說我有表1：在1列中映射2列到2行？
stuff | sender_id | recipient_id
，我有表2：
id | name
我要回：
stuff | sender_name | recipient_name
或
stuff | sender_id | sender_name | recipient_id | recipient_name
現在我正在尋找像這樣的東西： SELECT * FROM table1 INNER JOIN table2 ON id=sender_id OR id=recipient_id但是，作爲一個行滿足兩個條件返回的東西duplicates。
有沒有辦法在1個查詢中獲取我想要的信息？
的方法是使用內選擇這樣的：
SELECT stuff, sender_id
, (SELECT name FROM table2 WHERE id = sender_id) AS sender_name
, recipient_id
, (SELECT name FROM table2 WHERE id = recipient_id) AS recipient_name
FROM table1;
而另一種方法是使用雙重加入。
SELECT
t1.stuff,
tA.id sender_id,
tA.name sender_name,
tB.id recipient_id,
tB.name recipient_name
FROM
table1 t1
inner join table2 tA
on t1.sender_id = tA.id
inner join table2 tB
on t1.sender_id = tB.id
將是有趣的使用explain或analyze
解決方案的一個參考表2比較兩種查詢：
select t1.stuff
, max(case when t1.sender_id = t2.id then t2.name end) as sender_name
, max(case when t1.recipient_id = t2.id then t2.name end) as recipint_name
from t1
join t2
on t2.id in (t1.sender_id, t1.recipient_id)
group by t1.stuff;
這是一個有點亂，但也有它就會得心應手情況。
我創建的表和與萬行，每行（DB2 Express-C中，10.5修訂包1）釀它們：
db2 "create table t1 (stuff int not null primary key, sender_id int not null, recipient_id int not null)"
db2 "create table t2 (id int not null primary key, name varchar(10) not null);
db2 "insert into t1 with t (n) as (values 0 union all select n+1 from t where n+1 < 10000) select n, 2*n, 2*n+1 from t"
db2 "insert into t2 with t (n) as (values 0 union all select n+1 from t where n+1 < 10000) select 2*n, 'C' || rtrim(cast(2*n as char(10))) from t"
db2 runstats on table t1 with distribution and sampled detailed indexes all
db2 runstats on table t2 with distribution and sampled detailed indexes all
並檢查不同的查詢計劃。我添加了一個WHERE子句
兩個子選擇：
db2 "explain plan for SELECT stuff , (SELECT name FROM t2 WHERE id = sender_id) AS sender_name, (SELECT name FROM t2 WHERE id = recipient_id) AS recipient_name FROM t1 where t1.id between 500 and 600"
db2exfmt -d sample -g -1 -o sub.exfmt
兩個連接：
db2 "explain plan for SELECT t1.stuff, tA.name as sender_name, tB.name as recipient_name from t1 join t2 as tA on t1.sender_id = tA.id join t2 as tB on t1.sender_id = tB.id where t1.stuff between 500 and 600"
db2exfmt -d sample -g -1 -o dualjoin.exfmt
，最後用骨料和案例的變種：
db2 "explain plan for SELECT t1.stuff, max(case when t1.sender_id = t2.id then t2.name end) as sender_name, max(case when t1.recipient_id = t2.id then t2.name end) as recipint_name from t1 join t2 on t2.id in (t1.sender_id, t1.recipient_id) group by t1.stuff"
db2exfmt -d sample -g -1 -o singlejoin.exfmt
根據這一相當不科學的測試，@Juan Carlos Oropeza的解決方案是最便宜的：
Access Plan:
-----------
Total Cost: 132.657
Query Degree: 1
Rows
RETURN
( 1)
Cost
I/O
|
101.808
^NLJOIN
( 2)
132.657
53
/-------+--------\
101.808 1
TBSCAN FETCH
( 3) ( 7)
13.6735 13.6215
2 2
| /---+----\
101.808 1 10000
SORT IXSCAN TABLE: LELLE
( 4) ( 8) T2
13.6733 6.81423 Q1
2 1
| |
101.808 10000
FETCH INDEX: SYSIBM
( 5) SQL150906110744470
13.6625 Q1
2
/---+----\
101.808 10000
IXSCAN TABLE: LELLE
( 6) T1
6.84113 Q2
1
|
10000
INDEX: SYSIBM
SQL150906110646160
Q2
在@ shA.t使用兩個子選擇的更多的是有點貴：
Access Plan:
-----------
Total Cost: 251.679
Query Degree: 1
Rows
RETURN
( 1)
Cost
I/O
|
101.808
>^NLJOIN
( 2)
251.679
103.99
/-------+--------\
101.808 1
TBSCAN FETCH
( 3) ( 12)
132.695 13.6215
52.9898 2
| /---+----\
101.808 1 10000
SORT IXSCAN TABLE: LELLE
( 4) ( 13) T2
132.691 6.81423 Q1
52.9898 1
| |
101.808 10000
>^NLJOIN INDEX: SYSIBM
( 5) SQL150906110744470
132.67 Q1
52.9898
/-------+--------\
101.808 1
TBSCAN FETCH
( 6) ( 10)
13.6881 13.6215
2 2
| /---+----\
101.808 1 10000
SORT IXSCAN TABLE: LELLE
( 7) ( 11) T2
13.6839 6.81423 Q2
2 1
| |
101.808 10000
FETCH INDEX: SYSIBM
( 8) SQL150906110744470
13.6625 Q2
2
/---+----\
101.808 10000
IXSCAN TABLE: LELLE
( 9) T1
6.84113 Q3
1
|
10000
INDEX: SYSIBM
SQL150906110646160
Q3
我的解決方案是最昂貴的一個：
Access Plan:
-----------
Total Cost: 758.822
Query Degree: 1
Rows
RETURN
( 1)
Cost
I/O
|
10000
GRPBY
( 2)
758.139
124.996
|
20000
NLJOIN
( 3)
756.923
124.996
/----------+----------\
10000 2
FETCH FETCH
( 4) ( 6)
122.351 27.0171
49 3.96667
/---+----\ /---+----\
10000 10000 2 10000
IXSCAN TABLE: LELLE RIDSCN TABLE: LELLE
( 5) T1 ( 7) T2
58.1551 Q2 13.6291 Q1
21 2
| /-------+-------\
10000 1.0016 1.0016
INDEX: SYSIBM SORT SORT
SQL150906110646160 ( 8) ( 10)
Q2 6.81465 6.81465
1 1
| |
1.0016 1.0016
IXSCAN IXSCAN
( 9) ( 11)
6.81423 6.81423
1 1
| |
10000 10000
INDEX: SYSIBM INDEX: SYSIBM
SQL150906110744470 SQL150906110744470
Q1 Q1
雙內連接FTW，您將必須進行兩次連接。與測試很好的工作。 –
對於那些RDBMS是這樣的？請添加相關標籤 - 謝謝 –
由於您想從中獲得兩個不同的結果 –