Question regarding left join
2011-05-18 15:16 Tracy. 阅读(275) 评论(0) 编辑 收藏 举报You Asked
Hello All,Here's my problem:
I have a Rep table:
create table rep (id number, name varchar2(20)); insert into rep values (1,'X01'); insert into rep values (2,'X02');
And i have a Txn table
create table txn (id number, type varchar2(20), amount number, rep_id number); insert into txn values ('1', 'P', '4000', '1'); insert into txn values ('2', 'T', '2000', '1'); insert into txn values ('3', 'P', '700', '2'); insert into txn values ('4', 'P', '1200', '1'); insert into txn values ('5', 'T', '1300', '2');
I need to write a query that would return results like :
|REP|P_AMOUNT|T_AMOUNT
X01 5200 2000
X02 700 1300
I have come up with this
select r.name, p.P as p_amount, t.T as t_amount from rep r left join ( select rep_id, sum(amount) as P from txn where type='P' group by rep_id )p on p.rep_id = r.id left join ( select rep_id, sum(amount) as T from txn where type='T' group by rep_id )t on t.rep_id = r.id
Is it a good approach? The cost is low, but I am just not sure as to how this query would perform as compared to some other approaches.
Please advise.
Thanks in advance!
and we said...
ops$tkyte%ORA11GR2> select rep.id, rep.name, 2 sum( case when txn.type = 'P' then amount end ) p_amount, 3 sum( case when txn.type = 'T' then amount end ) t_amount 4 from rep, (select * from txn where type in ('P','T')) txn 5 where rep.id = txn.rep_id(+) 6 group by rep.id, rep.name 7 / ID NAME P_AMOUNT T_AMOUNT ---------- ------------------------------ ---------- ---------- 1 X01 5200 2000 2 X02 700 1300
would be the way I'd write it, the inline view to get only P's and T's is only necessary if there are other "types" in there (even then it is not necessary, just likely more efficient)
You can use the "left join" syntax if you prefer. I personally find it less readable.
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When I read this question I immediately thought of the new PIVOT syntax in 11g. I know there are many, many formulations of SQL to get an answer and the PIVOT formulation may not be the best performer but it's certainly worth consideration. I added three rows to demonstrate; 1) the condition of no transactions for a sales representative, and 2) only one type of transaction for a given sales representative. insert into rep values (3,'X03'); insert into rep values (4,'X04'); insert into txn values ('6', 'T', '6300', '3'); Here's the formulation with PIVOT and its results. ORA11GR2>r 1 SELECT name, p_amount, t_amount 2 FROM ( SELECT rep.name, txn.type, txn.amount 3 FROM rep, txn 4 WHERE rep.id = txn.rep_id (+) ) 5 PIVOT ( SUM( amount) AS amount 6 FOR ( type ) IN ( 'P' AS p, 'T' AS t ) ) 7* ORDER BY p_amount NULLS FIRST, t_amount NULLS FIRST, name NAME P_AMOUNT T_AMOUNT -------------------- ---------- ---------- X04 X03 6300 X02 700 1300 X01 5200 2000
本文来自博客园,作者:Tracy.,转载请注明原文链接:https://www.cnblogs.com/tracy/archive/2011/05/18/2050143.html