Is there any way to group by all the columns of a table without specifying the column names? Like:
select * from table group by *
nope. are you trying to do some aggregation? if so, you could do something like this to get what you need
;with a as
(
select sum(IntField) as Total
from Table
group by CharField
)
select *, a.Total
from Table t
inner join a
on t.Field=a.Field
Here is my suggestion:
DECLARE @FIELDS VARCHAR(MAX), @NUM INT
--DROP TABLE #FIELD_LIST
SET @NUM = 1
SET @FIELDS = ''
SELECT
'SEQ' = IDENTITY(int,1,1) ,
COLUMN_NAME
INTO #FIELD_LIST
FROM Req.INFORMATION_SCHEMA.COLUMNS
WHERE TABLE_NAME = N'new340B'
WHILE @NUM <= (SELECT COUNT(*) FROM #FIELD_LIST)
BEGIN
SET @FIELDS = @FIELDS + ',' + (SELECT COLUMN_NAME FROM #FIELD_LIST WHERE SEQ = @NUM)
SET @NUM = @NUM + 1
END
SET @FIELDS = RIGHT(@FIELDS,LEN(@FIELDS)-1)
EXEC('SELECT ' + @FIELDS + ', COUNT(*) AS QTY FROM [Req].[dbo].[new340B] GROUP BY ' + @FIELDS + ' HAVING COUNT(*) > 1 ')
He is trying find and display the duplicate rows in a table.
SELECT *, COUNT(*) AS NoOfOccurrences
FROM TableName GROUP BY *
HAVING COUNT(*) > 1
Do we have a simple way to accomplish this?
You can use Group by All but be careful as Group by All will be removed from future versions of SQL server.
I wanted to do counts and sums over full resultset. I achieved grouping by all with GROUP BY 1=1
.
SELECT DISTINCT * FROM MyFooTable;
If you group by all columns, you are just requesting that duplicate data be removed.
For example a table with the following data:
id | value
----+----------------
1 | foo
2 | bar
1 | foo
3 | something else
If you perform the following query which is essentially the same as SELECT * FROM MyFooTable GROUP BY *
if you are assuming * means all columns:
SELECT * FROM MyFooTable GROUP BY id, value;
id | value
----+----------------
1 | foo
3 | something else
2 | bar
It removes all duplicate values, which essentially makes it semantically identical to using the DISTINCT keyword with the exception of the ordering of results. For example:
SELECT DISTINCT * FROM MyFooTable;
id | value
----+----------------
1 | foo
2 | bar
3 | something else