Removing rows in SQL that have a duplicate column value - sql

I have looked high and low on SO for an answer over the last couple of hours (subqueries, CTE's, left-joins with derived tables) to this question but none of the solutions are really meeting my criteria..
I have a table with data like this :
COL1 COL2 COL3
1 A 0
2 A 1
3 A 1
4 B 0
5 B 0
6 B 0
7 B 0
8 B 1
Where column1 1 is the primary key and is an int. Column 2 is nvarchar(max) and column 3 is an int. I have determined that by using this query:
select name, COUNT(name) as 'count'
FROM [dbo].[AppConfig]
group by Name
having COUNT(name) > 3
I can return the total counts of "A, B and C" only if they have an occurrence of column C more than 3 times. I am now trying to remove all the rows that occur after the initial value of column 3. The sample table I provided would look like this now:
COL1 COL2 COL3
1 A 0
2 A 1
4 B 0
8 B 1
Could anyone assist me with this?

If all you want is the first row with a ColB-ColC combination, the following will do it:
select min(id) as id, colB, colC
from tbl
group by colB, colC
order by id
SQL Fiddle

This should work:
;WITH numbered_rows as (
SELECT
Col1,
Col2,
Col3,
ROW_NUMBER() OVER(PARTITION BY Col2, Col3 ORDER BY Col3) as row
FROM AppConfig)
SELECT
Col1,
Col2,
Col3
FROM numbered_rows
WHERE row = 1

SELECT DISTINCT MIN(COL1) AS COL1,COL2,COL3
FROM TABLE
GROUP BY COL2,COL3
ORDER BY COL1

Related

How can I find groups with more than one rows and list the rows in each such group?

I have a table "mytable" in a database.
Given a subset of the columns of the table, I would like to group by the subset of the columns, and find those groups with more than one rows:
For example, if the table is
col1 col2 col3
1 1 1
1 1 2
1 2 1
2 2 1
2 2 3
2 1 1
I am interested in finding groups by col1 and col2 with more than one rows, which are:
col1 col2 col3
1 1 1
1 1 2
and
col1 col2 col3
2 2 1
2 2 3
I was wondering how to write a SQL query for that purpose?
Is the following the best way to do that?
First get the col1 and col2 values of such groups:
SELECT col1 col2 COUNT(*)
FROM mytable
GROUP BY col1, col2
HAVING COUNT(*) > 1
Then based on the output of the previous query, manually write a query for each group:
SELECT *
FROM mytable
WHERE col1 = val1 AND col2 = val2
If there are many such groups, then I will have to manually write many queries, which can be a disadvantage.
I am using SQL Server.
Thanks.
This is a common problem. One solution is to get the "keys" in a derived table and join to that to get the rows.
declare #test as table (col1 int, col2 int, col3 int)
insert into #test values (1,1,1),(1,1,2),(1,2,1),(2,2,1),(2,2,3),(2,1,1)
select t.*
from #test t
inner join (
select col1, col2
from #test
group by col1, col2
having count(*) > 1
) k
on k.col1 = t.col1 and k.col2 = t.col2
col1 col2 col3
----------- ----------- -----------
1 1 1
1 1 2
2 2 1
2 2 3
The window function sum() over() may help here
Example
with cte as (
Select *
,Cnt = sum(1) over (partition by Col1,Col2)
From YourTable
)
Select *
From cte
Where Cnt>=2
Results
Another option (less performant)
Select top 1 with ties *
From YourTable
Order By case when sum(1) over (partition by Col1,Col2) > 1 then 1 else 2 end
Results

How to keep track of values which are present in a group as well as in all previous group in oracle SQL?

Let's say I have a table with col1 and col2
I group by col1 and order by col1
From the first group, I want to have all values of col2 but from the second group, I want to have only those values which were present in the first group and so on with the consecutive groups.
sample table
col1 col2
1 A
1 B
1 C
1 D
2 E
2 A
2 B
2 G
3 B
3 D
And the output should be
col1 col2
1 A
1 B
1 C
1 D
2 A
2 B
3 B
You can use window functions in order to avoid to read the same table twice:
Number the groups to make sure to have 1, 2, 3, ... without gaps.
Get a rolling count of col2, or in other words the cumulated numbers of their appearances.
Only show rows where the group number equals the count.
The query:
select col1, col2
from
(
select
col1, col2,
dense_rank() over (order by col1) as rn,
count(*) over (partition by col2 order by col1) as cnt
from mytable
) numbered_and_counted
where rn = cnt
order by col1, col2;
Demo: https://dbfiddle.uk/?rdbms=oracle_18&fiddle=f0cc6a211a1a4c767c9e3ce9deb8c28f

Oracle query - Selecting unique row number based on order of another column

I'm trying to find the best way to make a query with two columns, one is a number and the order a date:
Doing a select and ordering by the date column.
Table1:
col1 (NUMBER)
col2 (DATE)
1
02/2019
2
02/2019
3
02/2019
4
03/2019
2
04/2019
3
05/2019
I'm doing a query like this:
select col1, col2
from table1
order by col2 asc, col1 asc
fetch next 10;
The result I'm getting is also getting the next day's values, and repeating the value on col1 result like this:
col1 (NUMBER)
col2 (DATE)
1
02/2019
2
02/2019
3
02/2019
4
03/2019
2
04/2019
3
05/2019
But I would like a filter to limit to only a sequential col1 value like this:
col1 (NUMBER)
col2 (DATE)
1
02/2019
2
02/2019
3
02/2019
4
03/2019
ignoring values that would come in a "next batch" and not going through the risk of repeating col1 values, or getting col1 values that have a bigger col2 value than a previous result.
Any ideas on the best way to do this?
If I understand correctly, you can use a cumulative max():
select col1, col2
from (select t1.*,
max(col1) over (order by col2, col1 rows between unbounded preceding and 1 preceding) as running_max
from table1 t1
) t1
where running_max is null or col1 > running_max;
This returns rows whose value is greater than the values on the preceding rows.
EDIT:
If you want to return rows only up to the first time there is a decline, then:
select t1.*
from (select t1.*,
sum(case when prev_col1 > col1 then 1 else 0 end) over (order by col2, col1) as num_decreases
from (select t1.*,
lag(col1) over (order by col2, col1) as prev_col1
from table1 t1
) t1
where num_decreases = 0;

Count records in query in groups based on column value

Let's suppose a have a very simple query in SQL
SELECT Col1,Col2 From Table1
and it gives me result:
Col1 Col2
A 5
A 7
A 2
B 1
B 1
B 4
B 0
C 4
C 1
C 2
I want to count rows in groups made by Col1 and in order made by Col2. If values in Col2 for some rows in group are equal then they should have different numbers, as shown in example
So I want to have
Col1 Col2 Nr
A 5 2
A 7 3
A 2 1
B 0 1
B 1 2
B 1 3
B 4 4
C 4 3
C 1 1
C 2 2
Any ideas how to make it?
If your database supports window functions, use ROW_NUMBER
select col1,col2,row_number() over(partition by col1 order by col2) as nr
from tablename
If your database doesn't support window functions, use
select col1,col2,
(select count(*)+1 from tablename t1 where t1.col1=t.col1 and t1.col2<t.col2) as nr
from tablename t
You can use the row_number window function:
SELECT col1,
col2,
ROW_NUMBER() OVER (PARTITION BY col1 ORDER BY col2 ASC) AS Nr
FROM table1
ORDER BY 1, 2, 3

select query to fetch rows corresponding to all values in a column

Consider this example table "Table1".
Col1 Col2
A 1
B 1
A 4
A 5
A 3
A 2
D 1
B 2
C 3
B 4
I am trying to fetch those values from Col1 which corresponds to all values (in this case, 1,2,3,4,5). Here the result of the query should return 'A' as none of the others have all values 1,2,3,4,5 in Col2.
Note that the values in Col2 are decided by other parameters in the query and they will always return some numeric values. Out of those values the query needs to fetch values from Col1 corresponding to all in Col2. The values in Col2 could be 11,12,1,2,3,4 for instance (meaning not necessarily in sequence).
I have tried the following select query:
select distinct Col1 from Table1 where Col1 in (1,2,3,4,5);
select distinct Col1 from Table1 where Col1 exists (select distinct Col2 from Table1);
and its different variations. But the problem is that I need to apply an 'and' for Col2 not an 'or'.
like Return a value from Col1 where Col2 'contains' all values between 1 and 5.
Appreciate any suggestion.
You could use analytic ROW_NUMBER() function.
SQL FIddle for a setup and working demonstration.
SELECT col1
FROM
(SELECT col1,
col2,
row_number() OVER(PARTITION BY col1 ORDER BY col2) rn
FROM your_table
WHERE col2 IN (1,2,3,4,5)
)
WHERE rn =5;
UPDATE As requested by OP, some explanation about how the query works.
The inner sub-query gives you the following resultset:
SQL> SELECT col1,
2 col2,
3 row_number() OVER(PARTITION BY col1 ORDER BY col2) rn
4 FROM t
5 WHERE col2 IN (1,2,3,4,5);
C COL2 RN
- ---------- ----------
A 1 1
A 2 2
A 3 3
A 4 4
A 5 5
B 1 1
B 2 2
B 4 3
C 3 1
D 1 1
10 rows selected.
PARTITION BY clause will group each sets of col1, and ORDER BY will sort col2 in each group set of col1. Thus the sub-query gives you the row_number for each row in an ordered way. now you know that you only need those rows where row_number is at least 5. So, in the outer query all you need ot do is WHERE rn =5 to filter the rows.
You can use listagg function, like
SELECT Col1
FROM
(select Col1,listagg(Col2,',') within group (order by Col2) Col2List from Table1
group by Col1)
WHERE Col2List = '1,2,3,4,5'
You can also use below
SELECT COL1
FROM TABLE_NAME
GROUP BY COL1
HAVING
COUNT(COL1)=5
AND
SUM(
(CASE WHEN COL2=1 THEN 1 ELSE 0
END)
+
(CASE WHEN COL2=2 THEN 1 ELSE 0
END)
+
(CASE WHEN COL2=3 THEN 1 ELSE 0
END)
+
(CASE WHEN COL2=4 THEN 1 ELSE 0
END)
+
(CASE WHEN COL2=5 THEN 1 ELSE 0
END))=5