The confusing question is best asked through an example. Say we have the following result set:
What I want to do is count how many times one number appears from both columns.
So the returning data set might look like:
ID Counted
0 4
1 2
9 1
13 1
My original thought was to do some sort of addition between the counts on both IDs, but I'm not exactly sure how to GROUP them in SQL in a way that is working.
You can do this with a subquery, GROUP BY, and a UNION ALL, like this:
SELECT ID, COUNT(*)
FROM(
SELECT ID1 AS ID FROM MyTable
UNION ALL
SELECT ID2 AS ID FROM MyTable
) source
GROUP BY ID
ORDER BY ID ASC
Related
I am looking to combine these 2 statement into one to run as a stored procedure if possible.
I have not used temp tables in queries before and may have to with this, not sure asking advice.
I did not write the original queries and manually run the first one which returns a table listing ID's with duplicate data nad how many records. Then each record ID is put into the 2nd query to remove all but the TOP 1 based on additional filtering criteria.
I have looked at using CTE from SQL select into delete DIRECTLY but am stil at a loss on how to pass each result row ID value into the delete query.
The queries, edited for public consumption are
SELECT id, count() FROM [DEV].[dbo].[7dtest] where FileVer = 1 and CALC_DATE > FORMAT(DATEADD(DD,-7,GETDATE()), 'yyyy-MM-dd') group by id having count() > 1 order by count(*) desc
returns a table with id and number of duplicate rows
then take the id of each row and put into this delete statement
delete from [DEV].[dbo].[7dtest] where AutoID not in (
SELECT TOP 1 AutoID FROM [DEV].[dbo].[7dtest] where FileVer = 1 and id = '123' and CALC_DATE > FORMAT(DATEADD(DD,-7,GETDATE()), 'yyyy-MM-dd')
order by COMPLETED_DATE_CHECK_3 desc, COMPLETED_DATE_CHECK_2 desc, COMPLETED_DATE_CHECK_1 desc)
and FileVer = 1 and id = '123' and CALC_DATE > FORMAT(DATEADD(DD,-7,GETDATE()), 'yyyy-MM-dd')
Can this be done with CTE or do I need to create a temp table and some looping to get the ID one row at a time? Is there a better way I should be doing this?
TIA
I have a table format like below:
Id Code
1 A
1 B
2 A
3 A
3 C
4 A
4 B
I am trying to get count of code combinations like below:
Code Count
A,B 2 -- Row 1,2 and Row 6,7
A 1 -- Row 3
A,C 1 -- Row 4
I am unable to get the combination result. All I can do is group by but I am not getting count of IDs based in combinations.
You need to aggregate the rows, somehow, and do that twice. The code looks something like this:
select codes, count(*) as num_ids
from (select id, group_concat(code order by code) as codes
from t
group by id
) id
group by code;
group_concat() might be spelled listagg() or string_agg() depending on the database.
In SQL Server, use string_agg():
select codes, count(*) as num_ids
from (select id, string_agg(code, ',') within group (order by code) as codes
from t
group by id
) id
group by code;
Hi and thanks in advance.
Trying to simplify two SQL statements without much success.
Count unique records in a table (by unique, I mean the whole record is found only once in the table)
Select unique records in a table (by unique, I mean the whole record is found only once in the table)
Since the table has 30 columns, is there some way to simply specify ALL columns in the table rather than having to include all individually in the SQL?
I got this working where you spell out every column name (where 'col n name' refers to the last column) but it is not ideal since there are just so many columns …
SELECT col 1 name, col 2 name, col 3 name, …, col n name FROM table name
GROUP BY col 1 name, col 2 name, col 3 name, …, col n name
Having Count(*)=1
Thanks
deutz
create view tab_view1 as select DISTINCT * from tab;
select COUNT(*) from tab_view1; -- first desired result
select * from tab_view1; -- second desired result
First occurence is unique, second is not.
If you want to exclude any record that has duplicate:
create view tab_view2 as select tab.*, COUNT(*) AS occurs
from tab group by tab.*
having COUNT(*) = 1;
select COUNT(*) from tab_view2; -- first desired result
select * from tab_view2; -- second desired result
Let's say I have a table that looks like,
id
2
2
3
4
5
5
5
How do I get something like,
id count
2 2
3 1
4 1
5 3
where the count column is just the count of each id in the id column?
You want to use the GROUP BY operation
SELECT id, COUNT(id)
FROM table
GROUP BY id
select id, count(id) from table_name group by id
or
select id, count(*) from table_name group by id
This is your query:
SELECT id, COUNT(id)
FROM table
GROUP BY id
What GROUP BY clause does is this:
It will split your table based on ids i.e all your 1's are separated, then the 2's , 3's and so on. You can assume it like new tables are created where in one table all the 1's are stored, 2's in another , 3's in yet another and so on.
Then after that the SELECT query is applied on each of these separate tables and the result is returned for each of these "groups".
Good luck!
Kudos! :)
Does anyone happen to know a way of basically taking the 'Distinct' command but only using it on a single column. For lack of example, something similar to this:
Select (Distinct ID), Name, Term from Table
So it would get rid of row with duplicate ID's but still use the other column information. I would use distinct on the full query but the rows are all different due to certain columns data set. And I would need to output only the top most term between the two duplicates:
ID Name Term
1 Suzy A
1 Suzy B
2 John A
2 John B
3 Pete A
4 Carl A
5 Sally B
Any suggestions would be helpful.
select t.Id, t.Name, t.Term
from (select distinct ID from Table order by id, term) t
You can use row number for this
Select ID, Name, Term from(
Select ID, Name, Term, ROW_NUMBER ( )
OVER ( PARTITION BY ID order by Name) as rn from Table
Where rn = 1)
as tbl
Order by determines the order from which the first row will be picked.