我试图了解以下查询之间的区别

Question

我有一个销售人员和一个客户表，我正在尝试查找居住在客户居住的任何城市的所有销售人员（请注意，客户与特定的销售人员相关联，但现在我们只是比较城市）

创建表（ORACLE）

CREATE TABLE SALESMAN (  
    SALESMAN_ID INT CONSTRAINT SALESMAN_PK PRIMARY KEY,  
    NAME VARCHAR2(15),  
    CITY VARCHAR2(10),  
    COMMISSION DECIMAL(4,2))
;

INSERT ALL  
    INTO SALESMAN VALUES(5001,'JAMES HOOG','NEW YORK',0.15)  
    INTO SALESMAN VALUES(5002,'NAIL KNITE','PARIS',0.13)  
    INTO SALESMAN VALUES(5005,'PIT ALEX','LONDON',0.11)  
    INTO SALESMAN VALUES(5006,'MC LYON','PARIS',0.14)  
    INTO SALESMAN VALUES(5003,'LAUSON HEN','SAN JOSE',0.12)  
    INTO SALESMAN VALUES(5007,'PAUL ADAM','ROME',0.13)  
SELECT * FROM DUAL
;

CREATE TABLE CUSTOMER (   
    CUSTOMER_ID INT CONSTRAINT CUSTOMER_PK PRIMARY KEY,   
    CUST_NAME VARCHAR2(15),   
    CITY VARCHAR(10),   
    GRADE INT,   
    SALESMAN_ID INT,  
    CONSTRAINT FK_CUSTOMER_SALESMAN  
    FOREIGN KEY (SALESMAN_ID) REFERENCES SALESMAN (SALESMAN_ID)) 
;

INSERT ALL   
    INTO CUSTOMER VALUES (3002, 'NICK RIMANDO', 'NEW YORK', 100, 5001)   
    INTO CUSTOMER VALUES (3007, 'BRAD DAVIS', 'NEW YORK', 200, 5001)   
    INTO CUSTOMER VALUES (3005, 'GRAHAM ZUSI', 'CALIFORNIA', 200,5002)   
    INTO CUSTOMER VALUES (3008, 'JULIAN GREEN', 'LONDON', 300,5002)   
    INTO CUSTOMER VALUES (3004, 'FABIAN JOHSON', 'PARIS',300,5006)   
    INTO CUSTOMER VALUES (3009, 'GEOFF CAMEROON', 'BERLIN', 100,5003)   
    INTO CUSTOMER VALUES (3003, 'JOZY ALTIDOR', 'MOSCOW', 200,5007)   
    INTO CUSTOMER VALUES (3001, 'BRAD GUZAN', 'LONDON',NULL,5005)   
SELECT * FROM DUAL
;

SELECT * FROM SALESMAN;

SALESMAN_ID       NAME  CITY    COMMISSION
5001         JAMES HOOG NEW YORK    .15
5002         NAIL KNITE PARIS       .13
5005          PIT ALEX  LONDON      .11
5006          MC LYON   PARIS       .14
5003        LAUSON HEN  SAN JOSE    .12
5007          PAUL ADAM ROME        .13

SELECT * FROM CUSTOMER;

CUSTOMER_ID       CUST_NAME      CITY      GRADE    SALESMAN_ID
3002            NICK RIMANDO    NEW YORK    100       5001
3007             BRAD DAVIS     NEW YORK    200       5001
3005             GRAHAM ZUSI    CALIFORNIA  200       5002
3008            JULIAN GREEN    LONDON      300       5002
3004           FABIAN JOHSON    PARIS       300       5006
3009           GEOFF CAMEROON   BERLIN      100       5003
3003            JOZY ALTIDOR    MOSCOW      200       5007
3001            BRAD GUZAN      LONDON       -        5005

# EXPECTED OUTPUT
SALESMAN_ID    NAME     CITY      COMMISSION
5001        JAMES HOOG  NEW YORK    .15
5006         MC LYON    PARIS       .14
5005        PIT ALEX    LONDON      .11
5002       NAIL KNITE    PARIS      .13

# QUERY 1

SELECT DISTINCT
SLS.*
FROM SALESMAN SLS, CUSTOMER CUST
WHERE SLS.CITY = CUST.CITY

# QUERY 2
SELECT * FROM SALESMAN SLS  
WHERE EXISTS (SELECT SALESMAN_ID FROM CUSTOMER WHERE SLS.CITY = CUSTOMER.CITY)

# QUERY 3
SELECT * FROM SALESMAN SLS  
WHERE SLS.SALESMAN_ID IN  (SELECT DISTINCT CUST.SALESMAN_ID FROM CUSTOMER CUST WHERE SLS.CITY = CUST.CITY)
;

# OUTPUT FROM QUERY 3
SALESMAN_ID      NAME       CITY    COMMISSION
5001         JAMES HOOG   NEW YORK  .15
5006        MC LYON        PARIS    .14
5005        PIT ALEX      LONDON    .11

在上述三个查询中，查询 1 和 2 给出了预期的输出，但是查询 3 没有提供预期的输出。所有查询在 salesman 和 customer 表中的 city 之间都有相同的连接，但我不明白为什么查询 3 会给出不同的输出。

Answer 1

更新

基于cmets，第三个查询的正确形式应该只是检查客户表中是否存在销售员所在的城市：

SELECT * FROM SALESMAN SLS  
WHERE SLS.CITY IN  (SELECT DISTINCT CITY FROM CUSTOMER CUST)

这给出了与前两个查询相同的结果。

Demo on dbfiddle

原答案

您的第三个查询给出不同结果的原因是子查询only返回customer 中salesman 中每个城市的推销员ID 值，在本例中为5001 , 5005 和 5006。所以你不会在输出中得到 NAIL KNITE，因为巴黎、伦敦或纽约的客户都没有 NAIL 作为推销员。

请注意，查询 3 实际上返回正确的结果是有争议的，因为您的其他查询不考虑给定推销员在任何给定城市是否有任何客户。所以你的另外两个查询应该是：

查询 1：

SELECT DISTINCT
SLS.*
FROM SALESMAN SLS
JOIN CUSTOMER CUST ON SLS.CITY = CUST.CITY AND SLS.SALESMAN_ID = CUST.SALESMAN_ID
ORDER BY SLS.SALESMAN_ID

查询 2：

SELECT * FROM SALESMAN SLS  
WHERE EXISTS (SELECT * FROM CUSTOMER WHERE SLS.CITY = CUSTOMER.CITY AND CUSTOMER.SALESMAN_ID = SLS.SALESMAN_ID)
ORDER BY SALESMAN_ID

通过这些更改，所有三个查询都给出相同的结果：

SALESMAN_ID     NAME        CITY        COMMISSION
5001            JAMES HOOG  NEW YORK    .15
5005            PIT ALEX    LONDON      .11
5006            MC LYON     PARIS       .14

Demo on dbfiddle

Answer 2

在第三个查询中，您缺少join，并且您希望与第二个查询一样工作。这是您对join 的第三个查询。

SELECT * FROM SALESMAN SLS  
WHERE SLS.SALESMAN_ID 
IN  (
      SELECT 
        DISTINCT CUST.SALESMAN_ID 
      FROM CUSTOMER CUST 
      JOIN SALESMAN_ID SLS
      ON SLS.CITY = CUST.CITY
    )

当子查询结果很大时EXISTS比IN快很多，当子查询结果很小时IN比EXISTS快很多。