Rewrite the query with out joins and group by
Hi,
This was an interview question.
Table Names: bookshelf_checkout
bookshelf
And the join condition between these two tables is title
We need to rewrite below query without using join condition and group by clause ?
SELECT b.title,max(bc.returned_date - bc.checkout_date) "Most Days Out"
FROM bookshelf_checkout bc,bookshelf b
WHERE bc.title(+)=b.title
GROUP BY b.title;When I was in college, I read that most of the SELECT statements can be replaced by basic SQL operations (SET OPERATORS). Now I am trying to rewrite the query with SET operators but not able to get the exact result.
Kindly help me on this.
Thanks,
Suri
Something like this?
1 WITH books AS (
2 SELECT 'title 1' title FROM dual UNION ALL
3 SELECT 'title 2' FROM dual UNION ALL
4 SELECT 'title 3' FROM dual ),
5 bookshelf AS (
6 SELECT 'title 1' title, DATE '2012-05-01' checkout_date, DATE '2012-05-15' returned_date FROM dual UNION ALL
7 SELECT 'title 1' title, DATE '2012-05-16' checkout_date, DATE '2012-05-20' returned_date FROM dual UNION ALL
8 SELECT 'title 2' title, DATE '2012-04-01' checkout_date, DATE '2012-05-15' returned_date FROM dual )
9 SELECT bs.title, MAX(bs.returned_date - bs.checkout_date) OVER (PARTITION BY title) FROM bookshelf bs
10 UNION
11 (SELECT b.title, NULL FROM books b
12 MINUS
13* SELECT bs.title, NULL FROM bookshelf bs)
SQL> /
TITLE MAX(BS.RETURNED_DATE-BS.CHECKOUT_DATE)OVER(PARTITIONBYTITLE)
title 1 14
title 2 44
title 3Lukasz
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Msg 156, Level 15, State 1, Line 3
Incorrect syntax near the keyword 'INNER'.
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count(case
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CARR_CD FILE_NAME FILE_ID TOT_REC TOT_SUCC TOT_TRANS
LIBM CM_LIBM.TXT 12345678 5 4 2
LIBM CM_LIBM.TXT 12345677 10 0 0Using RANK can potentially produce multiple rows to be returned though your data may prevent this. ROW_NUMBER will always prevent duplicates. The ordering of the analytical function is irrelevant in your query if you use ROW_NUMBER. You can remove the outermost query and inspect the data returned by the inner query;
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fs.file_name,
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null
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row_number() over(partition by c.src_file_id order by null) rn
from file_status fs
left join comm c
on c.src_file_id = fs.file_id
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CARR_CD FILE_NAME FILE_ID TOT_REC TOT_SUCC TOT_TRANS RN
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Got the results only the matched records as per the selection field FLAG_TYPE.
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I've posted some sample data and queries below:
CREATE TABLE COMPLAINT
CNO NUMBER,
REASON VARCHAR2(15 BYTE),
TOTAL NUMBER
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(CNO, REASON, TOTAL)
Values
(1, 'edge', 250);
Insert into COMPLAINT
(CNO, REASON, TOTAL)
Values
(2, 'edge', 250);
Insert into COMPLAINT
(CNO, REASON, TOTAL)
Values
(3, 'brst', 300);
Insert into COMPLAINT
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(4, 'crea', 400);
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CREATE TABLE SCOTT.COMPLAINTROLL
CNO NUMBER,
ROLL VARCHAR2(15 BYTE)
SET DEFINE OFF;
Insert into COMPLAINTROLL
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Values
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Insert into COMPLAINTROLL
(CNO, ROLL)
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CNO REASON TOTAL
1 edge 250
2 edge 250
3 brst 300
4 crea 400
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1 roll1
1 roll2
1 roll3
1 roll4
2 roll22
-- total of reason code edge is 500
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from complaint c
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brst 300
crea 400
edge 500
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from complaint c,complaintroll cr
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brst 300
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edge 1250
{\code}
Thanks for reading this post.The problem that you described is simple. The outer join duplicates all the rows from the parent table (complaint). If you want to sum a column from the parent table, then this sum includes all the duplicated rows.
There are several solutions for this problem.
A) One had been shown already by Cenutil. Instead of doing an outer join, you can do a subquery in the select clause that delivers the additional information from the detail table.
SQL> select reason,
sum(total),
sum((select count(*) from complaintroll cr where c.cno=cr.cno)) cnt_rolls
from complaint c
group by c.reason;
REASON SUM(TOTAL) CNT_ROLLS
crea 400 0
brst 300 0
edge 500 5b) sum in two steps. First sum and count including the join criteria, then use this information to calculate the correct total sum.
SQL> select reason, sum(stotal), sum(stotal/scount), sum(scount), sum(cnt_rolls)
2 from (select reason, sum(total) stotal, count(*) scount, count(cr.cno) cnt_rolls
3 from complaint c
4 left join complaintroll cr on c.cno=cr.cno
5 group by reason, c.cno
6 )
7 group by reason;
REASON SUM(STOTAL) SUM(STOTAL/SCOUNT) SUM(SCOUNT) SUM(CNT_ROLLS)
crea 400 400 1 0
brst 300 300 1 0
edge 1250 500 5 5
sql> c) another option is to do the left join, but do the aggregation only one time for the parent table. Analytic functions are helpful for that. However since analytic fuinctions can't be used inside an aggregation function, we would again need an inline view.
SQL> select reason, sum(case when rn = 1 then total end) sum_total, count(*), count(crcno)
2 from (select row_number() over (partition by c.reason order by c.cno) rn,
3 c.*, cr.cno crcno
4 from complaint c
5 left join complaintroll cr on c.cno=cr.cno
6 )
7 group by reason;
REASON SUM_TOTAL COUNT(*) COUNT(CRCNO)
brst 300 1 0
crea 400 1 0
edge 250 5 5
SQL> Edited by: Sven W. on Feb 10, 2011 1:00 PM - formatting + column added to 2nd option -
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Anyway, I would delete the app and re-install. If you paid for the app in the first place you will be able to re-install free of charge. Also, the app should allow the in-app purchase to happen again as it should recognise that you have already purchased it previously.
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Dear All, can any one explain the following standard program and function module. i could not find any thing related this in web, could any one please explain them. the standard program is : /SCWM/SAPLSTOCK_OVERVIEW_MON and the standard function modu
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I recently created a calendar event on my iPhone 5. I invited an attendee who received the event too late (approximately 36 hours later). Is there a way to verify and view when I created the event? Also, is this a common problem? Why would this h