i want current CTC of each employee following is the design of my table
Ecode Implemented Date Salary
7654323 2010-05-20 350000
7654322 2010-05-17 250000
7654321 2003-04-01 350000
7654321 2004-04-01 450000
7654321 2005-04-01 750000
7654321 2007-04-01 650000
i want oracle query for following output
Ecode Salary
7654321 650000
7654322 250000
7654323 350000
thanks in advance
See also
Oracle Query for getting MAximum CTC (Salary) of Each Employee
If you want to keep the last salary for each ecode sorted by implemented_date:
SQL> WITH data AS (
2 SELECT 7654323 Ecode, '2010-05-20' Implemented_Date, 350000 Salary
3 FROM DUAL
4 UNION ALL SELECT 7654322, '2010-05-17', 250000 FROM DUAL
5 UNION ALL SELECT 7654321, '2003-04-01', 350000 FROM DUAL
6 UNION ALL SELECT 7654321, '2004-04-01', 450000 FROM DUAL
7 UNION ALL SELECT 7654321, '2005-04-01', 750000 FROM DUAL
8 UNION ALL SELECT 7654321, '2007-04-01', 650000 FROM DUAL
9 )
10 SELECT ecode,
11 MAX(salary)
12 KEEP (dense_rank FIRST ORDER BY Implemented_Date DESC) sal
13 FROM DATA
14 GROUP BY ecode;
ECODE SAL
---------- ----------
7654321 650000
7654322 250000
7654323 350000
SELECT *
FROM salary s
INNER JOIN
(SELECT ecode, MAX(implemented_date) as implemented_date
FROM salary GROUP BY ecode) curr
ON curr.ecode = s.ecode AND curr.implemented_date = s.implemented_date
I'd use analytical functions for that. You want to select the first value of salary for each combination of ecode and implementeddate ordered by the implementeddate to put the latest at the top.
select
distinct
first_value(ecode) OVER (PARTITION BY ecode ORDER BY IMPLEMENTEDDATE DESC NULLS LAST) Ecode,
first_value(implementeddate) OVER (PARTITION BY ecode ORDER BY IMPLEMENTEDDATE DESC NULLS LAST) ImplementedDate,
first_value(salary) OVER (PARTITION BY ecode ORDER BY IMPLEMENTEDDATE DESC NULLS LAST) Salary
from
tbl_Salary;
The "DISTINCT" will keep null rows at bay that would otherwise be returned for the other 3 versions of Ecode=7654321 that we're filtering out.
The result is:
ECODE IMPLEMENTEDDATE SALARY
----- --------------- ------
7654321 01/04/2007 650000
7654322 17/05/2010 250000
7654323 20/05/2010 350000
Related
I am using Oracle 11.
I have 2 tables
TblA with columns id, entity_id and effective_date.
TblADetail with columns id and value.
If Value = 0 for the effective date, I want to keep looking for the next effective date until I found value <> 0 anymore.
The below query only look for value on 3/10/21.
If value = 0, I want to look for value on 3/11/21. If that's not 0, I want to stop.
But, if that's 0, I want to look for value on 3/12/21. If that's not 0, I want to stop.
But, if that's 0, I want to keep looking until value is not 0.
How can I do that ?
SELECT SUM(pd.VALUE)
FROM TblA p,TblADetail pd
WHERE p.id = pd.id
AND p.effective_date = to_date('03/10/2021','MM/DD/YYYY')
AND TRIM (p.entity_id) = 123
Sample data:
TblA
id entity_id effective_date
1 123 3/10/21
2 123 3/11/21
3 123 3/12/21
TblADetail
id value
1 -136
1 136
2 2000
3 3000
In the above data, for entity_id 123, starting from effective_date 3/10/21, I would like to to return value 2000 (from TblADetail) effective_date 3/11/21.
So, starting from a certain date, I want the results from the minimum date that has non-zero values.
Thank you.
You can do what you need to do by grouping the sum on the effective date, and using the MIN analytic function to find the earliest date. Once you've done that, you simply need to select the date that matches the earliest date.
E.g.:
with tbla as (select 1 id, ' 123' entity_id, to_date('10/03/2021', 'dd/mm/yyyy') effective_date from dual union all
select 2 id, ' 123' entity_id, to_date('11/03/2021', 'dd/mm/yyyy') effective_date from dual union all
select 3 id, ' 123' entity_id, to_date('12/03/2021', 'dd/mm/yyyy') effective_date from dual),
tbla_detail as (select 1 id, -136 value from dual union all
select 1 id, 136 value from dual union all
select 2 id, 2000 value from dual union all
select 3 id, 3000 value from dual),
results as (select a.effective_date,
sum(ad.value) sum_value,
min(case when sum(ad.value) != 0 then a.effective_date end) over () min_effective_date
from tbla a
inner join tbla_detail ad on a.id = ad.id
where a.effective_date >= to_date('10/03/2021', 'dd/mm/yyyy')
and trim(a.entity_id) = '123'
group by a.effective_date)
select sum_value
from results
where effective_date = min_effective_date;
SUM_VALUE
----------
2000
Straightforward; read comments within code. Sample data in lines #1 - 13, query begins at line #14.
SQL> with
2 -- sample data
3 tbla (id, entity_id, effective_date) as
4 (select 1, 123, date '2021-03-10' from dual union all
5 select 2, 123, date '2021-03-11' from dual union all
6 select 3, 123, date '2021-03-12' from dual
7 ),
8 tblb (id, value) as
9 (select 1, -136 from dual union all
10 select 1, 136 from dual union all
11 select 2, 2000 from dual union all
12 select 3, 3000 from dual
13 ),
14 tblb_temp as
15 -- simple grouping per ID
16 (select id, sum(value) value
17 from tblb
18 group by id
19 )
20 -- return TBLA values whose ID equals TBLB_TEMP's minimum ID
21 -- whose value isn't zero
22 select a.id, a.entity_id, a.effective_date
23 from tbla a
24 where a.id = (select min(b.id)
25 from tblb_temp b
26 where b.value > 0
27 );
ID ENTITY_ID EFFECTIVE_
---------- ---------- ----------
2 123 03/11/2021
SQL>
Table COMPUTER:
Table SUPPLIER:
how to display the building location that has the most computers?
i Have been trying several ways include subquery, joins, max, count but all not working and error keeps happending
The result i pursueing is
SUPPID SNAME SADDRESS MAKE COUNT(*)
125 Apple Sdn.Bhd 18 Jalan Duta Apple 3
For example (where sample data is in lines #1 - 12; query you might be interested in begins at line #13):
SQL> with
2 -- sample data
3 computer (compid, make, suppid, locid) as
4 (select 13323, 'IBM' , 124, 333 from dual union all
5 select 13324, 'Apple', 125, 444 from dual union all
6 select 13325, 'Apple', 125, 444 from dual union all
7 select 13326, 'Apple', 125, 444 from dual
8 ),
9 supplier (suppid, sname, saddress) as
10 (select 124, 'IBM Sdn.Bhd' , '15 Jalan Duta' from dual union all
11 select 125, 'Apple Sdn.Bhd', '18 Jalan Duta' from dual
12 ),
13 comp_loc as
14 -- number of computers per location; RNK = 1 shows location with most computers
15 (select locid,
16 rank() over (order by count(*) desc) rnk,
17 count(*) cnt
18 from computer
19 group by locid
20 )
21 select distinct s.suppid, s.sname, s.saddress, c.make, l.cnt
22 from supplier s join computer c on c.suppid = s.suppid
23 join comp_loc l on l.locid = c.locid
24 where l.rnk = 1;
SUPPID SNAME SADDRESS MAKE CNT
---------- ------------- ------------- ----- ----------
125 Apple Sdn.Bhd 18 Jalan Duta Apple 3
SQL>
On Oracle 12 and newer
select s.suppid, s.sname, s.saddress, c.make, count(1)
from COMPUTER c
join SUPPLIER s
on c.suppid = s.suppid
group by s.suppid, s.sname, s.saddress, c.make
order by count(1) desc
fetch first 1 row only <-- this line will fetch you the top 1 line only
You might also use "fetch first 1 row with ties" to output all the top manufacturers if there are many of them having same "count". E.g If IBM and Appl were having same amount of lines
On Oracle version before 12 do the following:
select *
from (select s.suppid, s.sname, s.saddress, c.make, count(1)
from comps c
join suppls s
on c.suppid = s.suppid
group by s.suppid, s.sname, s.saddress, c.make
order by count(1) desc)
where rownum = 1; <-- this line will get you the top 1 manufacturer only
PS. version of the oracle database can be obtained for example using:
select version from v$instance;
i have below data.
table A
id
1
2
3
table B
id name data1 data2 datetime
1 cash 12345.00 12/12/2012 11:10:12
1 quantity 222.12 14/12/2012 11:10:12
1 date 20/12/2012 12/12/2012 11:10:12
1 date 19/12/2012 13/12/2012 11:10:12
1 date 13/12/2012 14/12/2012 11:10:12
1 quantity 330.10 17/12/2012 11:10:12
I want to retrieve data in one row like below:
tableA.id tableB.cash tableB.date tableB.quantity
1 12345.00 13/12/2012 330.10
I want to retrieve based on max(datetime).
The data model appears to be insane-- it makes no sense to join an ORDER_ID to a CUSTOMER_ID. It makes no sense to store dates in a VARCHAR2 column. It makes no sense to have no relationship between a CUSTOMER and an ORDER. It makes no sense to have two rows in the ORDER table with the same ORDER_ID. ORDER is also a reserved word so you cannot use that as a table name. My best guess is that you want something like
select *
from customer c
join (select order_id,
rank() over (partition by order_id
order by to_date( order_time, 'YYYYMMDD HH24:MI:SS' ) desc ) rnk
from order) o on (c.customer_id=o.order_id)
where o.rnk = 1
If that is not what you want, please (as I asked a few times in the comments) post the expected output.
These are the results I get with my query and your sample data (fixing the name of the ORDER table so that it is actually valid)
SQL> ed
Wrote file afiedt.buf
1 with orders as (
2 select 1 order_id, 'iphone' order_name, '20121201 12:20:23' order_time from dual union all
3 select 1, 'iphone', '20121201 12:22:23' from dual union all
4 select 2, 'nokia', '20110101 13:20:20' from dual ),
5 customer as (
6 select 1 customer_id, 'paul' customer_name from dual union all
7 select 2, 'stuart' from dual union all
8 select 3, 'mike' from dual
9 )
10 select *
11 from customer c
12 join (select order_id,
13 rank() over (partition by order_id
14 order by to_date( order_time, 'YYYYMMDD HH24:MI:SS' ) desc ) rnk
15 from orders) o on (c.customer_id=o.order_id)
16* where o.rnk = 1
SQL> /
CUSTOMER_ID CUSTOM ORDER_ID RNK
----------- ------ ---------- ----------
1 paul 1 1
2 stuart 2 1
Try something like
SELECT *
FROM CUSTOMER c
INNER JOIN ORDER o
ON (o.CUSTOMER_ID = c.CUSTOMER_ID)
WHERE TO_DATE(o.ORDER_TIME, 'YYYYMMDD HH24:MI:SS') =
(SELECT MAX(TO_DATE(o.ORDER_TIME, 'YYYYMMDD HH24:MI:SS')) FROM ORDER)
Share and enjoy.
Given this table:
CREATE TABLE positions
( "EMP_ID" CHAR(10 BYTE),
"GTYPE" NUMBER,
"AMT" NUMBER,
"START_DATE" DATE,
"STOP_DATE" DATE
)
and this data:
Insert Into positions (Emp_Id,Gtype,Amt,Start_Date,Stop_Date)
select 'XA0022',1,1000,'01-MAY-2010','08-MAY-2012' from dual union
Select 'XA0022',1,1000,'01-MAY-2010','31-DEC-2012' From Dual Union
Select 'XA0022',2,500,'03-APR-2012','15-JUL-2012' From Dual Union
Select 'XA0022',1,421,'01-MAY-2012','23-MAY-2012' From Dual Union
Select 'XA0022',1,1514,'09-MAY-2012','31-DEC-2012' From Dual union
select 'XA0022',1,600,'24-MAY-2012','24-MAY-2012' from dual;
How do I get to this:
from to type1 type2
01-May-2010 02-Apr-2012 2000 0
03-Apr-2012 30-Apr-2012 2000 500
01-May-2012 07-May-2012 2421 500
08-May-2012 08-May-2012 2421 500
09-May-2012 22-May-2012 2935 500
23-May-2012 23-May-2012 2935 500
24-May-2012 15-Jul-2012 3114 500
16-Jul-2012 31-Dec-2012 3014 0
Note: The amount is in effect on the start_date and is not in effect the day after the stop_date.
Any pointers gratefully received!
Use Oracle's pivot.
Oracle Pivot
select * from
(select emp_id, gtype, amt, start_date, stop_date
from positions)
pivot (sum(amt) as amt for (gtype) in (1 as "TYPE1", 2 as "TYPE2"))
order by emp_id, start_date;
With some more information (thanks), something like this?
select emp_id, gtype, start_date,
case when next_amt <> amt then next_start_date -1 end as to_date
from
(select emp_id, gtype, start_date, amt,
lead(start_date,1) over (order by emp_id, start_date) next_start_date,
lead(amt,1) over (order by emp_id, start_date) next_amt
from
positions)
I'm struggling with a subselect in oracle. I want to include the latest price from another table.
Here is my current attempt:
SELECT tab1.*
(select price from
old_prices
where part_no=tab1.article_no
order by valid_from desc) as old_price,
FROM articles tab1
order by article_no
The sub select returns several rows which I think is the problem. But I do not know how to limit the number of rows in Oracle.
SQL> create table articles (article_no,name)
2 as
3 select 1, 'PEN' from dual union all
4 select 2, 'PAPER' from dual
5 /
Table created.
SQL> create table old_prices (part_no,valid_from,price)
2 as
3 select 1, date '2008-01-01', 10 from dual union all
4 select 1, date '2009-01-01', 11 from dual union all
5 select 1, date '2010-01-01', 12 from dual union all
6 select 1, date '2011-01-01', 13 from dual union all
7 select 2, date '2010-01-01', 89.95 from dual union all
8 select 2, date '2011-01-01', 94.95 from dual union all
9 select 2, date '2012-01-01', 99.95 from dual
10 /
Table created.
SQL> select a.article_no
2 , max(a.name) keep (dense_rank last order by p.valid_from) name
3 , max(p.price) keep (dense_rank last order by p.valid_from) price
4 from articles a
5 , old_prices p
6 where a.article_no = p.part_no
7 group by a.article_no
8 /
ARTICLE_NO NAME PRICE
---------- ----- ----------
1 PEN 13
2 PAPER 99.95
2 rows selected.
Regards,
Rob.
If it's the latest price you're after:
SELECT tab1.*, p.price old_price
FROM articles tab1
, old_prices p
where p.part_no = tab1.article_no
and valid_from = (
select MAX(valid_from)
from old_prices p2
where p2.part_no = p.part_no
)
order by article_no
I want to include the lastest price
I presume you mean latest.
OK, well that's a bit of a problem to start with, there are several ways of doing this:
SELECT o.price
FROM old_prices o
WHERE o.part_no=&part_no
AND o.ondate=(SELECT MAX(o2.ondate)
FROM old_prices o2
WHERE o2.part_no=&part_no);
Seems the most obvious choice but its rather innefficient.
You could try....
SELECT ilv.price
FROM (SELECT o.price
FROM old_price o
WHERE o.part_no=&part_no
ORDER BY ondate DESC) ilv
WHERE rownum=1;
Or....
SELECT TO_NUMBER(
SUBSTR(
MAX(TO_CHAR(o.ondate, 'YYYYMMDDHH24MISS') || price)
, 15)
) as latest_price
FROM old_price o
WHERE o.part_no=&part_no;
To limit rows use ROWNUM < 10. This is a pseudocolumn returning the row number of each line of your resultset.
EDIT:
You need to add another subselect query (hope this is the right place for your need)
SELECT tab1.*
select (
(select price from old_prices
where part_no=tab1.article_no order by valid_from desc
) as x
where rownum = 1
) as old_price
FROM articles tab1
order by article_no
SELECT tab1.*
(select
price
from (
SELECT
part_no
, price
, row_number () over (partition by part_no order by valid_from desc ) rn
FROM
old_prices
) P
where rn =1
and tab1.article_no = P.part_no
) as old_price
FROM articles tab1
order by article_no
more efficient would be
SELECT
tab1.*
, P.price
FROM
articles tab1
, ( SELECT
part_no
, price
, row_number () over (partition by part_no order by valid_from desc ) rn
FROM
old_prices
) P
WHERE
P.part_no(+) = tab1.article_no
P.rn(+) = 1
;
with old_prices as(
select level * 15 price ,
mod (level ,5) part_no , --this is just to create a grouping type partno
(sysdate - level ) valid_from
from dual
connect by level < 100)
,
articles as(
select level ,
mod(level , 5 ) article_no ,
(sysdate + level) someOtherDateField
From dual
connect by level < 5
)
SELECT tab1.* ,
old_price.*
from articles tab1
left join
(
select price,
part_no ,
valid_from ,
rank() over(partition by part_no order by valid_from desc) rk
from old_prices
) old_price
on tab1.article_no = old_price.part_no
and old_price.rk = 1
order by article_no ;
Here's another way!
LEVEL ARTICLE_NO SOMEOTHERDATEFIELD PRICE PART_NO VALID_FROM RK
---------------------- ---------------------- ------------------------- ---------------------- ---------------------- ------------------------- ----------------------
1 1 25/05/11 07:30:54 15 1 23/05/11 07:30:54 1
2 2 26/05/11 07:30:54 30 2 22/05/11 07:30:54 1
3 3 27/05/11 07:30:54 45 3 21/05/11 07:30:54 1
4 4 28/05/11 07:30:54 60 4 20/05/11 07:30:54 1