Query to fetch the output in the given form.
Hi,
I have the following values in the table;
CURRENCY_MASTER -->
Curr_id Curr_name
1 Yen
2 Dollar
3 Rupee
4 Australian Dollar
5 Rubel
6 Taka
7 New Zealand dollar
8 Dinar
9 Euro
Output-
I want an output such that "Dollar" is at the top and the rest of them are arranged alphabetically.
Curr_name
Dollar
Australian Dollar
Dinar
Euro
New Zealand dollar
Rubel
Rupee
Taka
Yen
How to get this output.
Here's one way to do it:
test@ORA10G>
test@ORA10G> with t as (
2 select 1 as curr_id, 'Yen' as curr_name from dual union all
3 select 2 , 'Dollar' from dual union all
4 select 3 , 'Rupee' from dual union all
5 select 4 , 'Australian Dollar' from dual union all
6 select 5 , 'Rubel' from dual union all
7 select 6 , 'Taka' from dual union all
8 select 7 , 'New Zealand dollar' from dual union all
9 select 8 , 'Dinar' from dual union all
10 select 9 , 'Euro' from dual)
11 --
12 select
13 curr_id,
14 curr_name
15 from t
16 order by (case curr_name when 'Dollar' then '@'||curr_name else curr_name end);
CURR_ID CURR_NAME
2 Dollar
4 Australian Dollar
8 Dinar
9 Euro
7 New Zealand dollar
5 Rubel
3 Rupee
6 Taka
1 Yen
9 rows selected.
test@ORA10G>
test@ORA10G>pratz
Similar Messages
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Can you please explain how this query is fetching the rows?
here is a query to find the top 3 salaries. But the thing is that i am now able to understand how its working to get the correct data :How the data in the alias table P1 and P2 getting compared. Can you please explain in some steps.
SELECT MIN(P1.SAL) FROM PSAL P1, PSAL P2
WHERE P1.SAL >= P2.SAL
GROUP BY P2.SAL
HAVING COUNT (DISTINCT P1.SAL) <=3 ;
here is the data i used :
SQL> select * from psal;
NAME SAL
able 1000
baker 900
charles 900
delta 800
eddy 700
fred 700
george 700
george 700
Regards,
Renu... Please help me in understanding the query.
Your query looks like anything but a Top-N query.
If you run it in steps and analyze the output at the end of each step, then you should be able to understand what it does.
Given below is some brief information on the same:
test@ora>
test@ora> --
test@ora> -- Query 1 - using the non-equi (theta) join
test@ora> --
test@ora> with psal as (
2 select 'able' as name, 1000 as sal from dual union all
3 select 'baker', 900 from dual union all
4 select 'charles', 900 from dual union all
5 select 'delta', 800 from dual union all
6 select 'eddy', 700 from dual union all
7 select 'fred', 700 from dual union all
8 select 'george', 700 from dual union all
9 select 'george', 700 from dual)
10 --
11 SELECT p1.sal AS p1_sal, p1.NAME AS p1_name, p2.sal AS p2_sal,
12 p2.NAME AS p2_name
13 FROM psal p1, psal p2
14 WHERE p1.sal >= p2.sal;
P1_SAL P1_NAME P2_SAL P2_NAME
1000 able 1000 able
1000 able 900 baker
1000 able 900 charles
1000 able 800 delta
1000 able 700 eddy
1000 able 700 fred
1000 able 700 george
1000 able 700 george
900 baker 900 baker
900 baker 900 charles
900 baker 800 delta
900 baker 700 eddy
900 baker 700 fred
900 baker 700 george
900 baker 700 george
900 charles 900 baker
900 charles 900 charles
900 charles 800 delta
900 charles 700 eddy
900 charles 700 fred
900 charles 700 george
900 charles 700 george
800 delta 800 delta
800 delta 700 eddy
800 delta 700 fred
800 delta 700 george
800 delta 700 george
700 eddy 700 eddy
700 eddy 700 fred
700 eddy 700 george
700 eddy 700 george
700 fred 700 eddy
700 fred 700 fred
700 fred 700 george
700 fred 700 george
700 george 700 eddy
700 george 700 fred
700 george 700 george
700 george 700 george
700 george 700 eddy
700 george 700 fred
700 george 700 george
700 george 700 george
43 rows selected.
test@ora>
test@ora>This query joins PSAL with itself using a non equi-join. Take each row of PSAL p1 and see how it compares with PSAL p2. You'll see that:
- Row 1 with sal 1000 is >= to all sal values of p2, so it occurs 8 times
- Row 2 with sal 900 is >= to 9 sal values of p2, so it occurs 7 times
- Row 3: 7 times again... and so on.
- So, total no. of rows are: 8 + 7 + 7 + 5 + 4 + 4 + 4 + 4 = 43
test@ora>
test@ora> --
test@ora> -- Query 2 - add the GROUP BY
test@ora> --
test@ora> with psal as (
2 select 'able' as name, 1000 as sal from dual union all
3 select 'baker', 900 from dual union all
4 select 'charles', 900 from dual union all
5 select 'delta', 800 from dual union all
6 select 'eddy', 700 from dual union all
7 select 'fred', 700 from dual union all
8 select 'george', 700 from dual union all
9 select 'george', 700 from dual)
10 --
11 SELECT p2.sal AS p2_sal,
12 COUNT(*) as cnt,
13 COUNT(p1.sal) as cnt_p1_sal,
14 COUNT(DISTINCT p1.sal) as cnt_dist_p1_sal,
15 MIN(p1.sal) as min_p1_sal,
16 MAX(p1.sal) as max_p1_sal
17 FROM psal p1, psal p2
18 WHERE p1.sal >= p2.sal
19 GROUP BY p2.sal;
P2_SAL CNT CNT_P1_SAL CNT_DIST_P1_SAL MIN_P1_SAL MAX_P1_SAL
700 32 32 4 700 1000
800 4 4 3 800 1000
900 6 6 2 900 1000
1000 1 1 1 1000 1000
test@ora>
test@ora>Now, if you group by p2.sal in the output of query 1, and check the number of distinct p1.sal, min of p1.sal etc. you see that for p2.sal values - 800, 900 and 1000, there are 3 or less p1.sal values associated.
So, the last 3 rows are the ones you are interested in, essentially. As follows:
test@ora>
test@ora> --
test@ora> -- Query 3 - GROUP BY and HAVING
test@ora> --
test@ora> with psal as (
2 select 'able' as name, 1000 as sal from dual union all
3 select 'baker', 900 from dual union all
4 select 'charles', 900 from dual union all
5 select 'delta', 800 from dual union all
6 select 'eddy', 700 from dual union all
7 select 'fred', 700 from dual union all
8 select 'george', 700 from dual union all
9 select 'george', 700 from dual)
10 --
11 SELECT p2.sal AS p2_sal,
12 COUNT(*) as cnt,
13 COUNT(p1.sal) as cnt_p1_sal,
14 COUNT(DISTINCT p1.sal) as cnt_dist_p1_sal,
15 MIN(p1.sal) as min_p1_sal,
16 MAX(p1.sal) as max_p1_sal
17 FROM psal p1, psal p2
18 WHERE p1.sal >= p2.sal
19 GROUP BY p2.sal
20 HAVING COUNT(DISTINCT p1.sal) <= 3;
P2_SAL CNT CNT_P1_SAL CNT_DIST_P1_SAL MIN_P1_SAL MAX_P1_SAL
800 4 4 3 800 1000
900 6 6 2 900 1000
1000 1 1 1 1000 1000
test@ora>
test@ora>
test@ora>That's what you are doing in that query.
The thing is - in order to find out Top-N values, you simply need to scan that one table PSAL. So, joining it to itself is not necessary.
A much simpler query is as follows:
test@ora>
test@ora>
test@ora> --
test@ora> -- Top-3 salaries - distinct or not; using ROWNUM on ORDER BY
test@ora> --
test@ora> with psal as (
2 select 'able' as name, 1000 as sal from dual union all
3 select 'baker', 900 from dual union all
4 select 'charles', 900 from dual union all
5 select 'delta', 800 from dual union all
6 select 'eddy', 700 from dual union all
7 select 'fred', 700 from dual union all
8 select 'george', 700 from dual union all
9 select 'george', 700 from dual)
10 --
11 SELECT sal
12 FROM (
13 SELECT sal
14 FROM psal
15 ORDER BY sal DESC
16 )
17 WHERE rownum <= 3;
SAL
1000
900
900
test@ora>
test@ora>
test@ora>And for Top-3 distinct salaries:
test@ora>
test@ora> --
test@ora> -- Top-3 DISTINCT salaries; using ROWNUM on ORDER BY on DISTINCT
test@ora> --
test@ora> with psal as (
2 select 'able' as name, 1000 as sal from dual union all
3 select 'baker', 900 from dual union all
4 select 'charles', 900 from dual union all
5 select 'delta', 800 from dual union all
6 select 'eddy', 700 from dual union all
7 select 'fred', 700 from dual union all
8 select 'george', 700 from dual union all
9 select 'george', 700 from dual)
10 --
11 SELECT sal
12 FROM (
13 SELECT DISTINCT sal
14 FROM psal
15 ORDER BY sal DESC
16 )
17 WHERE rownum <= 3;
SAL
1000
900
800
test@ora>
test@ora>
test@ora>You may also want to check out the RANK and DENSE_RANK analytic functions.
RANK:
http://download.oracle.com/docs/cd/B19306_01/server.102/b14200/functions123.htm#SQLRF00690
DENSE_RANK:
http://download.oracle.com/docs/cd/B19306_01/server.102/b14200/functions043.htm#SQLRF00633
HTH
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Wrote file afiedt.buf
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15 )
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20 from (
21 select usr
22 ,case when transaction_type = 'b'
23 and lag(transaction_type) over (partition by usr order by transaction_date) = 'a'
24 then 1
25 else 0
26 end as sum_ab
27 from t
28 )
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SQL> /
USR SUM_AB
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This make no sense. We cannot guess you database structure.
Please post DDL (Create table script)
and DML (your current query)
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For more info, please have a look at this thread:
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"drop table T;
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, decode(grouping(symbol), 0, max(Cum_per_total), null) "% Cumm Total"
, decode(grouping(symbol), 0, max(rank), null) rank
from (
select name
, symbol
, dep
, amount
, per_total
, sum(per_total) over(order by rk) cum_per_total
, rank
, rk
from (
select name
, symbol
, dep
, sum(amount) amount
, round((sum(amount)/max(total_amount)) * 100,2) per_total
, dense_rank () over (order by sum(amount) desc) as rank
, row_number() over(order by sum(amount) desc) as rk
from (
select name
, symbol
, amount
, dep
, sum(amount) over() total_amount
, sum(amount) over ()
from t
group
by name, symbol, dep
group
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HR Amount, %HRTotal,$HR Cumm Total,
Finance Amount, %FinanceTotal,$Finance Cumm Total
Sales Amount, %SalesTotal,$Sales Cumm Total,
Marketing Amount, %MarketingTotal,$Marketing Cumm Total
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Thanks for your time and effort.
I am using the Oracle Database 11g Enterprise Edition Release 11.1.0.6.0 - Production
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2- %HRTotal
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5- %FinanceTotal
6- %Finance Cumm Total
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8- %SalesTotal
9- %Sales Cumm Total
10 -Marketing Amount
11- %MarketingTotal
12- %Marketing Cumm Total
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%HR Cumm Total (cumulative % based on the cumulative total of %HR Total) = Cumm % of amount case dep ='HR'
similarly rest of the column........
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Sundar
SundarHi All,
Thanks for your Prompt response. I tried the below queries, but still I don't have any luck. Actually the queries are returning the value before the condition met (say when the time difference is more than 06 hours). I want the
query to return exactly @ 06 hour difference or less than 06 hours,
Query 01: Select Top 1 VMName from TestTable where
convert(date,Exeutiontime)=convert(date,getdate())
and status='0'
and ExecutionTime >
dateadd(hour,-6,getdate())
Query 02: Select
Top 1 VMName from TestTable where
status='0'
and ExecutionTime >
dateadd(hour,-6,getdate())
Query 03: Select
Top 1 VMName from TestTable where status='0'
and ExecutionTime >
dateadd(hour,-6,ExecutionTime)
Can someone point out the mistake please.
Regards,
Sundar
Sundar
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