SQL select two same column on same table but with different ORDER - sql

Trainee here.
I need to create a list last names from the same table.
say we have a table named "sample", this table only consists of:
What I want to do here is both last name column will be selected but have different order, the first column would be ascending and the second column would be descending like the photo below

Here's one option: it splits first and last name into two subqueries which use row_number analytic function. It is then used to join rows.
Lines #1 - 6 represent your sample data. Query you really need begins at line #7.
SQL> with test (last_name, first_name) as
2 (select 'L1one' , 'F1one' from dual union all
3 select 'L2two' , 'F2two' from dual union all
4 select 'L3three', 'F3hthree' from dual union all
5 select 'L4four' , 'F4four' from dual
6 ),
7 ln as
8 (select last_name,
9 row_Number() over (order by last_name) rn
10 from test
11 ),
12 fn as
13 (select first_name,
14 row_number() over (order by first_name desc) rn
15 from test
16 )
17 select l.last_name, f.first_name
18 from ln l join fn f on f.rn = l.rn
19 order by l.last_name
20 /
LAST_NA FIRST_NA
------- --------
L1one F4four
L2two F3hthree
L3three F2two
L4four F1one
SQL>
[EDIT: both last names? I thought it was a typo]
If that's so, self-join is a better option:
SQL> with test (last_name, first_name) as
2 (select 'L1one' , 'F1one' from dual union all
3 select 'L2two' , 'F2two' from dual union all
4 select 'L3three', 'F3hthree' from dual union all
5 select 'L4four' , 'F4four' from dual
6 ),
7 temp as
8 (select last_name,
9 row_number() over (order by last_name asc) rna,
10 row_number() over (order by last_name desc) rnd
11 from test
12 )
13 select a.last_name, d.last_name
14 from temp a join temp d on a.rna = d.rnd
15 order by a.last_name;
LAST_NA LAST_NA
------- -------
L1one L4four
L2two L3three
L3three L2two
L4four L1one
SQL>

Here is one way to do it. In a single subquery, assign the ordinal number (rn) based on ascending order, but also keep track of the total row count. Then follow with a join.
with
test (last_name, first_name) as (
select 'L1one' , 'F1one' from dual union all
select 'L2two' , 'F2two' from dual union all
select 'L3three', 'F3hthree' from dual union all
select 'L4four' , 'F4four' from dual
)
, prep (last_name, rn, ct) as (
select last_name, row_number() over (order by last_name), count(*) over ()
from test
)
select a.last_name as last_name_asc, b.last_name as last_name_desc
from prep a inner join prep b on a.rn + b.rn = a.ct + 1
;
LAST_NAME_ASC LAST_NAME_DESC
-------------- --------------
L1one L4four
L2two L3three
L3three L2two
L4four L1one

Related

Hackerrank SQL problem to solve in Oracle's SQL version

Query the two cities in STATION with the shortest and longest CITY names, as well as their respective lengths (i.e.: number of characters in the name). If there is more than one smallest or largest city, choose the one that comes first when ordered alphabetically.
The STATION table is described as follows:
Sample Input
For example, CITY has four entries: DEF, ABC, PQRS and WXY.
Sample Output
ABC 3
PQRS 4
Explanation
When ordered alphabetically, the CITY names are listed as ABC, DEF, PQRS, and WXY, with lengths and . The longest name is PQRS, but there are options for shortest named city. Choose ABC, because it comes first alphabetically.
A little bit of analytic functions; sample data in lines #1 - 6; query begins at line #7.
SQL> with station (city) as
2 (select 'DEF' from dual union all
3 select 'ABC' from dual union all
4 select 'PQRS' from dual union all
5 select 'WXY' from dual
6 )
7 select city, len
8 from (select city,
9 length(city) len,
10 rank() over (partition by length(city) order by city) rn
11 from station
12 )
13 where rn = 1
14 order by city;
CITY LEN
---- ----------
ABC 3
PQRS 4
SQL>
Reading your comment, it seems you want something like this:
SQL> with station (city) as
2 (select 'DEF' from dual union all
3 select 'ABC' from dual union all
4 select 'PQRS' from dual union all
5 select 'WXY' from dual union all
6 select 'XX' from dual union all
7 select 'ABCDE' from dual
8 )
9 select city, len
10 from (select city,
11 length(city) len,
12 rank() over (order by length(city) , city) rna,
13 rank() over (order by length(city) desc, city) rnd
14 from station
15 )
16 where rna = 1
17 or rnd = 1
18 order by len, city;
CITY LEN
----- ----------
XX 2
ABCDE 5
SQL>
Try this SQL statement with the fetch first row only clause:
with station (city) as
(select 'DEF' from dual union all
select 'ABC' from dual union all
select 'PQRS' from dual union all
select 'WXY' from dual)
(select city,
length(city)
from station
order by 2, 1
fetch first row only)
union
(select city,
length(city)
from station
order by 2 desc, 1
fetch first row only);
I solved the question this way:
select min(tt.city), tt.city_length
from (select s.city, length(s.city) city_length
from station s
where length(s.city) = (select max(length(t.city)) from station t)
or length(s.city) = (select min(length(p.city)) from station p)
order by 2, 1) tt
group by tt.city_length;
You can use the ROW_NUMBER analytic function in the ORDER BY clause and then FETCH FIRST ROW WITH TIES:
SELECT city,
LENGTH(city) AS length
FROM station
ORDER BY
LEAST(
ROW_NUMBER() OVER ( ORDER BY LENGTH( city ) ASC, city ),
ROW_NUMBER() OVER ( ORDER BY LENGTH( city ) DESC, city )
)
FETCH FIRST ROW WITH TIES;
Which, for the sample data:
CREATE TABLE station ( city ) AS
SELECT 'ABC' FROM DUAL UNION ALL
SELECT 'DEF' FROM DUAL UNION ALL
SELECT 'PQRS' FROM DUAL UNION ALL
SELECT 'XYZ' FROM DUAL;
Outputs:
CITY | LENGTH
:--- | -----:
PQRS | 4
ABC | 3
db<>fiddle here
select min(city) || ' ' ||length(min(city)) from station
UNION
select max(city) || ' ' ||length(max(city)) from station;

Seat arrangement using Oracle SQL for examination

I was trying to arrange seat for an examination from the below dataset.
and the output dataset would be like the below(alternate department student one after another)
I am unable to get the desire output. Please help me on that. I am using the Oracle 11g express edition.
http://sqlfiddle.com/#!4/510071/1
Using ROW_NUMBER analytic function, create sort order for each department; then select values sorted by that number.
For example:
SQL> with test (roll_no, name, department) as
2 (select 1, 'anik', 'cse' from dual union all
3 select 2, 'sudipto', 'cse' from dual union all
4 select 3, 'injamam', 'cse' from dual union all
5 select 8, 'sajukta', 'ece' from dual union all
6 select 9, 'gourab', 'ece' from dual union all
7 select 10, 'soumenn', 'ece' from dual),
8 inter as
9 (select roll_no, name, department,
10 row_number() over (partition by department order by roll_no) rn
11 from test
12 )
13 select roll_no, name, department
14 from inter
15 order by rn, department;
ROLL_NO NAME DEP
---------- ------- ---
1 anik cse
8 sajukta ece
2 sudipto cse
9 gourab ece
3 injamam cse
10 soumenn ece
6 rows selected.
SQL>
You seem to want them interleaved. If so, use row_number() in the order by:
select s.*
from student s
order by row_number() over (partition by "department" order by "roll_no"),
"department";
Here is the SQL Fiddle.
Note: Don't wrap column names in double quotes. That means that the case of the identifier matters -- and just makes queries harder to write.

Oracle SQL to generate interleaved SQL results

Hi i am looking for a way to write a SQL statement which will come out with the following results :-
Lets say we have Dept & Emp ID i would like to generate like records from Dept 3 for the first two rows then followed by Dept 2 with one row then continue Dept 3 and so on :
DEPT EMPID
----- ------
3 1
3 2
2 3
3 7
3 8
2 9
Thank You.
You could use something like this
SELECT
DEPT,
EMPID
FROM (
SELECT
*,
ceil((row_number() OVER (PARTITION BY dept ORDER BY EMPID ))/ 2::numeric(5,2)) AS multiple_row_dept,
row_number() OVER (PARTITION BY dept ORDER BY EMPID ) AS single_row_dept
FROM
test_data2
) sub_query
ORDER BY
CASE
WHEN DEPT = 2 THEN single_row_dept
ELSE multiple_row_dept
END,
DEPT DESC,
EMPID
single_row_dept specifics which dept should appear only once, in this case its DEPT 2 followed by multiple other departments
First order a table by empid in a subquery,
then calculate a remainder of rowids divided by 3,
then depending on a result of calculation, return 2 or 3, using case expression,
like this
SELECT
CASE REMAINDER( rownum, 3 )
WHEN 0 THEN 2
ELSE 3
END As DeptId,
empid
FROM (
SELECT empid
FROM table1
ORDER BY empid
)
demo: http://sqlfiddle.com/#!4/bd1bb/3
The following code has some limitations:
1) There are only 2 DepId's in the source table
2) Ratio between records having different DepId is exactly 2:1
3) Ordering in one place of the query should be changed depending whether DeptId having more records is naturally greater then the other or not
with t as (
select 3 dept_id, 1 emp_id from dual
union all
select 3, 2 from dual
union all
select 3, 7 from dual
union all
select 3, 8 from dual
union all
select 2, 3 from dual
union all
select 2, 9 from dual)
select dept_id, emp_id
from (select dept_id,
emp_id,
dense_rank() over(partition by dept_id order by in_num + mod(in_num, out_num)) ord
from (select dept_id,
emp_id,
row_number() over(partition by dept_id order by emp_id) in_num,
/*in the following ORDER BY change to DESC or ASC depending on which dept_id has more records*/
dense_rank() over(order by dept_id) out_num
from t))
order by ord, dept_id desc;
DEPT_ID EMP_ID
---------- ----------
3 1
3 2
2 3
3 8
3 7
2 9

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.

In Oracle, how do I get a page of distinct values from sorted results?

I have 2 columns in a one-to-many relationship. I want to sort on the "many" and return the first occurrence of the "one". I need to page through the data so, for example, I need to be able to get the 3rd group of 10 unique "one" values.
I have a query like this:
SELECT id, name
FROM table1
INNER JOIN table2 ON table2.fkid = table1.id
ORDER BY name, id;
There can be multiple rows in table2 for each row in table1.
The results of my query look like this:
id | name
----------------
2 | apple
23 | banana
77 | cranberry
23 | dark chocolate
8 | egg
2 | yak
19 | zebra
I need to page through the result set with each page containing n unique ids. For example, if start=1 and n=4 I want to get back
2
23
77
8
in the order they were sorted on (i.e., name), where id is returned in the position of its first occurrence. Likewise if start=3 and n=4 and order = desc I want
8
23
77
2
I tried this:
SELECT * FROM (
SELECT id, ROWNUM rnum FROM (
SELECT DISTINCT id FROM (
SELECT id, name
FROM table1
INNER JOIN table2 ON table2.fkid = table1.id
ORDER BY name, id)
WHERE ROWNUM <= 4)
WHERE rnum >=1)
which gave me the ids in numerical order, instead of being ordered as the names would be.
I also tried:
SELECT * FROM (
SELECT DISTINCT id, ROWNUM rnum FROM (
SELECT id FROM (
SELECT id, name
FROM table1
INNER JOIN table2 ON table2.fkid = table1.id
ORDER BY name, id)
WHERE ROWNUM <= 4)
WHERE rnum >=1)
but that gave me duplicate values.
How can I page through the results of this data? I just need the ids, nothing from the "many" table.
update
I suppose I'm getting closer with changing my inner query to
SELECT id, name, rank() over (order by name, id)
FROM table1
INNER JOIN table2 ON table2.fkid = table1.id
...but I'm still getting duplicate ids.
You may need to debug it a little, but but it will be something like this:
SELECT * FROM (
SELECT * FROM (
SELECT id FROM (
SELECT id, name, row_number() over (partition by id order by name) rn
FROM table1
INNER JOIN table2 ON table2.fkid = table1.id
)
) WHERE rn=1 ORDER BY name, id
) WHERE rownum>=1 and rownum<=4;
It's a bit convoluted (and I would tend to suspect that it could be simplified) but it should work. You'd can put whatever start and end position you want in the WHERE clause-- I'm showing here with start=2 and n=4 are pulled from a separate table but you could simplify things by using a couple of parameters instead.
SQL> ed
Wrote file afiedt.buf
1 with t as (
2 select 2 id, 'apple' name from dual union all
3 select 23, 'banana' from dual union all
4 select 77, 'cranberry' from dual union all
5 select 23, 'dark chocolate' from dual union all
6 select 8, 'egg' from dual union all
7 select 2, 'yak' from dual union all
8 select 19, 'zebra' from dual
9 ),
10 x as (
11 select 2 start_pos, 4 n from dual
12 )
13 select *
14 from (
15 select distinct
16 id,
17 dense_rank() over (order by min_id_rnk) outer_rnk
18 from (
19 select id,
20 min(rnk) over (partition by id) min_id_rnk
21 from (
22 select id,
23 name,
24 rank() over (order by name) rnk
25 from t
26 )
27 )
28 )
29 where outer_rnk between (select start_pos from x) and (select start_pos+n-1 from x)
30* order by outer_rnk
SQL> /
ID OUTER_RNK
---------- ----------
23 2
77 3
8 4
19 5