Fetch a fixed number of rows in SQL query in Oracle [duplicate] - sql

This question already has answers here:
Paging with Oracle
(7 answers)
Closed 8 years ago.
Please help me to write an SQL query in the Oracle database. There is table called tbl and it has 12 rows. I want to select first 4 rows first then next 4 and the last 4 rows.
Can any anyone tell me how can I do this in Informix.

EDIT: now should be fixed with 3-level select:
select * from (
select q1.*, rownum as rn from ( --get correct rownum
select * from tbl order by column --get correct order
) q1
) q2
where q2.rn between 1 and 4; -- filter
for first part.
For second and third part:
where q2.rn between 5 and 8
where q2.rn between 9 and 12

There is nothing called as first rows, last rows, "n" rows unless you explicitly specify an ORDER BY and then select the required rows.
Top-n Row Limiting feature in Oracle 12c on ward:
SQL> select * from order_test order by val;
VAL
----------
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
20 rows selected.
First 4 rows :
SQL> SELECT val
2 FROM order_test
3 ORDER BY VAL
4 FETCH FIRST 4 ROWS ONLY;
VAL
----------
1
1
2
2
Next 4 rows(look at OFFSET) :
SQL> SELECT val
2 FROM order_test
3 ORDER BY VAL
4 OFFSET 4 ROWS FETCH NEXT 4 ROWS ONLY;
VAL
----------
3
3
4
4
Finally, next 4 rows with OFFSET 8 rows :
SQL> SELECT val
2 FROM order_test
3 ORDER BY VAL
4 OFFSET 8 ROWS FETCH NEXT 4 ROWS ONLY;
VAL
----------
5
5
6
6

You can use rownum:
select * from (select t.*, rownum rn from tbl t) where rn between 1 and 4;
/
select * from (select t.*, rownum rn from tbl t) where rn between 5 and 8;
/
select * from (select t.*, rownum rn from tbl t) where rn between 9 and 12;
/
If you're using order by clause then use row_number() (documentation)
select * from (select t.*, row_number() over (order by column_name) rn from tbl t) where rn between 1 and 4;
/
select * from (select t.*, row_number() over (order by column_name) rn from tbl t) where rn between 5 and 8;
/
select * from (select t.*, row_number() over (order by column_name) rn from tbl t) where rn between 9 and 12;
/

Related

Oralce sql:I want to select the TOP 3 Records [duplicate]

This question already has answers here:
How do I limit the number of rows returned by an Oracle query after ordering?
(14 answers)
Closed 8 months ago.
I want to select the TOP 3 Records ordered desc by 'cnt'
this is top 4
a b c cnt
99 YC 市購件異常 3
99 LY 漏油 2
99 QT16 其他異常 2
99 JGSH 機構損壞 1
then
select * from ()where rownum<= 3 order by cnt desc
get data
99 YC 市購件異常 3
99 LY 漏油 2
99 JGSH 機構損壞 1
i want to get
99 YC 市購件異常 3
99 LY 漏油 2
99 QT16 其他異常 2
Try this:
SELECT T.a, T.b, T.c, T.cnt
FROM
(
SELECT *, RANK() OVER(PARTITION BY a ORDER BY cnt DESC) RNK
FROM TEST_TBL
) T
WHERE T.RNK <= 3
It looks like you want to keep "duplicates" (in the cnt column) in the result.
In that case, I'd say that it is row_number analytic function that helps:
Sample data:
SQL> with test (a, b, cnt) as
2 (select 99, 'yc' , 3 from dual union all
3 select 99, 'ly' , 2 from dual union all
4 select 99, 'qt16', 2 from dual union all
5 select 99, 'jgsh', 1 from dual union all
6 --
7 select 99, 'abc' , 2 from dual --> yet another row with CNT = 2
8 ),
Query begins here: first rank rows (line #11), and then return the top 3 (line #15):
9 temp as
10 (select a, b, cnt,
11 row_number() over (partition by a order by cnt desc) rnk
12 from test
13 )
14 select * from temp
15 where rnk <= 3;
A B CNT RNK
---------- ---- ---------- ----------
99 yc 3 1
99 ly 2 2
99 abc 2 3
SQL>
Because, if you use rank analytic function (as Hana suggested), you might get more than desired 3 rows (see the rnk column's values) (depending on data you work with, of course; rank works with data you posted, but - if there are more rows that share the same cnt value, it won't work any more):
<snip>
9 temp as
10 (select a, b, cnt,
11 rank() over (partition by a order by cnt desc) rnk
12 from test
13 )
14 select * from temp
15 where rnk <= 3;
A B CNT RNK
---------- ---- ---------- ----------
99 yc 3 1
99 ly 2 2
99 abc 2 2
99 qt16 2 2
SQL>

LIMIT by GROUP in SQl (postgresql)

Below is my table students (having 100K+ rows in orignal, showing just a set):
id
name
class
marks
1
Ryan
5
8
2
Max
5
7
3
Max1
5
10
4
Ryan1
6
8
5
Max2
6
10
6
Ryan2
6
7
7
Ryan3
7
8
8
Max3
7
10
9
Ryan4
7
7
I want to fetch two rows per class ( 5 & 7) having marks <= 10 , also sorted by class, marks ASC
So, the expected result will be:-
id
name
class
marks
1
Ryan
5
8
3
Max1
5
10
7
Ryan3
7
8
8
Max3
7
10
To execute below I tried:-
SELECT DISTINCT t_out.class, t_top.marks, t_top.name
FROM students t_out
JOIN LATERAL (
select *
from students t_in
where t_in.class = t_out.class
ORDER BY t_in.id ASC
) t_top ON TRUE
WHERE t_top.marks <= 10
AND (t_out.class = 5 OR t_out.class = 7)
ORDER BY t_top.marks DESC LIMIT 2;
Result on original database:- it's loading since long time
Result on sample :- Error: near line 20: near "(": syntax error
Is 10 the highest marks?
You would use row_number():
select s.*
from (select s.*,
row_number() over (partition by class order by marks desc) as seqnum
from students s
where marks < 10 and class in (5, 7)
) s
where seqnum <= 2
order by class, marks;
Note: Your question is a little confusing. You seem to want two rows with the highest marks per class ordered in descending order by marks.
EDIT:
Based on your comment:
select s.*
from (select s.*,
row_number() over (partition by class order by marks desc) as seqnum,
count(*) over (partition by class) as cnt
from students s
where marks < 10
) s
where seqnum <= 2 and cnt >= 2
order by class, marks;

What is the most efficient SQL query to find the max N values for every entities in a table

I wrote these 2 queries, the first one is keeping duplicates and the second one is dropping them
Does anyone know a more efficient way to achieve this?
Queries are for MSSQL, returning the top 3 values
1-
SELECT TMP.entity_id, TMP.value
FROM(
SELECT TAB.entity_id, LEAD(TAB.entity_id, 3, 0) OVER(ORDER BY TAB.entity_id, TAB.value) AS next_id, TAB.value
FROM mytable TAB
) TMP
WHERE TMP.entity_id <> TMP.next_id
2-
SELECT TMP.entity_id, TMP.value
FROM(
SELECT TMX.entity_id, LEAD(TMX.entity_id, 3, 0) OVER(ORDER BY TMX.entity_id, TMX.value) AS next_id, TMX.value
FROM(
SELECT TAB.entity_id, LEAD(TAB.entity_id, 1, 0) OVER(ORDER BY TAB.entity_id, TAB.value) AS next_id, TAB.value, LEAD(TAB.value, 1, 0) OVER(ORDER BY TAB.entity_id, TAB.value) AS next_value
FROM mytable TAB
) TMX
WHERE TMP.entity_id <> TMP.next_id OR TMX.value <> TMX.next_value
) TMP
WHERE TMP.entity_id <> TMP.next_id
Example:
Table:
entity_id value
--------- -----
1 9
1 11
1 12
1 3
2 25
2 25
2 5
2 37
3 24
3 9
3 2
3 15
Result Query 1 (25 appears twice for entity_id 2):
entity_id value
--------- -----
1 9
1 11
1 12
2 25
2 25
2 37
3 9
3 15
3 24
Result Query 2 (25 appears only once for entity_id 2):
entity_id value
--------- -----
1 9
1 11
1 12
2 5
2 25
2 37
3 9
3 15
3 24
You can use the ROW_NUMBER which will allow duplicates as follows:
select entity_id, value from
(select t.*, row_number() over (partition by entity_id order by value desc) as rn
from your_Table) where rn <= 3
You can use the rank to remove the duplicate as follows:
select distinct entity_id, value from
(select t.*, rank() over (partition by entity_id order by value desc) as rn
from your_Table) where rn <= 3

pick all positive least numbers from data set [duplicate]

This question already has answers here:
Fetch the rows which have the Max value for a column for each distinct value of another column
(35 answers)
Oracle group using min date
(3 answers)
GROUP BY with MAX(DATE) [duplicate]
(6 answers)
Closed 3 years ago.
I have below data in a table
ID AMOUNT DAYS
1 10 1
1 20 2
1 30 3
1 1 4
2 34 1
2 234 2
2 234 3
2 34 4
3 3 1
3 3 2
3 23 3
3 20 4
I want below results as all amounts which have least days of a ID
ID AMOUNT DAYS
1 10 1
2 34 1
3 3 1
Please suggest a sql query to pick this desired output
For your example, you can simply do:
select t.*
from t
where t.days = 1;
If 1 is not fixed, then a correlated subquery is one method:
select t.*
from t
where t.days = (select min(t2.days) from t t2 where t2.id = t.id);
Another method is aggregation:
select t.id, min(t.days) as min_days,
min(t.amount) keep (dense_rank first order by t.days asc) as min_amount
from t
group by t.id;
Of course row_number()/rank() is another alternative.
With an index on (id, days) and a large table, one of the above methods may be faster in practice.
You can use rank() function
select ID, Amount, Days from
(
select rank() over (partition by ID order by days) as rn,
t.*
from tab t
)
where rn = 1;
Demo
First group by id to find the min days for each id and then join to the table
select t.*
from tablename t inner join (
select id, min(days) days
from tablename
group by id
) g on g.id = t.id and g.days = t.days

How to get change points in oracle select query?

How can I select change points from this data set
1 0
2 0
3 0
4 100
5 100
6 100
7 100
8 0
9 0
10 0
11 100
12 100
13 0
14 0
15 0
I want this result
4 7 100
11 12 100
This query based on analytic functions lag() and lead() gives expected output:
select id, nid, point
from (
select id, point, p1, lead(id) over (order by id) nid
from (
select id, point,
decode(lag(point) over (order by id), point, 0, 1) p1,
decode(lead(point) over (order by id), point, 0, 2) p2
from test)
where p1<>0 or p2<>0)
where p1=1 and point<>0
SQLFiddle
Edit: You may want to change line 3 in case there only one row for changing point:
...
select id, point, p1,
case when p1=1 and p2=2 then id else lead(id) over (order by id) end nid
...
It would be simple to use ROW_NUMBER analytic function, MIN and MAX.
This is a frequently asked question about finding the interval/series of values and skip the gaps. I like the word given to it as Tabibitosan method by Aketi Jyuuzou.
For example,
SQL> SELECT MIN(A),
2 MAX(A),
3 b
4 FROM
5 ( SELECT a,b, a-Row_Number() over(order by a) AS rn FROM t WHERE b <> 0
6 )
7 GROUP BY rn,
8 b
9 ORDER BY MIN(a);
MIN(A) MAX(A) B
---------- ---------- ----------
4 7 100
11 12 100
SQL>