I am selecting an int value from two different tables as shown below
select col1 from tablea
union
select col1 from tableb
The requirement is if result is found in first query, use that; otherwise, look in second table.
select top 1 * from (select col1 from tablea union select col1 from tableb) as a
Problem is the highest numerical value of the result set is returned as top 1 - not the first result found.
I don't care about the numerical value as in it's order - I just want to apply precedence to if find a value from select 1 don't bother to run second query.
Without the top 1 * I get returned 3 and 6. When I do top 1 * I get 6 and same result if I do the other select first.
Help!
When you UNION these two selects the order IS NOT DEFINED you can think of it as a RANDOM order. So you should define order to get right results. For example:
select top 1 C1 from
(select col1 as C1, 1 as c2 from tablea
union
select col1 as C1, 2 as c2 from tableb) as a
ORDER BY C2
Related
I am trying to find a way if data is not found based on col1 of a table then search with other column value
SELECT * FROM TABLE
WHERE COL1='123'
IF NULL
THEN
SELECT * FROM TABLE
WHERE COL2='ABC';
Thanks
This a typical SQL select statement involving an OR expression.
SELECT * from TABLE WHERE Col1 = '123' or Col2 = 'ABC';
You want all rows that satisfy the first condition - but if no row matches, then you want all rows that satisfy the second condition.
I would adress this with a row limiting clause (available starting version 12c):
select *
from mytable
where 'ABC' in (col1, col2)
order by rank() over(order by case col1 = 'ABC' then 1 else 2 end)
fetch first 1 row with ties
This is more efficient than union all because it does not require two scans on the table.
You can use exists with union all :
select t.*
from table t
where col1 = 123 union all
select t.*
from table t
where col2 = 'abc' and
not exists (select 1 from table t1 where t1.col1 = 123);
If you are expecting only one row, you can use:
SELECT t.*
FROM TABLE t
WHERE COL1 = '123' OR COL2 = 'ABC'
ORDER BY (CASE WHEN COL1 = '123' THEN 1 ELSE 2 END)
FETCH FIRST 1 ROW ONLY;
With multiple possible rows in the result set, I would go for:
SELECT t.*
FROM TABLE t
WHERE COL1 = '123' OR
(COL2 = 'ABC' AND
NOT EXISTS (SELECT 1 FROM TABLE t2 WHERE t2.COL1 = '123');
I have 2 tables and I want not equal rows to be fetched. How to write a query?
For example, table a contain 10 rows, table b contain 10 rows.
Equal rows in a and b is 5.
I want to take a not equal rows (not in b table)
How to fetch a table value which is not equal to b table ?
Result should be 5 record
To take rows in A but not in B:
select * from A minus select * from B
To take rows in A and B but not in both:
(select * from A union select * from B) minus (select * from A intersect select * from B)
This problem has been solved long ago. The optimal solution only reads each table once (unlike the "symmetric difference" solution which reads each table twice and does some additional work).
select 'A' as source, col1, col2, ...
from table_A
union all
select 'B' as source, col1, col2, ...
from table_B
group by col1, col2, ...
having count(*) = 1
;
If a row is present in both tables, then the count will be 2.
This assumes there are no duplicate rows in either table; if there may be duplicate rows, the HAVING condition can be modified, for example:
having count(case when source = 'A' then 1 end) = 0
or count(case when source = 'B' then 1 end) = 0
use EXCEPT
the syntax is similar to INTERSECT.
https://www.tutorialspoint.com/sql/sql-intersect-clause.htm
I tried the sql query given below:
SELECT * FROM (SELECT *
FROM TABLE_A ORDER BY COLUMN_1)DUMMY_TABLE
UNION ALL
SELECT * FROM TABLE_B
It results in the following error:
The ORDER BY clause is invalid in views, inline functions, derived
tables, subqueries, and common table expressions, unless TOP or FOR
XML is also specified.
I need to use order by in union all. How do I accomplish this?
SELECT *
FROM
(
SELECT * FROM TABLE_A
UNION ALL
SELECT * FROM TABLE_B
) dum
-- ORDER BY .....
but if you want to have all records from Table_A on the top of the result list, the you can add user define value which you can use for ordering,
SELECT *
FROM
(
SELECT *, 1 sortby FROM TABLE_A
UNION ALL
SELECT *, 2 sortby FROM TABLE_B
) dum
ORDER BY sortby
You don't really need to have parenthesis. You can sort directly:
SELECT *, 1 AS RN FROM TABLE_A
UNION ALL
SELECT *, 2 AS RN FROM TABLE_B
ORDER BY RN, COLUMN_1
Not an OP direct response, but I thought I would jimmy in here responding to the the OP's ERROR messsage, which may point you in another direction entirely!
All these answers are referring to an overall ORDER BY once the record set has been retrieved and you sort the lot.
What if you want to ORDER BY each portion of the UNION independantly, and still have them "joined" in the same SELECT?
SELECT pass1.* FROM
(SELECT TOP 1000 tblA.ID, tblA.CustomerName
FROM TABLE_A AS tblA ORDER BY 2) AS pass1
UNION ALL
SELECT pass2.* FROM
(SELECT TOP 1000 tblB.ID, tblB.CustomerName
FROM TABLE_B AS tblB ORDER BY 2) AS pass2
Note the TOP 1000 is an arbitary number. Use a big enough number to capture all of the data you require.
There will be times when you need to do something like this :
Pull top 5 from table 1 based on a sort
and bottom 5 from table 2 based on another sort
and union these together.
solution
select * from (
-- top 5 records
select top 5 col1, col2, col3
from table1
group by col1, col2
order by col3 desc ) z
union all
select * from (
-- bottom 5 records
select top 5 col1, col2, col3
from table2
group by col1, col2
order by col3 ) z
this was the only way i was able to get around the error and worked fine for me.
SELECT * FROM (SELECT *
FROM TABLE_A ORDER BY COLUMN_1)DUMMY_TABLE
UNION ALL
SELECT * FROM TABLE_B
ORDER BY 2;
2 is column number here .. In Oracle SQL you can use the column number by which you want to sort the data
This solved my SELECT statement:
SELECT * FROM
(SELECT id,name FROM TABLE_A
UNION ALL
SELECT id,name FROM TABLE_B ) dum
order by dum.id , dum.name
where id and name columns available in tables and you can use your columns .
Simply use that , no need parenthesis or anything else
SELECT *, id as TABLE_A_ID FROM TABLE_A
UNION ALL
SELECT *, id as TABLE_B_ID FROM TABLE_B
ORDER BY TABLE_A_ID, TABLE_B_ID
ORDER BY after the last UNION should apply to both datasets joined by union.
The solution shown below:
SELECT *,id AS sameColumn1 FROM Locations
UNION ALL
SELECT *,id AS sameColumn2 FROM Cities
ORDER BY sameColumn1,sameColumn2
select CONCAT(Name, '(',substr(occupation, 1, 1), ')') AS f1
from OCCUPATIONS
union
select temp.str AS f1 from
(select count(occupation) AS counts, occupation, concat('There are a total of ' ,count(occupation) ,' ', lower(occupation),'s.') As str from OCCUPATIONS group by occupation order by counts ASC, occupation ASC
) As temp
order by f1
I have a table with 10 columns.
I want to return all rows for which Col006 is distinct, but return all columns...
How can I do this?
if column 6 appears like this:
| Column 6 |
| item1 |
| item1 |
| item2 |
| item1 |
I want to return two rows, one of the records with item1 and the other with item2, along with all other columns.
In SQL Server 2005 and above:
;WITH q AS
(
SELECT *, ROW_NUMBER() OVER (PARTITION BY col6 ORDER BY id) rn
FROM mytable
)
SELECT *
FROM q
WHERE rn = 1
In SQL Server 2000, provided that you have a primary key column:
SELECT mt.*
FROM (
SELECT DISTINCT col6
FROM mytable
) mto
JOIN mytable mt
ON mt.id =
(
SELECT TOP 1 id
FROM mytable mti
WHERE mti.col6 = mto.col6
-- ORDER BY
-- id
-- Uncomment the lines above if the order matters
)
Update:
Check your database version and compatibility level:
SELECT ##VERSION
SELECT COMPATIBILITY_LEVEL
FROM sys.databases
WHERE name = DB_NAME()
The key word "DISTINCT" in SQL has the meaning of "unique value". When applied to a column in a query it will return as many rows from the result set as there are unique, different values for that column. As a consequence it creates a grouped result set, and values of other columns are random unless defined by other functions (such as max, min, average, etc.)
If you meant to say you want to return all rows for which Col006 has a specific value, then use the "where Col006 = value" clause.
If you meant to say you want to return all rows for which Col006 is different from all other values of Col006, then you still need to specify what that value is => see above.
If you want to say that the value of Col006 can only be evaluated once all rows have been retrieved, then use the "having Col006 = value" clause. This has the same effect as the "where" clause, but "where" gets applied when rows are retrieved from the raw tables, whereas "having" is applied once all other calculations have been made (i.e. aggregation functions have been run etc.) and just before the result set is returned to the user.
UPDATE:
After having seen your edit, I have to point out that if you use any of the other suggestions, you will end up with random values in all other 9 columns for the row that contains the value "item1" in Col006, due to the constraint further up in my post.
You can group on Col006 to get the distinct values, but then you have to decide what to do with the multiple records in each group.
You can use aggregates to pick a value from the records. Example:
select Col006, min(Col001), max(Col002)
from TheTable
group by Col006
order by Col006
If you want the values to come from a specific record in each group, you have to identify it somehow. Example of using Col002 to identify the record in each group:
select Col006, Col001, Col002
from TheTable t
inner join (
select Col006, min(Col002)
from TheTable
group by Col006
) x on t.Col006 = x.Col006 and t.Col002 = x.Col002
order by Col006
SELECT *
FROM (SELECT DISTINCT YourDistinctField FROM YourTable) AS A
CROSS APPLY
( SELECT TOP 1 * FROM YourTable B
WHERE B.YourDistinctField = A.YourDistinctField ) AS NewTableName
I tried the answers posted above with no luck... but this does the trick!
select * from yourTable where column6 in (select distinct column6 from yourTable);
SELECT *
FROM harvest
GROUP BY estimated_total;
You can use GROUP BY and MIN() to get more specific result.
Lets say that you have id as the primary_key.
And we want to get all the DISTINCT values for a column lets say estimated_total, And you also need one sample of complete row with each distinct value in SQL. Following query should do the trick.
SELECT *, min(id)
FROM harvest
GROUP BY estimated_total;
create table #temp
(C1 TINYINT,
C2 TINYINT,
C3 TINYINT,
C4 TINYINT,
C5 TINYINT,
C6 TINYINT)
INSERT INTO #temp
SELECT 1,1,1,1,1,6
UNION ALL SELECT 1,1,1,1,1,6
UNION ALL SELECT 3,1,1,1,1,3
UNION ALL SELECT 4,2,1,1,1,6
SELECT * FROM #temp
SELECT *
FROM(
SELECT ROW_NUMBER() OVER (PARTITION BY C6 Order by C1) ID,* FROM #temp
)T
WHERE ID = 1
How do I combine the resultsets to return a single result in SQL? For example -
SELECT * FROM Table1
SELECT * FROM Table2
I want to combine the two resultsets with the columns from the second resultset appended to the first.
Table 1 and Table 2 are not related to each other in any way. If Table 1 has 2 columns and Table 2 has 4 columns, I wanted 6 columns returned total in a single resultset. And if Table 1 has 4 rows and Table 2 has only 2 rows, I want NULLS in Table 2 rows for 3rd and 4th row.
Is it possible?
Edit: I do not know how many columns are present in Table1 and Table2, so cannot use UNION with nulls.
If your RDBMS supports ROW_NUMBER() you could do something like this.
WITH T1 AS
(
SELECT *, ROW_NUMBER() OVER ( ORDER BY T1id) AS RN1 FROM Table1
),
T2 AS
(
SELECT *, ROW_NUMBER() OVER ( ORDER BY T2id) AS RN2 FROM Table2
)
SELECT * FROM T1 FULL OUTER JOIN T2 ON RN1 =RN2
It's possible but it's probably quite a bad idea to do this. Why not just run two queries?
If you really want to do it, join the two result sets on a ROW_NUMBER() field.
Not a general solution, but works if you know your schema:
select a1, a2, null as b1, null as b2 from table1
union
select null as a1, null as a2, b1, b2 from table2