I have the following SQL to generate a row num for each record
MY_VIEW AS
( SELECT
my_id,
(case when col1 = 'A' then
1
when col1 = 'C' then
2
else
3
end) as rownum
from table_1
So I have data look like this:
my_id rownum
0001-A 1
0001-A 2
0001-B 2
Later, I want to use the smallest rownum for each unique "my_id" to do a inner join what another table_2. How should I proceed? This is what I have so far.
select * from table_2
inner join tabl_1
on table_2.my_id = table1.my_id
and row_num = (...the smallest from M_VIVE...)
In Postgres, I would recommend distinct on:
selecd distinct on (my_id) my_id
(case when col1 = 'A' then 1
when col1 = 'C' then 2
else 3
end) as rownum
from table_1
order by my_id, rownum;
However, you can just as easily do this using group by:
select my_id,
min(case when col1 = 'A' then 1
when col1 = 'C' then 2
else 3
end) as rownum
from table_1
group by my_id;
The distinct on approach allows you to include other columns. It might be a bit faster. On the downside, it is Postgres-specific.
You can use MIN() function for rownum against every my_id in table_1 and use that in the join.
You would need to make sure table_2 also has my_id field to make the join work.
select *
from
table_2
inner join
(select my_id, MIN(rownum) as minimum_rownum from tabl_1 group by my_id) t1
on table_2.my_id = t1.my_id;
Related
i am creating two tables with same type of attributes i have to join these tables using union in such a way that the resultant union set should have 1st table in ascending order and 2nd table in descending order
I think you want something like this:
select t.*
from ((select t1.*, 1 as which
from t1
) union all
(select t2.*, 2 as which
from t2
)
) t
order by which,
(case when which = 1 then col end) asc,
(case when which = 2 then col end) desc;
Need to return column 1 only if identical values are found in 2nd column of a repeating log. If any other value is seen exclude from result.
A 2
A 2
A 2
A 2
A 2
Exlude
B 2
B 1
B 2
B 3
B 2
select b. column1
from
( select *
from table
where column2 != 1
) b
where b.column2 = 2
Results:
A
You could use aggregation and HAVING:
SELECT col1
FROM tab
GROUP BY col1
HAVING COUNT(DISTINCT col2) = 1;
or if you need original rows:
SELECT s.*
FROM (SELECT t.*, COUNT(DISTINCT col2) OVER(PARTITION BY col1) AS cnt
FROM tab t) s
WHERE s.cnt = 1;
If you need the original rows, I would recommend not exists:
select t.*
from t
where not exists (select 1 from t t2 where t2.col1 = t.col1 and t2.col2 <> t.col2);
If you just want the col1 values (which makes sense to me), then I would phrase the aggregation as:
select col1
from t
group by col1
having min(col2) = max(col2);
If you want to include "all-null" as a valid option, then:
having min(col2) = max(col2) or min(col2) is null
Try this query
select column1 from (select column1,column2 from Test group by column1,column2) a group by column1 having count(column1)=1;
Let's say we have this table with columns RowID and Call:
RowID Call DesiredOut
1 A 0
2 A 0
3 B
4 A 1
5 A 0
6 A 0
7 B
8 B
9 A 2
10 A 0
I want to SQL query the last column DesiredOut as follows:
Each time Call is 'A' go back until 'A' is found again and count the number of records which are in between two 'A' entries.
Example: RowID 4 has 'A' and the nearest predecessor is in RowID 2. Between RowID 2 and RowID 4 we have one Call 'B', so we count 1.
Is there an elegant and performant way to do this with ANSI SQL?
I would approach this by first finding the rowid of the previous "A" value. Then count the number of values in-between.
The following query implements this logic using correlated subqueries:
select t.*,
(case when t.call = 'A'
then (select count(*)
from table t3
where t3.id < t.id and t3.id > prevA
)
end) as InBetweenCount
from (select t.*,
(select max(rowid)
from table t2
where t2.call = 'A' and t2.rowid < t.rowid
) as prevA
from table t
) t;
If you know that rowid is sequential with no gaps, you can just use subtraction instead of a subquery for the calculation in the outer query.
You could use a query to find the previous Call = A row. Then, you could count the number of rows between that row and the current row:
select RowID
, `Call`
, (
select count(*)
from YourTable t2
where RowID < t1.RowID
and RowID > coalesce(
(
select RowID
from YourTable t3
where `Call` = 'A'
and RowID < t1.RowID
order by
RowID DESC
limit 1
),0)
)
from YourTable t1
Example at SQL Fiddle.
Here is another solution using window functions:
with flagged as (
select *,
case
when call = 'A' and lead(call) over (order by rowid) <> 'A' then 'end'
when call = 'A' and lag(call) over (order by rowid) <> 'A' then 'start'
end as change_flag
from calls
)
select t1.rowid,
t1.call,
case
when change_flag = 'start' then rowid - (select max(t2.rowid) from flagged t2 where t2.change_flag = 'end' and t2.rowid < t1.rowid) - 1
when call = 'A' then 0
end as desiredout
from flagged t1
order by rowid;
The CTE first marks the start and end of each "A"-Block and the final select then uses these markers to get the difference between the start of one block and the end of the previous one.
If the rowid is not gapless, you can easily add a gapless rownumber inside the CTE to calculate the difference.
I'm not sure about the performance though. I wouldn't be surprised if Gordon's answer is faster.
SQLFiddle example: http://sqlfiddle.com/#!15/e1840/1
Believe it or not, this will be pretty fast if the two columns are indexed.
select r1.RowID, r1.CallID, isnull( R1.RowID - R2.RowID - 1, 0 ) as DesiredOut
from RollCall R1
left join RollCall R2
on R2.RowID =(
select max( RowID )
from RollCall
where RowID < R1.RowID
and CallID = 'A')
and R1.CallID = 'A';
Here is the Fiddle.
You could do something like that:
SELECT a.rowid - b.rowid
FROM table as a,
(SELECT rowid FROM table where rowid < a.rowid order by rowid) as b
WHERE <something>
ORDER BY a.rowid
As I cannot say which DBMS you are using this is more kind of pseudo code which could work based on your system.
i have 3 separate select statements that i need to union. but all of them need to be ordered by a different column.
i tried doing this
select * from(
select * from (select columns from table1 order by column1 ) A
UNION
select * from (select columns from table2 order by column2 ) B
UNION
select * from (select columns from table3 order by column3 ) C
) Table
but this doesn't work
does anyone have any experience with this?
You can do something like this:
select *
from((select columns, 'table1' as which from table1 )
UNION ALL
(select columns, 'table2' from table2 )
UNION ALL
(select columns, 'table3' from table3 )
) t
order by which,
(case when which = 'table1' then column1
when which = 'table2' then column2
when which = 'table3' then column3
end);
This assumes that the columns used for ordering are all of the same type.
Note that this query uses union all instead of union. I see no reason why you would want to eliminate duplicates if you want the results from the three subqueries ordered independently.
EDIT:
You can also express the order by separately for each table:
order by which,
(case when which = 'table1' then column1 end) ASC,
(case when which = 'table2' then column2 end) DESC
(case when which = 'table3' then column3 end)
You should separate these columns in the one common column and then order
SELECT * FROM
(
SELECT A.*,columnA as ORDER_COL FROM A
UNION ALL
SELECT B.*,columnB as ORDER_COL FROM B
UNION ALL
SELECT C.*,columnC as ORDER_COL FROM C
) as T1
ORDER BY ORDER_COL
You have to order it AFTER the UNION's.
You can "trick it" like this:
select Artificial, a,b,c from(
select 1 as Artificial, a,b,c from (select columns from table1 ) A
UNION
select 2 as Artificial,a,b,c from (select columns from table2 ) B
UNION
select 3 as Artificial,a,b,c from (select columns from table3 ) C
) derivedTable
order by Artificial, c,b,a
I want to get All records that has duplicate values for SOME of the fields (i.e. Key columns).
My code:
CREATE TABLE #TEMP (ID int, Descp varchar(5), Extra varchar(6))
INSERT INTO #Temp
SELECT 1,'One','Extra1'
UNION ALL
SELECT 2,'Two','Extra2'
UNION ALL
SELECT 3,'Three','Extra3'
UNION ALL
SELECT 1,'One','Extra4'
SELECT ID, Descp, Extra FROM #TEMP
;WITH Temp_CTE AS
(SELECT *
, ROW_NUMBER() OVER (PARTITION BY ID, Descp ORDER BY (SELECT 0))
AS DuplicateRowNumber
FROM #TEMP
)
SELECT * FROM Temp_cte
DROP TABLE #TEMP
The last column tells me how many times each row has appeared based on ID and Descp values.
I want that row but I ALSO need another column* that indicates both rows for ID = 1 and Descp = 'One' has showed up more than once.
So an extra column* (i.e. MultipleOccurances (bool)) which has 1 for two rows with ID = 1 and Descp = 'One' and 0 for other rows as they are only showing up once.
How can I achieve that? (I want to avoid using Count(1)>1 or something if possible.
Edit:
Desired output:
ID Descp Extra DuplicateRowNumber IsMultiple
1 One Extra1 1 1
1 One Extra4 2 1
2 Two Extra2 1 0
3 Three Extra3 1 0
SQL Fiddle
You say "I want to avoid using Count" but it is probably the best way. It uses the partitioning you already have on the row_number
SELECT *,
ROW_NUMBER() OVER (PARTITION BY ID, Descp
ORDER BY (SELECT 0)) AS DuplicateRowNumber,
CASE
WHEN COUNT(*) OVER (PARTITION BY ID, Descp) > 1 THEN 1
ELSE 0
END AS IsMultiple
FROM #Temp
And the execution plan just shows a single sort
Well, I have this solution, but using a Count...
SELECT T1.*,
ROW_NUMBER() OVER (PARTITION BY T1.ID, T1.Descp ORDER BY (SELECT 0)) AS DuplicateRowNumber,
CASE WHEN T2.C = 1 THEN 0 ELSE 1 END MultipleOcurrences FROM #temp T1
INNER JOIN
(SELECT ID, Descp, COUNT(1) C FROM #TEMP GROUP BY ID, Descp) T2
ON T1.ID = T2.ID AND T1.Descp = T2.Descp