I got table with columns [TagTime], [TagValue], [id]. Values in this table comming every 15 min, so there is multiple values with the same id's.
I want select (tagtime,tagvalue,id), that will return only first value in january 2017 for all the id's.
I tried with select top 1, then with group by, distinct... but it doesnt work. Have you some suggestions?
Try this (I assumed you are using MSSQL)
SELECT TOP 1 *
FROM [tbl1]
WHERE DATEPART(MONTH,TagTime) = '01' AND DATEPART(YEAR,TagTime) = '2017'
ORDER BY TagTime DESC
You can go this way
select id, tagvalue, tagtime
from
(select id, tagvalue, tagtime,
row_number() over (partition by id order by tagtime) [RowNum]
from YourTable
where DATEPART(year, tagtime) = 2017) src
where src.[RowNum] = 1
So, you having row number for each ID, ordered by tagtime, and then selecting just first of every ID
Related
I have a table with 32 ids and each id should have 31 rows (31 dates). I want to select the ids and all its dates only for ids with less than 31 rows (31 dates).
I've tried this and a few other things:
SELECT id, date
FROM daily_activity
WHERE id IN(SELECT id, COUNT(id) AS num_days
FROM daily_activity
GROUP by id
HAVING num_days < 31)"
ORDER BY id, date
Thanks
One solution would be to use count as a window function
with ids as (
select *, Count(*) over(partition by id) tot
from mytable
)
select id, date
from ids
where tot<31
You can try this maybe it'll work for you:
SELECT id, date
FROM daily_activity
where( select COUNT(daily_activity2.id)
FROM daily_activity daily_activity2
where daily_activity.id=daily_activity2.id
group by daily_activity2.id ) < 31
ORDER BY id, date
I have a table having two columns and I want to fetch data of 6 years with rules
The first row would be maximum date row that is available before and equals to input date (I will pass an input date)
From the second row till 6th row I need maximum(date row) that is earlier than previous row data selected data and there should not be 2 rows for same year i need only latest one according to the previous row but not in same year.
declare #tbl table (id int identity, marketdate date )
insert into #tbl (marketdate)
values('2018-05-31'),
('2017-06-01'),
('2017-05-28'),
('2017-04-28'),
('2016-05-26'),
('2015-04-18'),
('2015-04-20'),
('2015-03-18'),
('2014-05-31'),
('2014-04-18'),
('2013-04-15')
output:
id marketdate
1 2018.05.31
3 2017.05.28
5 2016.05.27
7 2015.04.20
9 2014.04.18
10 2013.04.15
Can't you do this with a simple order by/desc?
SELECT TOP 6 id, max(marketdate) FROM tbl
WHERE tbl.marketdate <= #date
GROUP BY YEAR(marketdate), id, marketdate
ORDER BY YEAR(marketdate) DESC
Based purely on your "Output" given your sample data, I believe the following is what you are after (The max date for each distinct year of data):
SELECT TOP 6
max(marketdate),
Year(marketDate) as marketyear
FROM #tbl
WHERE #tbl.marketdate <= getdate()
GROUP BY YEAR(marketdate)
ORDER BY YEAR(marketdate) DESC;
SQLFiddle of this matching your output
you can use row_number if you are using sql server
select top 6
id
, t.marketdate
from ( select rn = row_number() over (partition by year(marketdate)order by marketdate desc)
, id
, marketdate
from #tbl) as t
where t.rn = 1
order by t.marketdate desc
The following recursively searches for the next date, which must be at least one year earlier than the previous date.
Your parameterised start position goes where I chose 2018-06-01.
WITH
recursiveSearch AS
(
SELECT
id,
marketDate
FROM
(
SELECT
yourTable.id,
yourTable.marketDate,
ROW_NUMBER() OVER (ORDER BY yourTable.marketDate DESC) AS relative_position
FROM
yourTable
WHERE
yourTable.marketDate <= '2018-06-01'
)
search
WHERE
relative_position = 1
UNION ALL
SELECT
id,
marketDate
FROM
(
SELECT
yourTable.id,
yourTable.marketDate,
ROW_NUMBER() OVER (ORDER BY yourTable.marketDate DESC) AS relative_position
FROM
yourTable
INNER JOIN
recursiveSearch
ON yourTable.marketDate < DATEADD(YEAR, -1, recursiveSearch.marketDate)
)
search
WHERE
relative_position = 1
)
SELECT
*
FROM
recursiveSearch
WHERE
id IS NOT NULL
ORDER BY
recursiveSearch.marketDate DESC
OPTION
(MAXRECURSION 0)
http://sqlfiddle.com/#!18/56246/13
I'm working with Sql server 2008.i have a table contains following columns,
Id,
Name,
Date
this table contains more than one record for same id.i want to get distinct id having maximum date.how can i write sql query for this?
Use the ROW_NUMBER() function and PARTITION BY clause. Something like this:
SELECT Id, Name, Date FROM (
SELECT *, ROW_NUMBER() OVER (PARTITION BY Id ORDER BY Date desc) AS ROWNUM
FROM [MyTable]
) x WHERE ROWNUM = 1
If you need only ID column and other columns are NOT required, then you don't need to go with ROW_NUMBER or MAX or anything else. You just do a Group By over ID column, because whatever the maximum date is you will get same ID.
SELECT ID FROM table GROUP BY ID
--OR
SELECT DISTINCT ID FROM table
If you need ID and Date columns with maximum date, then simply do a Group By on ID column and select the Max Date.
SELECT ID, Max(Date) AS Date
FROM table
GROUP BY ID
If you need all the columns but 1 line having Max. date then you can go with ROW_NUMBER or MAX as mentioned in other answers.
SELECT *
FROM table AS M
WHERE Exists(
SELECT 1
FROM table
WHERE ID = M.ID
HAVING M.Date = Max(Date)
)
One way, using ROW_NUMBER:
With CTE As
(
SELECT Id, Name, Date, Rn = Row_Number() Over (Partition By Id
Order By Date DESC)
FROM dbo.TableName
)
SELECT Id --, Name, Date
FROM CTE
WHERE Rn = 1
If multiple max-dates are possible and you want all you could use DENSE_RANK instead.
Here's an overview of sql-server's ranking function: http://technet.microsoft.com/en-us/library/ms189798.aspx
By the way, CTE is a common-table-expression which is similar to a named sub-query. I'm using it to be able to filter by the row_number. This approach allows to select all columns if you want.
select Max(Date) as "Max Date"
from table
group by Id
order by Id
Try with Max(Date) and GROUP BY the other two columns (the ones with repeating data)..
SELECT ID, Max(Date) as date, Name
FROM YourTable
GROUP BY ID, Name
You may try with this
DECLARE #T TABLE(ID INT, NAME VARCHAR(50),DATE DATETIME)
INSERT INTO #T VALUES(1,'A','2014-04-20'),(1,'A','2014-04-28')
,(2,'A2','2014-04-22'),(2,'A2','2014-04-24')
,(3,'A3','2014-04-20'),(3,'A3','2014-04-28')
,(4,'A4','2014-04-28'),(4,'A4','2014-04-28')
,(5,'A5','2014-04-28'),(5,'A5','2014-04-28')
SELECT T.ID FROM #T T
WHERE T.DATE=(SELECT MAX(A.DATE)
FROM #T A
WHERE A.ID=T.ID
GROUP BY A.ID )
GROUP BY T.ID
select id, max(date) from NameOfYourTable group by id;
I have a table as below
ID Date
1 Null
1 Null
1 Null
1 02/02/2012
1 02/03/2012
1 02/04/2012
1 02/05/2012
I want to take a min date from the above table, that's result should be Null
I was trying to write
select min(date), Id from Table group by ID
then result is 02/02/2012, but I want Null.
Is there any otherway to pull Null value from the above table except the below method?
select top 1 date, ID from table order by date asc
Assuming that your dbms is SQL-Server.
If you want to group by id but select all fields anyway, you can use a cte with ROW_NUMBER function:
WITH cte AS(
SELECT x.*
, RN=ROW_NUMBER()OVER(Partition By id Order By date)
FROM Table x
)
SELECT * FROM cte
WHERE RN=1
http://sqlfiddle.com/#!3/cc2a4/7
By default the functions MAX and MIN do not count NULL in their evaluation of your data.
Try in this way, should do the trick :
SELECT
CASE WHEN MIN(COALESCE(Date, '19001231')) = '19001231' THEN NULL ELSE MIN(Date) END AS Date,
Id
FROM X
group by ID
I think some of them are given the answer. but the final result working with below query
SELECT CASE WHEN MIN(coalesce(date, '1900-01-01')) = '1900-01-01' THEN NULL ELSE MIN(date) END AS Date, ID
FROM table
GROUP BY ID
You can also do a simple check for 'NULL'. I didn't return an "ID" in the example since the "ID" seems meaningless in the example schema/query given.
IF (EXISTS(SELECT TOP 1 * FROM x WHERE date IS NULL)) SELECT NULL AS MinDate
ELSE SELECT MIN(date) AS MinDate FROM x;
http://sqlfiddle.com/#!3/d5fca/11
I did search around and I found this
SQL selecting rows by most recent date with two unique columns
Which is so close to what I want but I can't seem to make it work.
I get an error Column 'ID' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.
I want the newest row by date for each Distinct Name
Select ID,Name,Price,Date
From table
Group By Name
Order By Date ASC
Here is an example of what I want
Table
ID
Name
Price
Date
0
A
10
2012-05-03
1
B
9
2012-05-02
2
A
8
2012-05-04
3
C
10
2012-05-03
4
B
8
2012-05-01
desired result
ID
Name
Price
Date
2
A
8
2012-05-04
3
C
10
2012-05-03
1
B
9
2012-05-02
I am using Microsoft SQL Server 2008
Select ID,Name, Price,Date
From temp t1
where date = (select max(date) from temp where t1.name =temp.name)
order by date desc
Here is a SQL Fiddle with a demo of the above
Or as Conrad points out you can use an INNER JOIN (another SQL Fiddle with a demo) :
SELECT t1.ID, t1.Name, t1.Price, t1.Date
FROM temp t1
INNER JOIN
(
SELECT Max(date) date, name
FROM temp
GROUP BY name
) AS t2
ON t1.name = t2.name
AND t1.date = t2.date
ORDER BY date DESC
There a couple ways to do this. This one uses ROW_NUMBER. Just partition by Name and then order by what you want to put the values you want in the first position.
WITH cte
AS (SELECT Row_number() OVER (partition BY NAME ORDER BY date DESC) RN,
id,
name,
price,
date
FROM table1)
SELECT id,
name,
price,
date
FROM cte
WHERE rn = 1
DEMO
Note you should probably add ID (partition BY NAME ORDER BY date DESC, ID DESC) in your actual query as a tie-breaker for date
select * from (
Select
ID, Name, Price, Date,
Rank() over (partition by Name order by Date) RankOrder
From table
) T
where RankOrder = 1
I have found another memory efficient way (but probably crude way)that has worked for me in postgress. Order the query by the date desc, then select the first record of each distinct field.
SELECT distinct on (Name) ID, Price, Date from
table
order by Date desc
Use Distinct instead of Group By
Select Distinct ID,Name,Price,Date
From table
Order By Date ASC
http://technet.microsoft.com/en-us/library/ms187831.aspx