I have a table with two different stamps. Let's call them oristamp and tarstamp. I need to find only the records that for the same oristamp have different tarstamps. It is possible to do that with a simple query? I think should be used a cursor but I'm not familiar with that. Any help?
I hope I understand the question. I am assuming you want all rows where more than 1 distinct value for tarstamp exists for each oristamp.
DECLARE #t table(tarstamp int, oristamp int)
INSERT #t values
(1,1),
(1,1),
(1,2),
(2,2)
;WITH CTE as
(
SELECT *,
max(tarstamp) over (partition by oristamp) mx,
min(tarstamp) over (partition by oristamp) mn
FROM #t
)
SELECT *
FROM CTE
WHERE mx <> mn
Use a sub-query to find oristamp values having at least two different tarstamp values. Join with that sub-query:
select t1.*
from tablename t1
join (select oristamp from tablename
group by oristamp
having count(distinct tarstamp) >= 2) t2 on t1.oristamp = t2.oristamp
You could count the number of different oristamps per tarstamp and select the rows that have tarstamps with more than one oristamp, E.g.:
SELECT *
FROM mytable
WHERE tarstamp IN (SELECT tarstamp
FROM mytable
GROUP BY tarstamp
HAVING COUNT(DISTINCT oristamp) > 1)
In your case, using MIN and MAX aggregate functions will perform better as COUNT(DISTINCT oristamp) > 1 adds complexity to the execution plan:
SELECT tarstamp
FROM mytable
GROUP BY tarstamp
HAVING MIN(oristamp) != MAX(oristamp)
Try using a simple join:
DECLARE #t table(tarstamp int, oristamp int)
INSERT #t values
(1,1),
(1,1),
(1,2),
(2,2)
SELECT t1.*
FROM #t t1
INNER JOIN #t t2
ON t1.oristamp = t2.oristamp
WHERE t1.tarstamp <> t2.tarstamp
Try this query, which will return all those combination which contain different oristamp.
select distinct z.*
from
( select oristamp, count(*) ct
from mytable
Group by oristamp
)z,
( Select oristamp, tarstamp, count(*) ct
from mytable
Group by oristamp, tarstamp
)y
Where z.oristamp =y.oristamp
And z.ct != y.ct
Related
I have 2 tables, the first one has 10 distinct values:
,
each GlobalPnID has many values on the second table, I want to join 2 tables and select one random value of PortionKey of the second table that match the condition and move to the next GlobalPnID
SELECT TOP 10 gpnp.PortionKey, tt.GlobalPnID
from #TempTable tt
LEFT JOIN [dbo].[GlobalPartNumberPortions] gpnp ON gpnp.GlobalPnId = tt.GlobalPnID
-- tt is the first table
-- gpnp is the second
SELECT TT.GlobalPnID,X.PortionKey
FROM #TempTable AS TT
CROSS APPLY
(
SELECT TOP 1 R.PortionKey
FROM [dbo].[GlobalPartNumberPortions] AS R
WHERE R.GlobalPnId=TT.GlobalPnID
ORDER BY R.PortionID
)X
You could use Row_Number with a CTE and set the criteria you want, for example:
DECLARE #TempTable TABLE
(
globalpnid INT
)
DECLARE #GlobalPartNumberPortions TABLE
(
portionid INT,
portionkey NVARCHAR(10),
globalpnid INT
)
INSERT INTO #TempTable
(globalpnid)
VALUES (1),(2),(3),(4)
INSERT INTO #GlobalPartNumberPortions
(portionid,
portionkey,
globalpnid)
VALUES (1,'ABC',1),
(2,'XYZ',1),
(3,'AZZ',2),
(4,'QWE',3),
(5,'TYU',4);
WITH cteportion
AS (SELECT portionkey,
globalpnid,
rn = Row_number()
OVER (
partition BY globalpnid
ORDER BY RAND(CHECKSUM(NEWID()))))
FROM #GlobalPartNumberPortions)
SELECT gpnp.portionkey,
tt.globalpnid
FROM #TempTable tt
LEFT JOIN cteportion gpnp
ON tt.globalpnid = gpnp.globalpnid
AND gpnp.rn = 1
This will partition the second table by the globalpnid ordering on ORDER BY RAND(CHECKSUM(NEWID()))) and you can then use this in the join gpnp.rn = 1. In the example I've included, you'll see that GlobalPnID = 1 will alternate between ABC and XYZ.
Edit: as suggested by #Thorsten Kettner in the comment, you can order by RAND(CHECKSUM(NEWID())))
I am trying to sum two values across a UNION. Like:
SELECT
sum(target_value) FROM table
UNION ALL
SELECT
sum(target_value) FROM table_2
But rather than getting the 2 sum values, I want them to also be summed.
How should I go about doing this?
You can try using this in a subquery and calculate at the outer query. Ensure you use UNION ALL to evade the duplicate check
SELECT SUM(a) target_value
FROM
(SELECT
sum(target_value) a FROM table
UNION ALL
SELECT
sum(target_value) a FROM table_2) ;
You can use the WITH clause to do this:
WITH CTE AS (SELECT SUM(target_value) as FirstSum
FROM table
UNION
SELECT
SUM(target_value) as FirstSum
FROM table_2)
SELECT SUM(FirstSum) AS TotalSum FROM CTE
Please see the example below:
create table #temp (x int)
create table #temp2 (x2 int)
insert into #temp values (2)
insert into #temp values (3)
insert into #temp2 values (5)
insert into #temp2 values (6)
select t.col1,t.col2,t.col1+t.col2 as Total
from (
SELECT (select sum(x) FROM #temp) as col1,
(select sum(x2) FROM #temp2) as col2
) t
I would like to get 2 consecutive rows from an SQL table.
One of the columns storing UNIX datestamp and between 2 rows the difference only this value.
For example:
id_int dt_int
1. row 8211721 509794233
2. row 8211722 509794233
I need only those rows where dt_int the same (edited)
Do you want both lines to be shown?
A solution could be this:
with foo as
(
select
*
from (values (8211721),(8211722),(8211728),(8211740),(8211741)) a(id_int)
)
select
id_int
from
(
select
id_int
,id_int-isnull(lag(id_int,1) over (order by id_int) ,id_int-6) prev
,isnull(lead(id_int,1) over (order by id_int) ,id_int+6)-id_int nxt
from foo
) a
where prev<=5 or nxt<=5
We use lead and lag, to find the differences between rows, and keep the rows where there is less than or equal to 5 for the row before or after.
If you use 2008r2, then lag and lead are not available. You could use rownumber in stead:
with foo as
(
select
*
from (values (8211721),(8211722),(8211728),(8211740),(8211741)) a(id_int)
)
, rownums as
(
select
id_int
,row_number() over (order by id_int) rn
from foo
)
select
id_int
from
(
select
cur.id_int
,cur.id_int-prev.id_int prev
,nxt.id_int-cur.id_int nxt
from rownums cur
left join rownums prev
on cur.rn-1=prev.rn
left join rownums nxt
on cur.rn+1=nxt.rn
) a
where isnull(prev,6)<=5 or isnull(nxt,6)<=5
Assuming:
lead() analytical function available.
ID_INT is what we need to sort by to determine table order...
you may need to partition by some value lead(ID_int) over(partition by SomeKeysuchasOrderNumber order by ID_int asc) so that orders and dates don't get mixed together.
.
WITH CTE AS (
SELECT A.*
, lead(ID_int) over ([missing partition info] ORDER BY id_Int asc) - id_int as ID_INT_DIFF
FROM Table A)
SELECT *
FROM CTE
WHERE ID_INT_DIFF < 5;
You can try it. This version works on SQL Server 2000 and above. Today I don not a more recent SQL Server to write on.
declare #t table (id_int int, dt_int int)
INSERT #T SELECT 8211721 , 509794233
INSERT #T SELECT 8211722 , 509794233
INSERT #T SELECT 8211723 , 509794235
INSERT #T SELECT 8211724 , 509794236
INSERT #T SELECT 8211729 , 509794237
INSERT #T SELECT 8211731 , 509794238
;with cte_t as
(SELECT
ROW_NUMBER() OVER (ORDER BY id_int) id
,id_int
,dt_int
FROM #t),
cte_diff as
( SELECT
id_int
,dt_int
,(SELECT TOP 1 dt_int FROM cte_t b WHERE a.id < b.id) dt_int1
,dt_int - (SELECT TOP 1 dt_int FROM cte_t b WHERE a.id < b.id) Difference
FROM cte_t a
)
SELECT DISTINCT id_int , dt_int FROM #t a
WHERE
EXISTS(SELECT 1 FROM cte_diff b where b.Difference =0 and a.dt_int = b.dt_int)
I am trying to get a list of records showing changes in location and dates to display as one row for each record showing previous location.
Basically a query to take data like:
And display it like:
I tried using lag, but it mixes up some of the records. Would anyone be able suggest a good way to do this?
Thanks!
DECLARE #TABLE TABLE
(ID INT ,NAME VARCHAR(20), LOCATIONDATE DATETIME, REASON VARCHAR(20))
INSERT INTO #TABLE
(ID,NAME, LOCATIONDATE, REASON)
VALUES
( 1,'abc',CAST('2016/01/01' AS SMALLDATETIME),'move'),
( 2,'def',CAST('2016/02/01' AS SMALLDATETIME),'move'),
( 1,'abc',CAST('2016/06/01' AS SMALLDATETIME),'move'),
( 2,'def',CAST('2016/07/01' AS SMALLDATETIME),'move'),
( 1,'abc',CAST('2016/08/01' AS SMALLDATETIME),'move'),
( 3,'ghi',CAST('2016/08/01' AS SMALLDATETIME),'move')
select s.*
,t1.*
from
(
select t.*,
row_number() over (partition by id order by locationdate desc) rn
from #table t
) s
left join
(
select t.*,
row_number() over (partition by id order by locationdate desc) rn
from #table t
) t1
on t1.id = s.id and t1.rn = s.rn + 1
You can try it:
SELECT a.id,a.name,a.location as currentLocation,a.locationdatedate as currrentLocate,b.location as preLocation,b.locationdatedate as prevLocate,a.changereason
FROM test as a
JOIN test as b ON a.name = b.name
where a.locationdatedate > b.locationdatedate
group by a.name
Pleas try this one. it works here
SELECT l.id,
,l.name
,l.location as currentLocation
,l.locationdatedate as currrentLocate
,r.location as preLocation
,r.locationdatedate as prevLocate
,r.changereason
FROM tableName AS l inner join
tableName AS r ON r.id=l.id
WHERE l.locationdatedate !=r.locationdatedate AND l.locationdatedate > r.locationdatedate
What is the method in T-SQL to select the orginal values limited by a HAVING attribute. For example, if I have
A|B
10|1
11|2
10|3
How would I get all the values of B (Not An Average or some other summary stat), Grouped by A, having a Count (Occurrences of A) greater than or equal two 2?
Actually, you have several options to choose from
1. You could make a subquery out of your original having statement and join it back to your table
SELECT *
FROM YourTable yt
INNER JOIN (
SELECT A
FROM YourTable
GROUP BY
A
HAVING COUNT(*) >= 2
) cnt ON cnt.A = yt.A
2. another equivalent solution would be to use a WITH clause
;WITH cnt AS (
SELECT A
FROM YourTable
GROUP BY
A
HAVING COUNT(*) >= 2
)
SELECT *
FROM YourTable yt
INNER JOIN cnt ON cnt.A = yt.A
3. or you could use an IN statement
SELECT *
FROM YourTable yt
WHERE A IN (SELECT A FROM YourTable GROUP BY A HAVING COUNT(*) >= 2)
A self join will work:
select B
from table
join(
select A
from table
group by 1
having count(1)>1
)s
using(A);
You can use window function (no joins, only one table scan):
select * from (
select *, cnt=count(*) over(partiton by A) from table
) as a
where cnt >= 2