Always show a value highst when sorting - sql

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

Related

Compare before column in before row with next column in next row

My code is :
with x as
(
select 1 col from dual union all
select 2 col from dual union all
select 8 col from dual union all
select 4 col from dual union all
select 3 col from dual union all
select 2 col from dual
)
select col col1, col col2, col col3, rownum
from x
where col2.ROWNUM > col1.ROWNUM -1
and col2.ROWNUM > col3ROWNUM +1 ;
I want to compare col2.ROWNUM > col1.ROWNUM -1 and col2.ROWNUM > col3ROWNUM + 1 but that doesn't work and I got an error
ORA-01747: invalid user.table.column, table.column, or column specification
01747. 00000 - "invalid user.table.column, table.column, or column specification"
*Cause:
*Action:
Error at Line: 10 Column: 13
Please help me
It looks you got something wrong.
Result of that CTE is a single-column table whose only column is named col. There are no other columns.
SQL> with x as (
2 select 1 col from dual union all --> in UNION, all columns are
3 select 2 col from dual union all named by column name(s) from the
4 select 8 col from dual union all first SELECT statement
5 select 4 col from dual union all
6 select 3 col from dual union all
7 select 2 col from dual)
8 select x.*, rownum
9 from x;
COL ROWNUM
---------- ----------
1 1
2 2
8 3
4 4
3 5
2 6
6 rows selected.
SQL>
Therefore, where clause you wrote doesn't make any sense. Perhaps you should explain what you really have, rules that should be applied to source data and result you'd like to get.
Based on text you put into the title:
compare before column in before row with next column in next row
maybe you'd be interested in lag and lead analytic functions which then let you compare values in adjacent rows (pay attention to NULL values; I didn't). For example:
SQL> with x as (
2 select 1 col from dual union all
3 select 2 col from dual union all
4 select 8 col from dual union all
5 select 4 col from dual union all
6 select 3 col from dual union all
7 select 2 col from dual
8 ),
9 temp as
10 (select col,
11 rownum as rn
12 from x
13 ),
14 temp2 as
15 (select
16 rn,
17 col as this_row,
18 lag(col) over (order by rn) as previous_row,
19 lead(col) over (order by rn) as next_row
20 from temp
21 )
22 select this_row,
23 previous_row,
24 next_row,
25 --
26 case when this_row < previous_row then 'This < previous'
27 when this_row < next_row then 'This < next'
28 else 'something else'
29 end as result
30 from temp2
31 order by rn;
Result:
THIS_ROW PREVIOUS_ROW NEXT_ROW RESULT
---------- ------------ ---------- ---------------
1 2 This < next
2 1 8 This < next
8 2 4 something else
4 8 3 This < previous
3 4 2 This < previous
2 3 This < previous
6 rows selected.
SQL>
Use lead or lag functions. But, please, do not use rownum for such purposes.
Rownum indicates simply the order in which a row was found in the database and cannot be used for other purposes except limiting the number of rows fetched, like where rownum<=1 to be certain you won't get a too_many_rows exception, for instance. Still, if in a query you do fetch the pseud-column rownum, give it an alias so that you may use that value later on.
Moreover, what is supposed to mean col2.ROWNUM or col1.ROWNUM? That is not clear. col1 and col2 are two columns, which do not have the attribute rownum.
Something that may help in the future for analytic queries:
https://oracle-base.com/articles/misc/lag-lead-analytic-functions
And, if you wish to get a working SQL, please explain clearly what you wish to achieve, for I haven't really understood what that code is intended to do.
A way you may use rownum without getting errors:
with x as (
select 1 col from dual union all
select 2 col from dual union all
select 8 col from dual union all
select 4 col from dual union all
select 3 col from dual union all
select 2 col from dual)
,x2 as (
select col col1 ,col col2, col col3 ,rownum rn
from x
)
select *
from x2
where rn between 2 and 3 --- rownum cannot be used in such a
condition!!!
;
Or, to be certain you get only the first row from a table satisfying a given condition:
select x_col1, x_col2 into v_col1, v_col2
from x_table
where ... --- logical conditions
and rownum<=1; --- rownum <= 1 avoids too_many_rows_exception if several rows satisfy the logical conditions given before
In Oracle, results sets have a non-deterministic order (i.e. they are unordered) unless you use an ORDER BY clause. Therefore, if you have a physical table, you need another column to provide the order (rather than relying on the ROWNUM pseudo-column, which may result in unexpected behaviour):
CREATE TABLE x (order_id, col) AS
SELECT 1, 1 FROM DUAL UNION ALL
SELECT 2, 2 FROM DUAL UNION ALL
SELECT 3, 8 FROM DUAL UNION ALL
SELECT 4, 4 FROM DUAL UNION ALL
SELECT 5, 3 FROM DUAL UNION ALL
SELECT 6, 2 FROM DUAL;
If you want to find the rows that go up in succession, then you can use MATCH_RECOGNIZE for row-by-row pattern matching:
SELECT *
FROM x
MATCH_RECOGNIZE(
ORDER BY order_id
MEASURES
any_row.col AS col1,
FIRST(up.col) AS col2,
LAST(up.col) AS col3,
FIRST(order_id) AS start_order_id
PATTERN ( any_row up{2} )
DEFINE up AS ( col > PREV(col) )
)
or the LEAD analytic function:
SELECT *
FROM (
SELECT col AS col1,
LEAD(col, 1) OVER (ORDER BY order_id) AS col2,
LEAD(col, 2) OVER (ORDER BY order_id) AS col3,
order_id
FROM x
)
WHERE col2 > col1
AND col3 > col2;
Which both output:
COL1
COL2
COL3
START_ORDER_ID
1
2
8
1
fiddle
It looks like you want to find the rows where the value of the column is bigger than it is in both - the previous and next row. If so, you could try this:
WITH
tbl (ID, COL) AS -- Sample data (ID column is just to preserve order of the rows)
(
Select 1, 1 From Dual Union All
Select 2, 2 From Dual Union All
Select 3, 8 From Dual Union All
Select 4, 4 From Dual Union All
Select 5, 3 From Dual Union All
Select 6, 2 From DUAL
)
Select ID, COL, CASE WHEN COL > LAG(COL, 1) OVER(Order By ID) And COL > LEAD(COL, 1) OVER(Order By ID) THEN 'YES' END "BIGGER_THAN_PREV_AND_NEXT"
From tbl
Order By ID
ID COL BIGGER_THAN_PREV_AND_NEXT
---------- ---------- -------------------------
1 1
2 2
3 8 YES
4 4
5 3
6 2
... with a bit different sample data this will find the other row(s) that satisfy the condition ...
WITH
tbl (ID, COL) AS -- Sample data (ID column is just to preserve order of the rows)
(
Select 1, 1 From Dual Union All
Select 2, 2 From Dual Union All
Select 3, 8 From Dual Union All
Select 4, 4 From Dual Union All
Select 5, 5 From Dual Union All -- value of COL changed from 3 to 5
Select 6, 2 From DUAL
)
Select ID, COL, CASE WHEN COL > LAG(COL, 1) OVER(Order By ID) And COL > LEAD(COL, 1) OVER(Order By ID) THEN 'YES' END "BIGGER_THAN_PREV_AND_NEXT"
From tbl
Order By ID
ID COL BIGGER_THAN_PREV_AND_NEXT
---------- ---------- -------------------------
1 1
2 2
3 8 YES
4 4
5 5 YES
6 2
OR without ID - using ROWNUM (as in your question), - not adviseable, though...
WITH
tbl (COL) AS -- Sample data (without ID column)
(
Select 1 From Dual Union All
Select 2 From Dual Union All
Select 8 From Dual Union All
Select 4 From Dual Union All
Select 5 From Dual Union All
Select 2 From DUAL
)
Select COL, CASE WHEN COL > LAG(COL, 1) OVER(Order By ROWNUM) And COL > LEAD(COL, 1) OVER(Order By ROWNUM) THEN 'YES' END "BIGGER_THAN_PREV_AND_NEXT"
From tbl
COL BIGGER_THAN_PREV_AND_NEXT
---------- -------------------------
1
2
8 YES
4
5 YES
2
Any Order By clause added to the query could change the ROWNUM values and the result...

Count occurences of values in table, when I treat one value as occurence of all other values

I have a table with one column (just to simplify the problem) with values 0-23 or *.
I want to count occurrences of each value 0-23, but treat * as occurrence of all other values
for example:
column_name
-------------
3
4
5
6
7
*
4
4
3
*
I want to get something like that:
column_name | count
--------------------
1 | 2
2 | 2
3 | 4
4 | 5
5 | 3
6 | 3
7 | 3
.....
I tried experimenting with different count and "group by" methods, but always getting very strange results. Basically the main problem here is to how count rows when I need to have one value in all other groups.
You could use analytic function that counts values where * is replaced by actual value between 0 and 23:
SELECT DISTINCT n.RN "COL_1", Count(REPLACE(t.COL_1, '*', n.RN)) OVER(Partition By n.RN) "CNT"
FROM tbl t
INNER JOIN ( Select To_Char(LEVEL - 1) "RN" From Dual Connect By LEVEL <=24 ) n ON(n.RN = REPLACE(t.COL_1, '*', n.RN))
WHERE n.RN IN(SELECT COL_1 FROM tbl)
ORDER BY To_Number(n.RN)
which with your sample data:
WITH
tbl (COL_1) AS
(
Select '3' From Dual Union All
Select '4' From Dual Union All
Select '5' From Dual Union All
Select '6' From Dual Union All
Select '7' From Dual Union All
Select '*' From Dual Union All
Select '4' From Dual Union All
Select '4' From Dual Union All
Select '3' From Dual Union All
Select '*' From Dual Union All
Select '3' From Dual
)
... results as:
COL_1 CNT
---------------------------------------- ----------
3 5
4 5
5 3
6 3
7 3
... and if you exclude the Where clause you will get all the rows (0 - 23) with number of occurances counted by REPLACE of * with any of the numbers
COL_1 CNT
---------------------------------------- ----------
0 2
1 2
2 2
3 5
4 5
5 3
6 3
7 3
8 2
9 2
10 2
11 2
12 2
13 2
14 2
15 2
16 2
17 2
18 2
19 2
20 2
21 2
22 2
23 2
You can do it using successive WITH's :
First one to calculate number of occurrence of *.
And the second is to calculate number of occurrence of each number.
with cte as (
select count(1) as c
from mytable
where column_name = '*'
),
cte2 as (
select column_name, count(1) as c
from mytable, cte
group by column_name
)
select column_name, cte.c + cte2.c
from cte2, cte;
You can with nested statements too,
SELECT ID,(count_ + (
SELECT COUNT(ID) FROM sql_test_a
WHERE ID = '*')) as count_
FROM (
SELECT ID,COUNT(ID) as count_
FROM sql_test_a WHERE ID != '*' GROUP BY ID);

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;

Query to delete duplicate records by keeping original in oracle

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.

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.