Query to delete duplicate records by keeping original in oracle - sql

This is the table.
Id. Name
1 A
1 A
2 B
2 C
1 A
2 B
2 D
The output should be
Id. Name
1 A
2 B
2 C
2 D


please try
Select distinct id, name
from <name of you table>
order by name
Check this link.


Sample data:
create table demo (id, name) as
select 1, 'A' from dual union all
select 1, 'A' from dual union all
select 2, 'B' from dual union all
select 2, 'C' from dual union all
select 1, 'A' from dual union all
select 2, 'B' from dual union all
select 2, 'D' from dual;
select * from demo order by 1,2;
ID NAME
---------- ------------------------------
1 A
1 A
1 A
2 B
2 B
2 C
2 D
7 rows selected
Delete all but the first row in each (id, name) group:
delete demo where rowid in
( select lag(rowid) over (partition by id, name order by null) from demo );
3 rows deleted
select * from demo order by 1,2
ID N
---------- -
1 A
2 B
2 C
2 D
4 rows selected.

Related

select only those users whose contacts length is not 5

I have table like this:
id
name
contact
1
A
65489
1
A
1
A
45564
2
B
3
C
12345
3
C
1234
4
D
32
4
D
324
I only want users who have no contact or the contact length is not five.
If the user has two or more contacts and the length of one of them is five and the rest is not, then such users should not be included in the table.
so,If the customer has at least one contact length of five, I do not want that.
so, i want table like this:
id
name
contact
2
B
4
D
32
4
D
324
Can you halp me?

You could actually do a range check here:
SELECT id, name, contact
FROM yourTable t1
WHERE NOT EXISTS (
SELECT 1
FROM yourTable t2
WHERE t2.id = t1.id AND TO_NUMBER(t2.contact) BETWEEN 10000 AND 99999
);
Note that if contact already be a numeric column, then just remove the calls to TO_NUMBER above and compare directly.

Yet another option:
SQL> with test (id, name, contact) as
2 (select 1, 'a', 65879 from dual union all
3 select 1, 'a', null from dual union all
4 select 1, 'a', 45564 from dual union all
5 select 2, 'b', null from dual union all
6 select 3, 'c', 12345 from dual union all
7 select 3, 'c', 1234 from dual union all
8 select 4, 'd', 32 from dual union all
9 select 4, 'd', 324 from dual
10 )
11 select *
12 from test a
13 where exists (select null
14 from test b
15 where b.id = a.id
16 group by b.id
17 having nvl(max(length(b.contact)), 0) < 5
18 );
ID N CONTACT
---------- - ----------
2 b
4 d 32
4 d 324
SQL>

COUNT analytic function can also be used to get the job done.
select id, name, contact
from (
select id, name, contact
, count( decode( length(contact), 5, 1, null ) ) over( partition by id, name ) cnt
from YourTable
)
where cnt = 0
demo

Oracle - generate a running number by group

I need to generate a running number / group sequence inside a select statement for a group of data.
For example
Group Name Sequence
1 a 1
1 b 2
1 c 3
2 d 1
2 e 2
2 f 3
So for each group the sequence should be a running number starting with 1 depending on the order of column"Name".
I already pleayed around with Row_Number() and Level but I couldn't get a solution.
Any idea how to do it?

Analytic functions help.
SQL> with test (cgroup, name) as
2 (select 1, 'a' from dual union all
3 select 1, 'b' from dual union all
4 select 1, 'c' from dual union all
5 select 2, 'd' from dual union all
6 select 2, 'e' from dual union all
7 select 2, 'f' from dual
8 )
9 select cgroup,
10 name,
11 row_number() over (partition by cgroup order by name) sequence
12 from test
13 order by cgroup, name;
CGROUP N SEQUENCE
---------- - ----------
1 a 1
1 b 2
1 c 3
2 d 1
2 e 2
2 f 3
6 rows selected.
SQL>

Try this
SELECT
"Group",
Name,
DENSE_RANK() OVER (PARTITION BY "Group" ORDER BY Name) AS Sequence
FROM table;

Always show a value highst when sorting

I Oracle, I have a table with following values
1
2
4
10
I always want 2 to show up highest following by all other values in DESCending order, as follows :
2
10
4
1

You can order by a value you build with a case; for example:
with tab(col) as (
select 1 from dual union all
select 2 from dual union all
select 4 from dual union all
select 10 from dual
)
select col
from tab
order by case when col = 2 then 1 else 2 end asc,
col desc
gives:
COL
----------
2
10
4
1

try like below if column is not null
with tab(col) as (
select 1 from dual union all
select 2 from dual union all
select 4 from dual union all
select 10 from dual
)
select col
from tab
ORDER BY NULLIF(col, 2) desc NULLS FIRST
output
COL
2
10
4
1
demo link

How to combine and count two columns in oracle?

My table seems like this;
A B
1 100
1 102
1 105
2 100
2 105
3 100
3 102
I want output like this:
A Count(B)
1 3
1,2 2
1,2,3 3
2 2
3 2
2,3 2
How can i do this?
I try to use listagg but it didn't work.

I suspect that you want to count the number of sets of A that are in the data -- and that your sample results are messed up.
If so:
select grp, count(*)
from (select listagg(a, ',') within group (order by a) as grp
from t
group by b
) b;
This gives you the counts for the full combinations present in the data. The results would be:
1,2,3 1
1,3 1
1,2 1
You can get the original number of rows by doing:
select grp, sum(cnt)
from (select listagg(a, ',') within group (order by a) as grp, count(*) as cnt
from t
group by b
) b;

Oracle Setup:
CREATE TABLE table_name ( A, B ) AS
SELECT 1, 100 FROM DUAL UNION ALL
SELECT 1, 102 FROM DUAL UNION ALL
SELECT 1, 105 FROM DUAL UNION ALL
SELECT 2, 100 FROM DUAL UNION ALL
SELECT 2, 105 FROM DUAL UNION ALL
SELECT 3, 100 FROM DUAL UNION ALL
SELECT 3, 102 FROM DUAL;
Query:
SELECT A,
COUNT(B)
FROM (
SELECT SUBSTR( SYS_CONNECT_BY_PATH( A, ',' ), 2 ) AS A,
B
FROM table_name
CONNECT BY PRIOR B = B
AND PRIOR A + 1 = A
)
GROUP BY A
ORDER BY A;
Output:
A COUNT(B)
----- ----------
1 3
1,2 2
1,2,3 1
2 2
2,3 1
3 2

How to do select count(*) group by and select * at same time?

For example, I have table:
ID | Value
1 hi
1 yo
2 foo
2 bar
2 hehe
3 ha
6 gaga
I want my query to get ID, Value; meanwhile the returned set should be in the order of frequency count of each ID.
I tried the query below but don't know how to get the ID and Value column at the same time:
SELECT COUNT(*) FROM TABLE group by ID order by COUNT(*) desc;
The count number doesn't matter to me, I just need the data to be in such order.
Desire Result:
ID | Value
2 foo
2 bar
2 hehe
1 hi
1 yo
3 ha
6 gaga
As you can see because ID:2 appears most times(3 times), it's first on the list,
then ID:1(2 times) etc.

you can try this -
select id, value, count(*) over (partition by id) freq_count
from
(
select 2 as ID, 'foo' as value
from dual
union all
select 2, 'bar'
from dual
union all
select 2, 'hehe'
from dual
union all
select 1 , 'hi'
from dual
union all
select 1 , 'yo'
from dual
union all
select 3 , 'ha'
from dual
union all
select 6 , 'gaga'
from dual
)
order by 3 desc;

select t.id, t.value
from TABLE t
inner join
(
SELECT id, count(*) as cnt
FROM TABLE
group by ID
)
x on x.id = t.id
order by x.cnt desc

How about something like
SELECT t.ID,
t.Value,
c.Cnt
FROM TABLE t INNER JOIN
(
SELECT ID,
COUNT(*) Cnt
FROM TABLE
GROUP BY ID
) c ON t.ID = c.ID
ORDER BY c.Cnt DESC
SQL Fiddle DEMO

I see the question is already answered, but since the most obvious and most simple solution is missing, I'm posting it anyway. It doesn't use self joins nor subqueries:
SQL> create table t (id,value)
2 as
3 select 1, 'hi' from dual union all
4 select 1, 'yo' from dual union all
5 select 2, 'foo' from dual union all
6 select 2, 'bar' from dual union all
7 select 2, 'hehe' from dual union all
8 select 3, 'ha' from dual union all
9 select 6, 'gaga' from dual
10 /
Table created.
SQL> select id
2 , value
3 from t
4 order by count(*) over (partition by id) desc
5 /
ID VALU
---------- ----
2 bar
2 hehe
2 foo
1 yo
1 hi
6 gaga
3 ha
7 rows selected.