distinct t.personid, val, rank
from (select t.*,
(select COUNT(distinct val) from t t2 where t2.personid = t.personid and t2.val >= t.val
) as rank
from t
) t
where rank in (1, 2, 3)
However, each database that you mention (and I note, Hadoop is not a database) has a better way of doing this. Unfortunately, none of them are standard SQL.
Here is an example of it working in SQL Server:
with t as (
select 1 as personid, 5 as val union all
select 1 as personid, 6 as val union all
select 1 as personid, 6 as val union all
select 1 as personid, 7 as val union all
select 1 as personid, 8 as val
)
select distinct t.personid, val, rank
from (select t.*,
(select COUNT(distinct val) from t t2 where t2.personid = t.personid and t2.val >= t.val
) as rank
from t
) t
where rank in (1, 2, 3);
Using GROUP_CONCAT and FIND_IN_SET you can do that.Check SQLFIDDLE.
SELECT *
FROM tbl t
WHERE FIND_IN_SET(t.value,(SELECT
SUBSTRING_INDEX(GROUP_CONCAT(t1.value ORDER BY VALUE DESC),',',3)
FROM tbl t1
WHERE t1.name = t.name
GROUP BY t1.name)) > 0
ORDER BY t.name,t.value desc
If your result set is not so heavy, you can write a stored procedure (or an anonymous PL/SQL-block) for that problem which iterates the result set and finds the bigges three by a simple comparing algorithm.
Try this -
CREATE TABLE #list ([name] [varchar](100) NOT NULL, [value] [int] NOT NULL)
INSERT INTO #list VALUES ('John', 20), ('John', 7), ('John', 7), ('John', 7), ('John', 4);
WITH cte
AS (
SELECT NAME
,value
,ROW_NUMBER() OVER (
PARTITION BY NAME ORDER BY (value) DESC
) RN
FROM #list
)
SELECT NAME
,value
FROM cte
WHERE RN < 4
ORDER BY value DESC
This works for MS SQL. Should be workable in any other SQL dialect that has the ability to assign row numbers in a group by or over clause (or equivelant)
if object_id('tempdb..#Data') is not null drop table #Data;
GO
create table #data (name varchar(25), value integer);
GO
set nocount on;
insert into #data values ('John', 20);
insert into #data values ('John', 7);
insert into #data values ('John', 7);
insert into #data values ('John', 7);
insert into #data values ('John', 5);
insert into #data values ('Jack', 5);
insert into #data values ('Jane', 30);
insert into #data values ('Jane', 21);
insert into #data values ('John', 5);
insert into #data values ('John', -1);
insert into #data values ('John', -1);
insert into #data values ('Jane', 18);
set nocount off;
GO
with D as (
SELECT
name
,Value
,row_number() over (partition by name order by value desc) rn
From
#Data
)
SELECT Name, Value
FROM D
WHERE RN <= 3
order by Name, Value Desc
Name Value
Jack 5
Jane 30
Jane 21
Jane 18
John 20
John 7
John 7
I have this data:
ID NAME DATE
3 JOHN 2011-08-08
2 YOKO 2010-07-07
1 JOHN 2009-06-06
Code (for SQL Server 2005):
DECLARE #TESTABLE TABLE (id int, name char(4), date smalldatetime)
INSERT INTO #TESTABLE VALUES (3, 'JOHN', '2011-08-08')
INSERT INTO #TESTABLE VALUES (2, 'YOKO', '2010-07-07')
INSERT INTO #TESTABLE VALUES (1, 'JOHN', '2009-06-06')
I want to get, for each NAME, the ID that has the most recent DATE. Like this:
3 JOHN 2011-08-08
2 YOKO 2010-07-07
What is the most elegant way of accomplishing this?
;WITH x AS
(
SELECT ID, NAME, [DATE],
rn = ROW_NUMBER() OVER
(PARTITION BY NAME ORDER BY [DATE] DESC)
FROM #TESTABLE
)
SELECT ID, NAME, [DATE] FROM x WHERE rn = 1
ORDER BY [DATE] DESC;
Try to avoid reserved words (and vague column names) like [DATE]...
SELECT <fields>
FROM SourceTable st
INNER JOIN (SELECT name, MAX(Datefield) as Datefield
FROM SourceTable
GROUP BY name) x
ON x.Name = st.name
AND x.datefield = st.datefield
below is a possible solution:
Select c.CustomerID, c.CustomerName, c.CustomerOrder, c.CustomerOrderDate, c.CustomerQty
from tblCustomer c
inner join (select c2.CustomerName, MAX(c2.CustomerOrderDate) as MaxDate from tblCustomer c2 group by c2.CustomerName) c2
on c.CustomerName = c2.CustomerName
where c.CustomerOrderDate = c2.MaxDate