使用EXIST命令创建一个sql视图语句

Question

我必须在sql中使用一个exist命令编写一个创建视图语句。我试着在网上查找，但我有一些困难。我尽我所能写了创建视图文件，但它现在还没有工作。我知道我需要在我的声明中使用关键字EXIST。我想创建的语句是

Write a query that shows returns the name and city of the university that has no people in database 
that are associated with it. 

，我至今写的代码是这样的

CREATE VIEW exist AS 
SELECT a.university_name, a.city 
FROM lab5.university as a 
INNER JOIN lab5.person as b 
ON a.uid = b.uid 
WHERE b.uid NOT EXIST 

的表，我使用的

  Table "table.university" 
    Column  |   Type   |      Modifiers  
-----------------+-----------------------+-------------------------------------- 
uid    | integer    | not null default nextval('university_uid_seq'::regclass) 
university_name | character varying(50) | 
city   | character varying(50) | 

和

  Table "table.person" 
Column |   Type   |      Modifiers      
--------+-----------------------+----------------------------------------------- 
pid | integer    | not null default nextval('person_pid_seq'::reg class) 
uid | integer    | 
fname | character varying(25) | not null 
lname | character varying(25) | not null 

Answer 1

嗨使用下面的代码，让我知道反馈

CREATE VIEW checkuni AS SELECT a.university_name，a.city FROM 大学作为一个 WHERE NOT EXISTS （SELECT p.uid从一个人如对 WHERE p.uid =一个。 UID）

Answer 2

EXISTS是谓词所以你的查询应该是这个样子：

SELECT a.university_name, a.city 
FROM lab5.university as a 
WHERE NOT EXISTS (
    SELECT 1 
    FROM lab5.person as b 
    WHERE a.uid = b.uid 
); 

您也能表达次e。使用外连接和IS NULL谓词同样的事情：

SELECT a.university_name, a.city 
FROM lab5.university as a 
LEFT JOIN lab5.person as b 
    ON a.uid = b.uid 
WHERE b.uid IS NULL 

在您可能需要使用一个SELECT DISTINCT某些情况下，但我想这不会是必要的，你的情况。