Question

我有一个表格，其主要组合名称，年龄，地址基于我们必须从DUMMY_TEST表中找到id的组合。

对于DUMMY_TEST表中的id，必须获取应与DUMMY_ID_VALUE表映射的所有名称，年龄，地址组合，以获得Valuee列结果。

需要以下给定的预期结果而不加入同一个表DUMMY_TEST两次，因为它实时有数百万条记录会产生性能问题。

任何人都可以帮助创建一个性能良好的查询，而无需使用相同的表两次。

--table creation
CREATE TABLE DUMMY(NAME VARCHAR2(100),AGE VARCHAR2(100),ADDRESS VARCHAR2(100), ID VARCHAR2(100));
--data loading
INSERT INTO DUMMY VALUES('SAM','30','ITALY','100');
INSERT INTO DUMMY VALUES('RAGHU','20','VENICE','300');
INSERT INTO DUMMY VALUES('TOM','40','JAPAN','200');

CREATE TABLE DUMMY_TEST(NAME VARCHAR2(100),AGE VARCHAR2(100),ADDRESS VARCHAR2(100), ID VARCHAR2(100));
INSERT INTO DUMMY_TEST VALUES('SAM','30','ITALY','100');
INSERT INTO DUMMY_TEST VALUES('TOM','40','JAPAN','200');
INSERT INTO DUMMY_TEST VALUES('BROSNAN','20','INDIA','100');
INSERT INTO DUMMY_TEST VALUES('ARJUN','30','AMERICA','200');
INSERT INTO DUMMY_TEST VALUES('SAMUEL','40','BERLIN','100');
INSERT INTO DUMMY_TEST VALUES('RAM','60','GERMANY','200');

--table creation
CREATE TABLE DUMMY_ID_VALUE(ID VARCHAR2(100),VALUEE VARCHAR2(100));
--data loading
INSERT INTO DUMMY_ID_VALUE VALUES(100,'INCLUDED');
INSERT INTO DUMMY_ID_VALUE VALUES(200,'exclueded');
INSERT INTO DUMMY_ID_VALUE VALUES(300,'PARTIAL');

查询

SELECT DT.NAME,DT.AGE,DT.ADDRESS,DT.ID,DUMMY_ID_VALUE.VALUEE 
FROM DUMMY 
  LEFT OUTER JOIN DUMMY_TEST A
  ON DUMMY.NAME = A.NAME AND DUMMY.AGE = A.AGE AND DUMMY.ADDRESS = A.ADDRESS 
  LEFT OUTER JOIN DUMMY_TEST DT
  ON DT.ID = A.ID
  INNER JOIN DUMMY_ID_VALUE
  ON DUMMY_ID_VALUE.ID = DT.ID;

实际结果

NAME   AGE ADDRESS ID  VALUEE
---------------------------------
SAM        30  ITALY   100 INCLUDED
BROSNAN    20  INDIA   100 INCLUDED
SAMUEL 40  BERLIN  100 INCLUDED

预期结果：

NAME   AGE ADDRESS ID  VALUEE
---------------------------------
SAM        30  ITALY   100 INCLUDED
BROSNAN    20  INDIA   100 INCLUDED
SAMUEL 40  BERLIN  100 INCLUDED
RAGHU  20  VENICE  300 PARTIAL

Answer 1

所以，这里有两个问题，一个（避免双重加入）很容易解决：

select A.NAME, A.AGE, A.ADDRESS, A.ID, DUMMY_ID_VALUE.VALUEE 
FROM DUMMY
    left OUTER JOIN DUMMY_TEST A ON (DUMMY.NAME = A.NAME AND DUMMY.AGE = A.AGE AND DUMMY.ADDRESS = A.ADDRESS) or (DUMMY.ID = A.ID)
    INNER JOIN DUMMY_ID_VALUE on DUMMY_ID_VALUE.ID = A.ID

但我不确定它是否会改善表现另一个是你得到的结果不符合你的期望，假设你在查询中输入的结果应该是6行。

Answer 2

您的查询没问题，join除外。我会去：

SELECT DT.NAME, DT.AGE, DT.ADDRESS, DT.ID, DUMMY_ID_VALUE.VALUEE 
FROM DUMMY LEFT OUTER JOIN
     DUMMY_TEST A
     ON DUMMY.NAME = A.NAME AND DUMMY.AGE = A.AGE AND DUMMY.ADDRESS = A.ADDRESS LEFT OUTER JOIN
     DUMMY_TEST DT
     ON DT.ID = A.ID LEFT JOIN
     DUMMY_ID_VALUE
     ON DUMMY_ID_VALUE.ID = DT.ID;

最终加入必须是LEFT JOIN。为了性能，您的查询没问题。 DUMMY_TEST(NAME, AGE, ADDRESS)上的索引将有助于您的表现。

避免相同的表连接

2 个答案: