说我有表1：
Information:Gradle tasks [:app:generateDebugSources,:app:generateDebugAndroidTestSources]:app:preBuild UP-TO-DATE :app:preDebugBuild UP-TO-DATE :app:checkDebugManifest :app:preReleaseBuild UP-TO-DATE :app:prepareComAndroidSupportAppcompatV72300Library UP-TO-DATE :app:prepareComAndroidSupportSupportV42300Library UP-TO-DATE :app:prepareDebugDependencies :app:compileDebugAidl :app:compileDebugRenderscript :app:generateDebugBuildConfig UP-TO-DATE :app:generateDebugAssets UP-TO-DATE :app:mergeDebugAssets UP-TO-DATE :app:generateDebugResValues UP-TO-DATE :app:generateDebugResources :app:mergeDebugResources :app:processDebugManifest UP-TO-DATE :app:processDebugResources C:\Users\hp\AndroidStudioProjects\MyApplication\app\src\main\res\layout\activity_main.xml Error:(7) Error parsing XML: not well-formed (invalid token) Error:Execution failed for task ':app:processDebugResources'. > com.android.ide.common.process.ProcessException: org.gradle.process.internal.ExecException: Process 'command 'C:\Users\hp\AppData\Local\Android\sdk\build-tools\23.0.0\aapt.exe'' finished with non-zero exit value 1
Information:BUILD FAILED
Information:Total time: 37.338 secs
Information:2 errors
Information:0 warnings
Information:See complete output in console`
console :
Executing tasks: [:app:generateDebugSources, :app:generateDebugAndroidTestSources] Configuration on demand is an incubating feature.
:app:preBuild UP-TO-DATE
:app:preDebugBuild UP-TO-DATE
:app:checkDebugManifest
:app:preReleaseBuild UP-TO-DATE
:app:prepareComAndroidSupportAppcompatV72300Library UP-TO-DATE
:app:prepareComAndroidSupportSupportV42300Library UP-TO-DATE
:app:prepareDebugDependencies
:app:compileDebugAidl
:app:compileDebugRenderscript
:app:generateDebugBuildConfig UP-TO-DATE
:app:generateDebugAssets UP-TO-DATE
:app:mergeDebugAssets UP-TO-DATE
:app:generateDebugResValues UP-TO-DATE
:app:generateDebugResources
:app:mergeDebugResources
:app:processDebugManifest UP-TO-DATE
:app:processDebugResources
AGPBI: {"kind":"error","text":"Error parsing XML: not well-formed (invalid token)","sources":[{"file":"C:\\Users\\hp\\AndroidStudioProjects\\MyApplication\\app\\src\\main\\res\\layout\\activity_main.xml","position":{"startLine":6}}],"original":""}
FAILED
FAILURE: Build failed with an exception.
* What went wrong:
Execution failed for task ':app:processDebugResources'. com.android.ide.common.process.ProcessException: org.gradle.process.internal.ExecException: Process 'command 'C:\Users\hp\AppData\Local\Android\sdk\build-tools\23.0.0\aapt.exe'' finished with non-zero exit value 1
BUILD FAILED
Total time: 37.338 secs`
build.gradle file :
com.android.tools.build:gradle:1.3.0
我有表2：
stuff | sender_id | recipient_id
我想回来：
id | name
或
stuff | sender_name | recipient_name
现在我正在寻找类似的东西：
stuff | sender_id | sender_name | recipient_id | recipient_name但是这会返回重复的东西，因为一行填充了两个条件。
有没有办法在1个查询中获取我想要的信息？
答案 0 :(得分：1)
一种方法是使用内部选择，如下所示：
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;
答案 1 :(得分：1)
一个对table2的引用的解决方案：
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;
这有点乱，但有些情况会很方便。
我创建了表并用10000行填充它们（db2 express-c，10.5 fixpak 1）：
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
并检查了不同查询的计划。我添加了一个where子句
两个子选择：
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
两个加入：
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
最后是带有聚合和案例的变体：
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
根据这个相当不科学的测试，@Juan Carlos Oropeza的解决方案是最便宜的：
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中使用两个子选项有点贵：
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
我的解决方案是最昂贵的解决方案：
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
答案 2 :(得分：0)
另一种方式是使用双连接。
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
使用explain或analyze